The Solution to the Frame Quantum Detection Problem

Abstract

We will give a complete solution to the frame quantum detection problem. We will solve both cases of the problem: the quantum injectivity problem and quantum state estimation problem. We will answer the problem in both the real and complex cases and in both the finite dimensional and infinite dimensional cases. Finite Dimensional Case:

• (1) We give two complete classifications of the sets of vectors which solve the injectivity problem - for both the real and complex cases. We also give methods for constructing them.

• (2) We show that the frames which solve the injectivity problem are open and dense in the family of all frames.

• (3) We show that the Parseval frames which give injectivity are dense in the Parseval frames.

• (4) We classify all frames for which the state estimation problem is solvable, and when it is not solvable, we give the best approximation to a solution.

Infinite Dimensional Case:

• (1) We give a classification of all frames which solve the injectivity problem and give methods for constructing solutions.

• (2) We show that the frames solving the injectivity problem are neither open nor dense in all frames.

• (3) We give necessary and sufficient conditions for a frame to solve the state estimation problem for all measurements in $\ell _1$ and show that there is no injective frame for which the state estimation problem is solvable for all measurements in $\ell _2$.

• (4) When the state estimation problem does not have an exact solution, we give the best approximation to a solution.

1 Introduction and Preliminaries

In this paper we will give a complete answer to the frame quantum detection problem including the injectivity problem and state estimation problem. We will answer the problem in both the real and complex cases and in both the finite dimensional and infinite dimensional cases.

Important Notation Throughout the paper we will let $\{e_i\}_{i=1}^n$ be the canonical orthonormal basis of $\mathbb {R}^n$ or $\mathbb {C}^n$ and $\{e_i\}_{i=1}^{\infty }$ will denote the canonical orthonormal basis of real or complex $\ell _2$. Also, $\iota$ will be used to denote the complex unit.

For a vector $x_k$ in $\mathbb {R}^n$ or $\mathbb {C}^n$, we denote its coordinates as

$$\begin{aligned} x_k=(x_{k1}, x_{k2}, \ldots , x_{kn}). \end{aligned}$$

Similarly, for $x_k$ belonging to $\ell _2$, we write

$$\begin{aligned} x_k=(x_{k1}, x_{k2}, \ldots , x_{ki}, \ldots ). \end{aligned}$$

To explain exactly what we will solve, we need to introduce the basics of quantum detection. Let $L^{\infty }(\mathbb {H})$ be the space of bounded linear operators on a finite or infinite dimensional (real or complex) Hilbert space $\mathbb {H}$. Let $\{e_i\}_{i\in I}$ be an orthonormal basis for $\mathbb {H}$. For an operator $T\in L_0(\mathbb {H})$, the finite rank operators on $\mathbb {H}$, the trace of T is given by: $\mathrm {tr}(T)= \sum _{i\in I}\langle Te_i,e_i\rangle$, which is finite and independent of the orthonormal basis. The trace induces a scalar product by $\langle T,S\rangle _{HS}= \mathrm {tr}(TS^*)$. The closure of $L_0(\mathbb {H})$ with respect to this scalar product, denoted $L^2(\mathbb {H})$ is the space of the Hilbert-Schmidt operators on $\mathbb {H}$. For any $T\in L^{\infty }(\mathbb {H})$ we denote by $|T|=\sqrt{TT^*}$, the positive square root of $TT^*$. We say that T is a trace class operator if $\mathrm {tr}(|T|)<\infty$. The set of all trace class operators is denoted by $L^1(\mathbb {H})$ and forms a Banach space under the trace norm$\Vert T\Vert _1=\mathrm {tr}(|T|)$.

Let

$$\begin{aligned} Sym(\mathbb {H}) = \{T:T\in L^{\infty }(\mathbb {H}),\ T=T^*\}, \end{aligned}$$

denote the real Banach space of self-adjoint operators on $\mathbb {H}$ and let

$$\begin{aligned} Sym^+(\mathbb {H})= \{T =T^*\ge 0\}, \end{aligned}$$

denote the real cone of positive self-adjoint operators on $\mathbb {H}$. The main objects to analyze these operators are the positive operator-valued measures.

1.1 Positive Operator-Valued Measures

In quantum mechanics, the definition of a von Neumann measurement can be generalized using positive operator-valued measures (POVMs) [9, 11, 14]. The advantage of this is that it allows one to distinguish more accurately among elements of a set of non-orthogonal quantum states.

Let X denote a set of outcomes (e.g. this could be a finite or infinite subset of $\mathbb {Z}^d$ or $\mathbb {R}^d$). Let $\beta$ denote a sigma algebra of subsets of X.

Definition 1.1

A positive operator-valued measure (POVM) is a function $\Pi :\beta \rightarrow Sym^+(\mathbb {H})$ satisfying:

• (1) $\Pi (\emptyset )=0$ (the zero operator).

• (2) For every at most countable disjoint family $\{U_i\}_{i\in I}\subset \beta$, $x,y \in \mathbb {H}$ we have

  $$\begin{aligned} \left\langle \Pi \left( \cup _{i\in I}U_i\right) x,y \right\rangle = \sum _{i\in I}\langle \Pi (U_i)x,y\rangle . \end{aligned}$$

• (3) $\Pi (X)= I$ (the identity operator).

1.2 Quantum Systems

A quantum system is defined as a von Neumann algebra $\mathcal {A}$ of operators acting on $\mathbb {H}$. The set of states on $\mathbb {H}$ is

$$\begin{aligned} \mathcal {S}(\mathbb {H})=\{T\in L^1(\mathbb {H}),\ T=T^*\ge 0,\ \mathrm {tr}(T)=1\}, \end{aligned}$$

and it represents the reservoir of quantum states for any quantum system.

The set of quantum states$\mathcal {S}(\mathcal {A})$ associated to a quantum system $\mathcal {A}$ is obtained by identifying states that differ by a null state with respect to $\mathcal {A}$. Thus, the set of quantum states are in one-to-one correspondence with the linear functionals on $\mathcal {A}$ of the form:

$$\begin{aligned} \rho :\mathcal {A}\rightarrow \mathbb {C}, \text{ for } \text{ some } S\in \mathcal {S}(\mathbb {H}),\ \rho (T)=\mathrm {tr}(TS),\ \text{ for } \text{ every } T\in \mathcal {A}. \end{aligned}$$

A quantum state $\rho \in \mathcal {S}(\mathcal {A})$ is called a pure state if it is an extreme point in the convex $weak^*$ compact set of quantum states $\mathcal {S}(\mathcal {A})$. We say a POVM $\Pi$ is associated to a von Neumann algebra$\mathcal {A}$ if $\Pi :\beta \rightarrow \mathcal {A}\cap Sym^+(\mathbb {H})$.

Given a quantum state $\rho$, the quantum measurement performed by the POVM $\Pi$ is the map $p: \beta \rightarrow \mathbb {R}$ defined by $p(U)=\rho (\Pi (U))=\mathrm {tr}(\Pi (U)T)$, where $T\in \mathcal {S}(\mathbb {H})$ is in the equivalence class associated to $\rho$.

1.3 The Quantum Detection Problem

Let $L(\beta ,\mathbb {R})$ denote the set of real-valued bounded functions defined on $\beta$. Given a POVM $\Pi$ associated to a von Neumann algebra $\mathcal {A}$, the quantum detection problem is formulated as follows.

Quantum Detection Problem Is there a unique quantum state $\rho \in \mathcal {S}(\mathcal {A})$ compatible with the set of quantum measurements performed by the POVM $\Pi$?

Specifically, the quantum detection problem asks two questions:

• (1) Injectivity, or state separability Is the following map injective

  $$\begin{aligned} \mathbb {M}: \mathcal {S}( \mathcal {A})\rightarrow L(\beta ,\mathbb {R}),\ \ \ \mathbb {M}(\rho )(U)=\rho (\Pi (U))? \end{aligned}$$

• (2) Range analysis, or state estimation Assume $\mathbb {M}$ is injective. Then, given a map $p \in L(\beta ,\mathbb {R})$, determine if p is in the range of $\mathbb {M}$, hence is of the form $p=\mathbb {M}(\rho )$ for some unique $\rho \in \mathcal {S}( \mathcal {A})$. If not, find a quantum state $\rho$ that best approximates p in some sense (e.g. robustness to noise).

We point out that in the context of quantum detection in quantum mechanics, a significant amount of work has been put into computing the probability of detection error [10, 13, 14, 17, 19]. We will not address this question here.

1.4 Frame POVMs

In this section we introduce the Hilbert space frame version of the quantum detection problem. For a background on frame POVMs we recommend [1, 11, 12, 16]. For a background on Hilbert space frame theory we recommend [4, 6, 8].

Definition 1.2

A family of vectors $\{x_k\}_{k\in I}$ is a frame for a real or complex, finite or infinite dimensional Hilbert space $\mathbb {H}$ if there are constants $0<A\le B<\infty$ satisfying:

$$\begin{aligned} A\Vert x\Vert ^2 \le \sum _{k\in I}|\langle x, x_k\rangle |^2 \le B\Vert x\Vert ^2, \text{ for } \text{ all } x\in \mathbb {H}. \end{aligned}$$

We have

• (1) A, B are the lower and upper frame bounds of the frame.

• (2) If $A=B$ this is a tight frame. If $A=B=1$ this is a Parseval frame.

• (3) If we only assume we have $0<B<\infty$, this is called a B-Bessel sequence. Note that $\Vert x_k\Vert ^2 \le B$, for all $k\in I$.

We define the analysis operator of the frame as $T:\mathbb {H}\rightarrow \ell _2(I)$ by

$$\begin{aligned} T(x)=(\langle x,x_1\rangle ,\langle x,x_2\rangle ,\ldots ) =\sum _{k\in I}\langle x,x_k\rangle e_k. \end{aligned}$$

The synthesis operator$T^*$ is given by:

$$\begin{aligned} T^*\left( \{a_k\}_{k\in I}\right) = \sum _{k\in I}a_kx_k. \end{aligned}$$

The frame operator is $S=T^*T$. This is a positive, self-adjoint invertible operator on $\mathbb {H}$ satisfying:

$$\begin{aligned} S(x)= \sum _{k\in I}\langle x,x_k\rangle x_k. \end{aligned}$$

It is known that for any frame $\{x_k\}_{k\in I}$, $\{S^{-1/2}x_k\}_{k\in I}$ is a Parseval frame. It is also known that a frame is Parseval if and only if its frame operator is the identity operator.

Definition 1.3

A frame $\{x_k\}_{k\in I}$ is said to be bounded if there is a constant $C>0$ such that

$$\begin{aligned} \Vert x_k\Vert \ge C, \text{ for } \text{ all } k\in I. \end{aligned}$$

If $\{x_k\}_{k\in I}$ is a Parseval frame for a Hilbert space $\mathbb {H}$, it naturally induces a POVM $\Pi$ on $X=I$ with $\beta =2^{I}$ (the power set of I):

$$\begin{aligned} \Pi (U) =\sum _{k\in U}x_kx_k^*, \text{ where } x_k^*:\mathbb {H}\rightarrow \mathbb {C},\ x_k^*(x)=\langle x,x_k\rangle , \end{aligned}$$

with strong convergence for any $U\subset I$.

Given a state $T\in \mathcal {S}(\mathbb {H})$ (i.e. a unit-trace, trace class, positive, self-adjoint operator on $\mathbb {H}$), the frame induced quantum measurement is given by the function

$$\begin{aligned} p:\beta \rightarrow \mathbb {R},\ \ p(U)=\sum _{k\in U}\mathrm {tr}(Tx_kx_k^*) = \sum _{k\in U}\langle Tx_k,x_k\rangle . \end{aligned}$$

For the von Neumann algebra $\mathcal {A}=L^{\infty }(\mathbb {H})$, the quantum states coincide with the convex set of states $\mathcal {S}(\mathbb {H})$. In this case, the injectivity problem and the state estimation problem ask:

Injectivity Problem Is there a Parseval frame $\chi =\{x_k\}_{k\in I}$ so that the map $\mathbb {M}:\mathcal {S}(\mathbb {H})\rightarrow L(\beta ,\mathbb {R})$ defined by $\mathbb {M}(T)(U)=\sum _{k\in U}\langle Tx_k,x_k\rangle$ for $U\subset I$ is injective?

State Estimation Problem Given an injective Parseval frame $\{x_k\}_{k\in I}$ and a function $p: \beta \rightarrow \mathbb {R}$, is there any $T\in \mathcal {S}(\mathbb {H})$ so that $\mathbb {M}(T)=p$? If not, find a quantum state T that best approximates p.

1.5 Generalizing Quantum Detection

We will work on a much more general quantum detection problem. In particular, we will work with

• (1) Self-adjoint operators which may not be positive.

• (2) Operators which are not trace one but are Hilbert Schmidt.

• (3) Frames which are not Parseval.

We will see that solving the problem in this more general form will also solve the original problem.

First, we need a definition.

Definition 1.4

A family of vectors $\mathcal {X}=\{x_k\}_{k\in I}$ in a Hilbert space $\mathbb {H}$ is said to be injective if whenever a Hilbert Schmidt self-adjoint operator T satisfies

$$\begin{aligned} \langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k\in I, \end{aligned}$$

then $T=0$.

Now we will show that we do not need to find Parseval frames for the quantum detection problem. If we have a frame giving injectivity, then its canonical Parseval frame is injective.

Proposition 1.5

Let $\{x_k\}_{k\in I}$ be a frame for $\mathbb {H}$ which gives injectivity. If F is a bounded invertible operator on $\mathbb {H}$, then $\{Fx_k\}_{k\in I}$ also gives injectivity.

Proof

Let T be a Hilbert Schmidt self-adjoint operator such that

$$\begin{aligned} \langle TFx_k, Fx_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Then $\langle F^*TFx_k, x_k\rangle =0$, for all k. Note that $F^*TF$ is also a Hilbert Schmidt self-adjoint operator. Therefore, $F^*TF=0$ and hence $T=0$. $\square$

Corollary 1.6

Let $\{x_{k}\}_{k\in I}$ be a frame with frame operator S. If $\{x_{k}\}_{k\in I}$ gives injectivity, then the canonical Parseval frame $\{S^{-1/2}x_k\}_{k\in I}$ also gives injectivity.

2 The Solution for the Finite Dimensional Case

In this section we will solve the finite dimensional injectivity problem and the state estimation problem for both the real and complex cases. These problems were originally solved by Scott [18] (See also [2]) where the solutions are called informationally complete quantum measurements. We will have to redo this here since we need much more information about the solutions and need proofs in a format that will easily generalize to infinite dimensions.

2.1 Solution to the Injectivity Problem

First, we will see that we do not need to work with positive operators via the following theorem.

Theorem 2.1

Given a family of vectors $\mathcal {X}=\{x_k\}_{k=1}^m$ in $\mathbb {H}^n$, the following are equivalent:

• (1) Whenever T, S are positive and self-adjoint, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=S$.

• (2) Whenever T, S are self-adjoint, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=S$.

• (3) $\mathcal {X}$ is injective.

Proof

$(1)\Rightarrow (2)$: Let T, S be self-adjoint operators such that

$$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } k. \end{aligned}$$

Set

$$\begin{aligned} m_1:=\inf _{\Vert x\Vert =1}\langle T x, x\rangle , \quad m_2:=\inf _{\Vert x\Vert =1}\langle S x, x\rangle \end{aligned}$$

then $m_1, m_2\in \mathbb {R}.$ Set $m=\min \{m_1, m_2\}$.

Now let $P=T-mI$, $Q=S-mI$. Then for any $x\in \mathbb {H}, \Vert x\Vert =1$, we have

$$\begin{aligned} \langle Px, x\rangle =\langle (T-mI)x, x\rangle =\langle Tx, x\rangle - m\ge 0. \end{aligned}$$

Hence, P is positive. Similarly, Q is positive.

We have

$$\begin{aligned} \langle Px_k, x_k\rangle&=\langle (T-mI)x_k, x_k\rangle \\&=\langle Tx_k, x_k\rangle - m\Vert x_k\Vert ^2\\&=\langle Sx_k, x_k\rangle - m\Vert x_k\Vert ^2\\&=\langle Qx_k, x_k\rangle . \end{aligned}$$

By (1) we get $P=Q$ and therefore $T=S$.

$(2)\Rightarrow (3)$: Let T be any self-adjoint operator such that

$$\begin{aligned} \langle Tx_k,x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Then

$$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } k, \end{aligned}$$

where $S=0$. It follows that $T=0$.

$(3)\Rightarrow (1)$: Let any positive self-adjoint operators T, S satisfy

$$\begin{aligned} \langle Tx_k,x_k\rangle =\langle Sx_k,x_k\rangle , \text{ for } \text{ all } k. \end{aligned}$$

Then

$$\begin{aligned} \langle (T-S)x_k,x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since $T-S$ is a self-adjoint operator, $T=S$ by (3). $\square$

Remark 2.2

If we further require that the operators are trace one, then to prove injectivity, we only need to show that if T is trace zero and $\langle Tx_k,x_k\rangle =0$ for all $k=1, 2, \ldots$, then $T=0$. Since if T, S are trace one and

$$\begin{aligned} \langle Tx_k,x_k\rangle =\langle Sx_k,x_k\rangle , \text{ for } \text{ all } k. \end{aligned}$$

then

$$\begin{aligned} \langle (T-S)x_k,x_k\rangle =0, \text{ for } \text{ all } k \text{ and } \mathrm {tr}(T-S)=0. \end{aligned}$$

2.1.1 The Real Case

We start with a proposition which shows where our classification of the quantum detection problem comes from.

Proposition 2.3

Given a self-adjoint operator $T=(a_{ij})_{i,j=1}^n$ on $\mathbb {R}^n$ and a vector $x=(x_1,x_2,\ldots ,x_n)\in \mathbb {R}^n$, we have

$$\begin{aligned} \langle Tx,x\rangle = \sum _{i=1}^n \sum _{j=1}^n a_{ij}x_ix_j= \sum _{i=1}^na_{ii}x_i^2 + 2\sum _{i=1}^n\sum _{j=i+1}^na_{ij}x_ix_j. \end{aligned}$$

Proof

First we compute:

$$\begin{aligned} Tx = \left( \sum _{j=1}^na_{1j}x_j,\sum _{j=1}^na_{2j}x_j,\ldots , \sum _{j=1}^n a_{nj}x_j\right) . \end{aligned}$$

So,

$$\begin{aligned} \langle Tx,x\rangle = \sum _{j=1}^na_{1j}x_1x_{j} + \sum _{j=1}^na_{2j}x_2x_j+\cdots + \sum _{j=1}^n a_{nj}x_nx_j. \end{aligned}$$

Using the fact that T is self-adjoint:

$$\begin{aligned} \langle Tx,x\rangle&= \sum _{i=1}^n \sum _{j=1}^n a_{ij}x_ix_j\\&= \sum _{i=1}^na_{ii}x_i^2 + \sum _{1\le i<j\le n}a_{ij}x_ix_j + \sum _{1\le j<i\le n}a_{ij}x_ix_j\\&= \sum _{i=1}^na_{ii}x_i^2 + \sum _{1\le i<j\le n}a_{ij}x_ix_j + \sum _{1\le i<j\le n}a_{ji}x_jx_i\\&= \sum _{i=1}^na_{ii}x_i^2 + 2\sum _{1\le i<j\le n}a_{ij}x_ix_j. \end{aligned}$$

$\square$

This proposition leads us to the following definition:

Definition 2.4

To a vector $x=(x_1, x_2, \ldots , x_n)\in \mathbb {R}^n$ we associate a vector $\tilde{x}$ in $\mathbb {R}^{\frac{n(n+1)}{2}}$ by:

$$\begin{aligned} \tilde{x}=(x_1x_1, x_1x_2, \ldots , x_1x_n; x_2x_2, x_2x_3, \ldots , x_2x_n;\ldots ; x_{n-1}x_{n-1}, x_{n-1}x_n; x_nx_n). \end{aligned}$$

To a self-adjoint operator $T=(a_{ij})_{i,j=1}^n$ on $\mathbb {R}^n$, we associate a vector $\tilde{T}$ in $\mathbb {R}^{\frac{n(n+1)}{2}}$ by:

$$\begin{aligned} \tilde{T}= (a_{11}, 2a_{12},\ldots , 2a_{1n}; a_{22}, 2a_{23}, \ldots , 2a_{2n}; \ldots ; a_{(n-1)(n-1)}, 2a_{(n-1)n}; a_{nn}). \end{aligned}$$

Proposition 2.3 now becomes:

Corollary 2.5

Given a self-adjoint operator $T=(a_{ij})_{i,j=1}^n$ on $\mathbb {R}^n$ and a vector $x=(x_1,x_2,\ldots ,x_n)\in \mathbb {R}^n$, we have

$$\begin{aligned} \langle Tx,x\rangle = \langle \tilde{T},\tilde{x}\rangle . \end{aligned}$$

We are now able to give a classification of the frames $\chi$ which give injectivity for the quantum detection problem.

Theorem 2.6

Let $\chi =\{x_k\}_{k=1}^m$ be a frame for $\mathbb {R}^n$. The following are equivalent:

• (1) $\chi$ gives injectivity.

• (2) We have that $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathcal {K}:=\mathbb {R}^\frac{n(n+1)}{2}$.

Proof

$(1)\Rightarrow (2)$: Let a vector

$$\begin{aligned} a=(a_{11}, a_{12},\ldots , a_{1n}; a_{22}, a_{23}, \ldots , a_{2n}; \ldots ; a_{(n-1)(n-1)}, a_{(n-1)n}; a_{nn})\in \mathcal {K} \end{aligned}$$

be such that $\langle a, \tilde{x}_k\rangle =0$ for all k.

Define an operator $T=(b_{ij})_{i,j=1}^n$ on $\mathbb {R}^n$, where $b_{ii}=a_{ii}$ for $i=1, 2, \ldots , n$ and $b_{ij}=b_{ji}=\dfrac{1}{2}a_{ij}$ for $i<j$. Then T is a self-adjoint operator.

For any $x=(x_1, x_2, \ldots , x_n)\in \mathbb {R}^n$ we have

$$\begin{aligned} \langle Tx,x\rangle&= \sum _{i=1}^{n}\sum _{j=1}^{n}b_{ij}x_ix_j\\&=\sum _{i=1}^{n}b_{ii}x_i^2+2\sum _{1\le i<j\le n}b_{ij}x_ix_j\\&=\sum _{i=1}^{n}a_{ii}x_i^2+\sum _{1\le i<j\le n}a_{ij}x_ix_j\\&=\langle a, \tilde{x}\rangle . \end{aligned}$$

Therefore, $\langle Tx_k,x_k\rangle =\langle a, \tilde{x}_k\rangle =0$ for all k. This implies $T=0$ and hence $a=0$.

$(2)\Rightarrow (1)$: Let $T=(a_{ij})_{i,j=1}^n$ be a self-adjoint operator such that

$$\begin{aligned} \langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Then by Corollary 2.5,

$$\begin{aligned} \langle \tilde{T}, \tilde{x}_k\rangle =\langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathcal {K}$, we have that $\tilde{T}=0$ and so $T=0$. $\square$

The theorem gives a lower limit on the number of vectors needed to achieve injectivity.

Corollary 2.7

If a frame $\mathcal {X}=\{x_k\}_{k=1}^m$ gives injectivity in $\mathbb {R}^n$, then $m\ge \dfrac{n(n+1)}{2}$.

As a consequence (See the related [7]):

Corollary 2.8

Given a frame $\{x_k\}_{k=1}^m$ for $\mathbb {R}^n$, the following are equivalent:

• (1) The family $\{x_kx_k^*\}_{k=1}^m$ spans the class of self-adjoint operators.

• (2) The family of vectors $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathbb {R}^{\frac{n(n+1)}{2}}$.

Proof

This is immediate since for every $x\in \mathbb {R}^n$ and self-adjoint operator T, we have

$$\begin{aligned} \langle T, xx^*\rangle =\mathrm {tr}(Txx^*)=\langle Tx, x\rangle . \end{aligned}$$

$\square$

Remark 2.9

For any of the frames $\{x_k\}_{k=1}^m$ giving injectivity, if S is the frame operator, then $\{S^{-1/2}x_k\}_{k=1}^m$ is a Parseval frame giving injectivity by Corollary 1.6.

Normally in the frame quantum detection problem, there is the added assumption that the trace of the operators is one. We will now see that with this assumption, we can eliminate one measurement. We start with a simple example.

Example 2.10

Let $\mathcal {X}=\{(1,0), (1,1)\}$ in $\mathbb {R}^2$. Then $\mathcal {X}$ gives injectivity in $\mathbb {R}^2$ for all self-adjoint operators of trace one.

Indeed, let

$$\begin{aligned} T=\begin{bmatrix} a&b\\ b&c \end{bmatrix} \end{aligned}$$

be a self-adjoint matrix of trace zero such that

$$\begin{aligned} \langle T(1,0),(1,0)\rangle = \langle T(1,1), (1,1)\rangle =0. \end{aligned}$$

Then

$$\begin{aligned} a= \langle T(1,0),(1,0)\rangle = 0. \end{aligned}$$

Since $a+c=0$ and

$$\begin{aligned} \langle T(1,1),(1,1)\rangle = \langle (a+b,b+c), (1,1)\rangle = a+2b+c, \end{aligned}$$

then $b=c=0$. So $T=0$.

For the classification of all frames which give injectivity with this added assumption, we will need:

Definition 2.11

Let $x=(x_1, x_2, \ldots , x_n)\in \mathbb {R}^n$. Define

$$\begin{aligned} \tilde{x}=(x_1x_2, \ldots , x_1x_n; x_2^2-x_1^2, x_2x_3, \ldots , x_2x_n;\ldots ; x_{n-1}^2-x_1^2, x_{n-1}x_n; x_n^2-x_1^2). \end{aligned}$$

Now we can prove the trace one version of our classification.

Theorem 2.12

Let $\mathcal {X}=\{x_k\}_{k=1}^m$ be a frame for $\mathbb {R}^n$. The following are equivalent:

• (1) $\mathcal {X}$ gives injectivity for all self-adjoint operators of trace one.

• (2) We have that $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathcal {K}:=\mathbb {R}^{\frac{n(n+1)}{2}-1}$.

Proof

Note that we are trying to show that when two positive, self-adjoint operators T, S of trace one satisfy

$$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } k=1,2,\ldots ,m, \end{aligned}$$

then $T=S$. This is clearly equivalent to showing that if T is a self-adjoint operator of trace zero and $\langle Tx_k,x_k\rangle =0$ for all $k=1,2,\ldots ,$ then $T=0$.

$(1)\Rightarrow (2)$: Let a vector

$$\begin{aligned} a=(a_{12}, \ldots , a_{1n}; a_{22}, \ldots , a_{2n}; \ldots ; a_{(n-1)(n-1)}, a_{(n-1)n}; a_{nn})\in \mathcal {K} \end{aligned}$$

be such that $\langle a, \tilde{x}_k\rangle =0$ for all k.

Define an operator $T=(b_{ij})_{i,j=1}^n$, where $b_{11}=-\sum _{i=2}^na_{ii}, b_{ii}=a_{ii}$ for $i=2, 3, \ldots , n$ and $b_{ij}=b_{ji}=\dfrac{1}{2}a_{ij}$ for $i<j$. Then T is self-adjoint and $\mathrm {tr}(T)=0$.

For any $x=(x_1, x_2, \ldots , x_n)\in \mathbb {R}^n$ we have

$$\begin{aligned} \langle Tx,x\rangle&= \sum _{i=1}^{n}\sum _{j=1}^{n}b_{ij}x_ix_j\\&=\sum _{i=1}^{n}b_{ii}x_i^2+2\sum _{1\le i<j\le n}b_{ij}x_ix_j\\&=\left( -\sum _{i=2}^{n}a_{ii}\right) x_1^2+\sum _{i=2}^{n}a_{ii}x_i^2+\sum _{1\le i<j\le n}a_{ij}x_ix_j\\&=\langle a, \tilde{x}\rangle . \end{aligned}$$

Therefore, $\langle Tx_k,x_k\rangle =\langle a, \tilde{x}_k\rangle =0$ for all k. This implies $T=0$ and hence $a=0$.

$(2)\Rightarrow (1)$: Let $T=(a_{ij})_{i,j=1}^n$ be a self-adjoint operator with $\mathrm {tr}(T)=0$ and such that $\langle Tx_k, x_k\rangle =0$ for all k. Then $a_{11}=-\sum _{i=2}^{n}a_{ii}.$

Define

$$\begin{aligned} \tilde{T}\,=\,(2a_{12}, 2a_{13}, \ldots , 2a_{1n}; a_{22}, 2a_{23}, \ldots , 2a_{2n}; \ldots ; a_{(n-1)(n-1)}, 2a_{(n-1)n}; a_{nn}). \end{aligned}$$

Then $\tilde{T}\in \mathcal {K}$ and

$$\begin{aligned} \langle \tilde{T}, \tilde{x}_k\rangle =\langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathcal {K}$, then $\tilde{T}=0$. Hence $T=0$. $\square$

2.1.2 The Complex Case

We need to adjust some of the definitions for the real case so they will apply to the complex case.

Definition 2.13

Given $x=(x_1,x_2,\ldots ,x_n) \in \mathbb {C}^n$, define

$$\begin{aligned} \tilde{x}&= (|x_1|^2,\mathrm {Re}(\bar{x}_1x_2), \mathrm {Im}(\bar{x}_1x_2),\ldots ,\mathrm {Re}(\bar{x}_1x_n),\mathrm {Im}(\bar{x}_1x_n);\\&|x_2|^2,\mathrm {Re}(\bar{x}_2x_3),\mathrm {Im}(\bar{x}_2x_3), \ldots ,\mathrm {Re}(\bar{x}_2x_n),\mathrm {Im}(\bar{x}_2x_n);\ldots ;\\&|x|^2_{n-1},\mathrm {Re}(\bar{x}_{n-1}x_n), \mathrm {Im}(\bar{x}_{n-1}x_n); |x_n|^2)\in \mathbb {R}^{n^2}. \end{aligned}$$

Now we can give our classification theorem for injectivity in the quantum detection problem for the complex case.

Theorem 2.14

Let $\mathcal {X}=\{x_k\}_{k=1}^m$ be a frame for $\mathbb {C}^n$. The following are equivalent:

• (1) $\mathcal {X}$ gives injectivity.

• (2) We have that $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathbb {R}^{n^2}$.

Proof

$(1)\Rightarrow (2)$: Let a be any vector

$$\begin{aligned}&a= (a_{11},u_{12},v_{12},\ldots ,u_{1n},v_{1n}; a_{22},u_{23},v_{23}, \ldots ,u_{2n},v_{2n};\ldots ;\\&\quad a_{(n-1)(n-1)},u_{(n-1)n},v_{(n-1)n}; a_{nn})\in \mathbb {R}^{n^2} \end{aligned}$$

such that $\langle a,\tilde{x}_k\rangle =0$ for all k.

Define an operator $T=(b_{ij})_{i,j=1}^n$ with $b_{ii}=a_{ii}$ for $i=1, 2, \ldots , n$ and $b_{ij}=\bar{b}_{ji}=\dfrac{1}{2}(u_{ij}-\iota v_{ij})$ for $i<j$. Then T is a self-adjoint operator.

For any $x=(x_1, x_2, \ldots , x_n)\in \mathbb {C}^n$ we have

$$\begin{aligned} \langle Tx,x\rangle&= \sum _{i=1}^n \sum _{j=1}^n b_{ij}\bar{x}_ix_j\\&= \sum _{i=1}^nb_{ii}|x_i|^2 + \sum _{1\le i<j\le n}b_{ij}\bar{x}_ix_j + \sum _{1\le j<i\le n}b_{ij}\bar{x}_ix_j\\&= \sum _{i=1}^nb_{ii}|x_i|^2 +\sum _{1\le i<j\le n}b_{ij}\bar{x}_ix_j + \sum _{1\le j<i\le n}\bar{b}_{ji}\bar{x}_ix_j\\&= \sum _{i=1}^nb_{ii}|x_i|^2 +\sum _{1\le i<j\le n}b_{ij}\bar{x}_ix_j + \sum _{1\le i<j\le n}\bar{b}_{ij}\bar{x}_jx_i\\&=\sum _{i=1}^nb_{ii}|x_i|^2 + 2\sum _{1\le i<j\le n}\mathrm {Re}( b_{ij}\bar{x}_ix_j)\\&=\sum _{i=1}^nb_{ii}|x_i|^2 + 2\sum _{1\le i<j\le n}(\mathrm {Re}( b_{ij})\mathrm {Re}(\bar{x}_ix_j)-\mathrm {Im}( b_{ij})\mathrm {Im}(\bar{x}_ix_j))\\&=\sum _{i=1}^{n}a_{ii}|x_i|^2+\sum _{1\le i<j\le n}\left( u_{ij}\mathrm {Re}(\bar{x}_ix_j)+v_{ij}\mathrm {Im}(\bar{x}_ix_j)\right) \\&=\langle a,\tilde{x}\rangle . \end{aligned}$$

Therefore, $\langle Tx_k,x_k\rangle =\langle a, \tilde{x}_k\rangle =0$ for all k. This implies $T=0$ and hence $a=0$.

$(2)\Rightarrow (1)$: Let $T=(a_{ij})_{i,j=1}^n$ be a self-adjoint operator such that

$$\begin{aligned} \langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Define

$$\begin{aligned}&\tilde{T} = (a_{11},2\mathrm {Re}(a_{12}), -2\mathrm {Im}(a_{12}),\ldots ,2\mathrm {Re}(a_{1n}),-2\mathrm {Im}(a_{1n});\\&\quad a_{22},2\mathrm {Re}(a_{23}),-2\mathrm {Im}(a_{23}), \ldots ,2\mathrm {Re}(a_{2n}),-2\mathrm {Im}(a_{2n});\ldots ; \\&\quad a_{(n-1)(n-1)},2\mathrm {Re}(a_{(n-1)n}), -2\mathrm {Im}(a_{(n-1)n}); a_{nn})\in \mathbb {R}^{n^2}. \end{aligned}$$

Then we have

$$\begin{aligned} \langle \tilde{T}, \tilde{x}_k\rangle =\langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathcal {K}$ we have that $\tilde{T}=0$ and so $T=0$; i.e. $\{x_k\}_{k=1}^m$ gives injectivity. $\square$

Corollary 2.15

If a frame $\mathcal {X}=\{x_k\}_{k=1}^m$ gives injectivity in $\mathbb {C}^n$, then $m\ge n^2$.

Similar to the real case, we have a classification for injectivity for positive self-adjoint operators of trace one in a complex Hilbert space. This requires another definition to fit this case.

Definition 2.16

Given $x=(x_1,x_2,\ldots ,x_n) \in \mathbb {C}^n$, define

$$\begin{aligned}&\tilde{x} = (\mathrm {Re}(\bar{x}_1x_2), \mathrm {Im}(\bar{x}_1x_2),\ldots ,\mathrm {Re}(\bar{x}_1x_n),\mathrm {Im}(\bar{x}_1x_n);\\&\quad |x_2|^2-|x_1|^2,\mathrm {Re}(\bar{x}_2x_3),\mathrm {Im}(\bar{x}_2x_3), \ldots ,\mathrm {Re}(\bar{x}_2x_n),\mathrm {Im}(\bar{x}_2x_n);\ldots ;\\&\quad |x_{n-1}|^2-|x_1|^2,\mathrm {Re}(\bar{x}_{n-1}x_n), \mathrm {Im}(\bar{x}_{n-1}x_n); |x_n|^2-|x_1|^2)\in \mathbb {R}^{n^2-1}. \end{aligned}$$

Now we classify the frames which give injectivity in the complex case for operators of trace one.

Theorem 2.17

Let $\mathcal {X}=\{x_k\}_{k=1}^m$ be a frame for $\mathbb {C}^n$. The following are equivalent:

• (1) $\mathcal {X}$ gives injectivity for all self-adjoint operators of trace one.

• (2) We have that $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathbb {R}^{n^2-1}$.

Proof

$(1)\Rightarrow (2)$: Let a vector

$$\begin{aligned}&a= (u_{12},v_{12},\ldots ,u_{1n},v_{1n}; a_{22},u_{23},v_{23}, \ldots ,u_{2n},v_{2n};\ldots ;\\&\quad a_{(n-1)(n-1)},u_{(n-1)n},v_{(n-1)n}; a_{nn})\in \mathbb {R}^{n^2-1} \end{aligned}$$

be such that $\langle a,\tilde{x}_k\rangle =0$ for all k.

Define an operator $T=(b_{ij})_{i,j=1}^n$ with $b_{11}=-\sum _{i=2}^{n}a_{ii}$, $b_{ii}=a_{ii}$ for $i= 2, \ldots , n$ and $b_{ij}=\bar{b}_{ji}=\dfrac{1}{2}(u_{ij}-\iota v_{ij})$ for $i<j$. Then T is a self-adjoint operator and $\mathrm {tr}(T)=0$.

For any $x=(x_1, x_2, \ldots , x_n)\in \mathbb {C}^n$ we have

$$\begin{aligned} \langle Tx,x\rangle&=\sum _{i=1}^nb_{ii}|x_i|^2 + 2\sum _{1\le i<j\le n}(\mathrm {Re}( b_{ij})\mathrm {Re}(\bar{x}_ix_j)-\mathrm {Im}( b_{ij})\mathrm {Im}(\bar{x}_ix_j))\\&=\left( -\sum _{i=2}^{n}a_{ii}\right) |x_1|^2+\sum _{i=2}^{n}a_{ii}|x_i|^2+\sum _{1\le i<j\le n}\left( u_{ij}\mathrm {Re}(\bar{x}_ix_j)+v_{ij}\mathrm {Im}(\bar{x}_ix_j)\right) \\&=\langle a,\tilde{x}\rangle . \end{aligned}$$

Therefore, $\langle Tx_k,x_k\rangle =\langle a, \tilde{x}_k\rangle =0$ for all k. This implies $T=0$ and hence $a=0$.

$(2)\Rightarrow (1)$: Let $T=(a_{ij})_{i,j=1}^n$ be a self-adjoint operator such that $\mathrm {tr}(T)=0$ and $\langle Tx_k, x_k\rangle =0$ for all k. Then $a_{11}=-\sum _{i=2}^{n}a_{ii}.$

Define

$$\begin{aligned} \tilde{T} =&(2\mathrm {Re}(a_{12}), -2\mathrm {Im}(a_{12}),\ldots ,2\mathrm {Re}(a_{1n}),-2\mathrm {Im}(a_{1n});\\&a_{22},2\mathrm {Re}(a_{23}),-2\mathrm {Im}(a_{23}), \ldots ,2\mathrm {Re}(a_{2n}),-2\mathrm {Im}(a_{2n});\ldots ; \\&a_{(n-1)(n-1)},2\mathrm {Re}(a_{(n-1)n}), -2\mathrm {Im}(a_{(n-1)n}); a_{nn})\in \mathbb {R}^{n^2-1}. \end{aligned}$$

Then we have that

$$\begin{aligned} \langle \tilde{T}, \tilde{x}_k\rangle =\langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since $\{\tilde{x}_k\}_{k=1}^m$ spans $\mathbb {R}^{n^2-1}$ then $\tilde{T}=0$. Hence $T=0$. $\square$

Now we will give another classification of injectivity for the quantum detection problem. This classification has the disadvantage that the requirements are quite complex and difficult to verify in practice. The advantage here is in the other direction. That is, if a frame gives injectivity in the quantum detection problem, then it must satisfy these complex requirements.

Theorem 2.18

Let $\mathcal {X}=\{x_k\}_{k=1}^m$ be a frame for a real or complex Hilbert space $\mathbb {H}^n$. The following are equivalent:

• (1) $\mathcal {X}$ gives injectivity.

• (2) For every orthonormal basis $\mathcal {E}=\{e_j\}_{j=1}^n$ for $\mathbb {H}^n$ we have:

  $$\begin{aligned} H(\mathcal {E})=:\mathrm {span}\{(|\langle x_k,e_1\rangle |^2,|\langle x_k,e_2\rangle |^2, \ldots , |\langle x_k,e_n\rangle |^2):k=1,2,\ldots ,m\}=\mathbb {R}^n. \end{aligned}$$

Proof

$(1)\Rightarrow (2)$: We prove the contrapositive. Suppose that (2) fails. Then there is an orthonormal basis $\mathcal {E}=\{e_j\}_{j=1}^n$ so that $H(\mathcal {E})\not = \mathbb {R}^n$. Hence there is a non-zero vector $\lambda =(\lambda _1,\lambda _2,\ldots ,\lambda _n)\in \mathbb {R}^n$ such that $\lambda \perp H(\mathcal {E})$.

Define an operator on $\mathbb {H}^n$ by

$$\begin{aligned} Te_j = \lambda _je_j, j=1, 2, \ldots , n. \end{aligned}$$

Then T is a non-zero self-adjoint operator and satisfies $\langle Tx_k,x_k\rangle =0$, for all $k=1, 2,\ldots ,m$, which is a contradiction.

$(2)\Rightarrow (1)$: Let T be a self-adjoint operator such that $\langle Tx_k,x_k\rangle =0$, for all k. Let $\mathcal {E}= \{e_j\}_{j=1}^n$ be an eigenbasis for T with respective eigenvalues $\{\lambda _j\}_{j=1}^n$. Then for every $k=1,2,\ldots ,m$ we have

$$\begin{aligned} \langle Tx_k,x_k\rangle = \sum _{j=1}^n\lambda _j|\langle x_k, e_j\rangle |^2 =0. \end{aligned}$$

That is,

$$\begin{aligned} (\lambda _1,\lambda _2,\ldots ,\lambda _n) \perp H(\mathcal {E}) = \mathbb {R}^n \text{ by } \text{ assumption } \text{(2) }. \end{aligned}$$

Therefore, $\lambda _j=0$ for all $j=1,2,\ldots ,n$ and so $T=0$. $\square$

Finally in this subsection, we notice that if a family of vectors gives injectivity in a Hilbert space $\mathbb {H}^n$, then it is a frame for $\mathbb {H}^n$.

Proposition 2.19

Let $\{x_k\}_{k=1}^m$ be a family of vectors in $\mathbb {H}^n$ which is injective. Then $\mathrm {span}\{x_k\}_{k=1}^m=\mathbb {H}^n$.

Proof

Suppose by contradiction that $W:=\mathrm {span}\{x_k\}_{k=1}^m\not =\mathbb {H}^n$. Let P be the orthogonal projection onto $W^\perp$. Then $\langle Px_k, x_k\rangle =0$ for all k, but $P\not =0$, a contradiction.

$\square$

2.2 Constructing the Solutions to the Injectivity Problem

In this subsection, we will construct large classes of frames which give injectivity for the quantum detection problem in both the real and complex cases.

Theorem 2.20

Let $\{x_k\}_{k=1}^n$ be a linearly independent set in $\mathbb {R}^n$ such that the first coordinates of these vectors are non-zero. Now choose $(n-1)$ linearly independent vectors $\{x_k\}_{k=n+1}^{2n-1}$ in $\mathbb {R}^n$ such that each vector is zero in the first coordinate and is non-zero in the second coordinate. Continuing this procedure we get a frame $\{x_k\}_{k=1}^{\frac{n(n+1)}{2}}$ which gives injectivity.

Proof

We will show that $\{\tilde{x}_k\}_{k=1}^{\frac{n(n+1)}{2}}$ is a basis for $\mathbb {R}^{\frac{n(n+1)}{2}}$.

Indeed, suppose that $\sum _{k=1}^{\frac{n(n+1)}{2}}\alpha _k\tilde{x}_k=0$ for some scalars $\{\alpha _k\}$. Since after n, all tilde vectors are zero in the first coordinate, then we get

$$\begin{aligned} \sum _{k=1}^n\alpha _kx_{k1}x_k=0. \end{aligned}$$

Since $\{x_k\}_{k=1}^n$ are linearly independent, $\alpha _kx_{k1}=0$ for all k and since $x_{k1}\not =0$, $\alpha _k=0$ for $k=1,2,\ldots ,n.$

Now do this argument for the next $(n-1)$ vectors and continue we get $\alpha _k=0$ for all $k=1, 2, \ldots , \frac{n(n+1)}{2}$. $\square$

A simple example satisfying the construction is the following.

Example 2.21

The frame

$$\begin{aligned} \{e_i\}_{i=1}^n\cup \{e_i+e_j : i<j\}_{i,j=1}^n \end{aligned}$$

gives injectivity.

For the complex case, we have the following construction. The proof is as in the real case.

Theorem 2.22

Let $\{x_k\}_{k=1}^{2n-1}$ be a basis for $\mathbb {R}^{2n-1}$, where

$$\begin{aligned} x_k=(u_{k1}, u_{k2}, v_{k2}, \ldots , u_{kn}, v_{kn}) \end{aligned}$$

and $u_{k1}\not =0$, $k=1,\ldots , 2n-1.$

Define $(2n-1)$ vectors $\{z_k\}_{k=1}^{2n-1}$ in $\mathbb {C}^n$ by

$$\begin{aligned} z_k=(u_{k1}, u_{k2}+\iota v_{k2}, \ldots , u_{kn}+\iota v_{kn}). \end{aligned}$$

Now let $\{x_k\}_{k=2n}^{4n-4}$ be a basis for $\mathbb {R}^{2n-3}$, where

$$\begin{aligned} x_k=(u_{k2}, u_{k3}, v_{k3}, \ldots , u_{kn}, v_{kn}) \end{aligned}$$

and $u_{k2}\not =0$, $k=2n, \ldots , 4n-4$.

Define $(2n-3)$ vectors $\{z_k\}_{k=2n}^{4n-4}$ in $\mathbb {C}^n$ by

$$\begin{aligned} z_k=(0, u_{k2}, u_{k3}+\iota v_{k3}, \ldots , u_{kn}+\iota v_{kn}). \end{aligned}$$

Continuing this procedure we get $n^2$ vectors $\{z_k\}_{k=1}^{n^2}$ in $\mathbb {C}^n$ and they give injectivity.

As we have seen, we can get Parseval frames giving injectivity by taking $\{S^{-1/2}x_k\}_{k=1}^m$, where $\{x_k\}_{k=1}^m$ gives injectivity and has frame operator S. But the above construction can be adjusted to directly construct Parseval frames giving injectivity.

Theorem 2.23

Let $\{\lambda _{ij}\}_{i=1,j=i}^{\ n\ \ \ n}$ be non-negative numbers satisfying:

• (1) $\lambda _{ij}=0 \text{ if } \text{ and } \text{ only } \text{ if } j<i.$

• (2) For each $j=1,2,\ldots ,n$ we have $\sum _{i=1}^n \lambda _{ij}=1.$

Let $\mathcal {E}=\{e_j\}_{j=1}^n$ be the canonical basis of $\mathbb {R}^n$. Let $\{x_k\}_{k=1}^{\frac{n(n+1)}{2}}$ be vectors in $\mathbb {R}^n$ which satisfy:

• (1) $\{x_k\}_{k=1}^n$ is a linearly independent set with $x_{k1}\not = 0$ for all $k=1,\ldots ,n$ and it has frame operator $S_1$ with eigenvectors $\mathcal {E}$ and respective eigenvalues $\{\lambda _{1j}\}_{j=1}^n$ (See [5].)

• (2) $\{x_k\}_{k=n+1}^{2n-1}$ is a linearly independent set with $x_{k1}=0$, for all k, $x_{k2}\not = 0$ for all k, and it has frame operator $S_2$ with eigenvectors $\mathcal {E}$ and respective eigenvalues $\{\lambda _{2j}\}_{j=1}^n$.

• (3) continue.

Then the vectors $\{x_k\}_{k=1}^{\frac{n(n+1)}{2}}$ form a Parseval frame for $\mathbb {R}^n$ which is injective.

Proof

This is injective by Theorem 2.20. To see that it is Parseval, observe that the frame operator of this frame is $\sum _{i=1}^nS_i$. Now, let $y\in \mathbb {R}^n$ and compute:

$$\begin{aligned} \sum _{i=1}^nS_iy&= \sum _{i=1}^n S_i\left( \sum _{j=1}^n \langle y,e_j\rangle e_j\right) \\&= \sum _{i=1}^n\sum _{j=1}^n \langle y,e_j\rangle S_ie_j\\&= \sum _{i=1}^n \sum _{j=1}^n \langle y,e_j\rangle \lambda _{ij}e_j\\&= \sum _{j=1}^n \langle y, e_j\rangle e_j \sum _{i=1}^n \lambda _{ij}\\&= \sum _{j=1}^n \langle y,e_j\rangle e_j = y. \end{aligned}$$

$\square$

Similarly, we have the following theorem for the complex case.

Theorem 2.24

Fix $\{\lambda _{ij}\}_{i=1,j=i}^{ n\ n}$ be non-negative numbers satisfying:

• (1) $\lambda _{ij}=0 \text{ if } \text{ and } \text{ only } \text{ if } j<i.$

• (2) For each $j=1,2,\ldots ,n$ we have $\sum _{i=1}^n \lambda _{ij}=1.$

Let $\mathcal {E}=\{e_j\}_{j=1}^n$ be the canonical basis of $\mathbb {C}^n$. Let $\{z_k\}_{k=1}^{n^2}$ be vectors in $\mathbb {C}^n$ which satisfy:

• (1) For each $k=1, \ldots , 2n-1$, $z_k$ has the form

  $$\begin{aligned} z_k=(u_{k1}, u_{k2}+\iota v_{k2}, \ldots , u_{kn}+\iota v_{kn}), \end{aligned}$$

  where $u_{k1}\not =0$ and the set $\{(u_{k1}, u_{k2}, v_{k2}, \ldots , u_{kn}, v_{kn})\}_{k=1}^{2n-1}$ is linearly independent in $\mathbb {R}^{2n-1}$. Moreover $\{z_k\}_{k=1}^{2n-1}$ has frame operator $S_1$ with eigenvectors $\mathcal {E}$ and respective eigenvalues $\{\lambda _{1j}\}_{j=1}^n$.

• (2) For each $k=2n, \ldots , 4n-4$, $z_k$ has the form

  $$\begin{aligned} z_k=(0, u_{k2}, u_{k3}+\iota v_{k3}, \ldots , u_{kn}+\iota v_{kn}), \end{aligned}$$

  where $u_{k2}\not =0$ and the set $\{(u_{k2}, u_{k3}, v_{k3}, \ldots , u_{kn}, v_{kn})\}_{k=2n}^{4n-4}$ is linearly independent in $\mathbb {R}^{2n-3}$. Moreover $\{z_k\}_{k=2n}^{4n-4}$ has frame operator $S_2$ with eigenvectors $\mathcal {E}$ and respective eigenvalues $\{\lambda _{2j}\}_{j=1}^n$.

• (3) continue.

Then the vectors $\{z_k\}_{k=1}^{n^2}$ form a Parseval frame for $\mathbb {C}^n$ which is injective.

Remark 2.25

We can easily vary the above construction to find frames which give injectivity and have any previously prescribed eigenvalues for their frame operators.

We recall:

Definition 2.26

Two orthonormal bases $\{x_k\}_{k=1}^n$ and $\{y_k\}_{k=1}^n$ are mutually unbiased if

$$\begin{aligned} |\langle x_k,y_j\rangle |= \frac{1}{\sqrt{n}}, \text{ for } \text{ all } i,j=1,2,\ldots ,n. \end{aligned}$$

A family of orthonormal bases is mutually unbiased if each pair is mutually unbiased.

It is known that the maximal number of mutually unbiased bases in $\mathbb {H}^n$ is n+1 and this is rarely achieved. It holds if $n=p^m$ for a prime p. It is observed in [18] and [1] that a maximal family of mutually unbiased bases will give injectivity in the quantum detection problem.

2.3 The Solutions are Open and Dense

In this section we will show that the family of m-element frames which solve the quantum detection injectivity problem is open and dense in the family of all m-element frames. For this, we need to measure the distance between m-element frames. There is a standard metric measuring the distance between frames.

Definition 2.27

Given frames $\mathcal {X}=\{x_k\}_{k=1}^m$ and $\mathcal {Y}=\{y_k\}_{k=1}^m$ for a real or complex Hilbert space $\mathbb {H}^n$, the distance between them is

$$\begin{aligned} d(\mathcal {X},\mathcal {Y})^2 = \sum _{k=1}^m\Vert x_k-y_k\Vert ^2. \end{aligned}$$

Theorem 2.28

The set of all m-element frames on $\mathbb {H}^n$ that give injectivity in the frame quantum detection problem is dense in the space of all m-element frames on $\mathbb {H}^n$.

Proof

We will prove the real case. The complex case is similar.

Let a frame $\{x_k\}_{k=1}^{\frac{n(n+1)}{2}}\subset \mathbb {R}^n$ give injectivity. By Theorem 2.6, this is equivalent to the determinant of the matrix whose rows are $\tilde{x}_k$, $k=1, 2, \ldots , \frac{n(n+1)}{2}$ being non-zero.

The determinant of this matrix is a polynomial of $\frac{n^2(n+1)}{2}$ variables $x_{ki}$ for $1 \le k \le \frac{n(n+1)}{2}$ and $1 \le i \le n$. Since the complement of the zero set of this polynomial is dense in $\mathbb {R}^{\frac{n^2(n+1)}{2}}$, the set of all $\frac{n(n+1)}{2}$-element frames which give injectivity is dense in the space of all $\frac{n(n+1)}{2}$-element frames on $\mathbb {R}^n$.

Now let any m-element frame $\{x_k\}_{k=1}^m$ in $\mathbb {R}^n$ with $m\ge \frac{n(n+1)}{2}$ and $\delta >0$. Then there exists a subframe containing $\frac{n(n+1)}{2}$ vectors. We can assume that this subframe is $\{x_k\}_{k=1}^{\frac{n(n+1)}{2}}$. By denseness above, there is an injective frame $\{y_k\}_{k=1}^{\frac{n(n+1)}{2}}$ such that

$$\begin{aligned} \sum _{k=1}^{n(n+1)/2}\Vert x_k-y_k\Vert ^2<\delta . \end{aligned}$$

Now define a new frame $\{\phi _k\}_{k=1}^m$, where $\phi _k=y_k$ for $k=1, \ldots , \frac{n(n+1)}{2}$ and $\phi _k=x_k$ for $k>\frac{n(n+1)}{2}$. Then the frame $\{\phi _k\}_{k=1}^{m}$ is injective and

$$\begin{aligned} \sum _{k=1}^m\Vert x_k-\phi _k\Vert ^2<\delta . \end{aligned}$$

The conclusion of the theorem then follows. $\square$

Remark 2.29

In the real case it is known that the complement of the zero set of a nontrivial polynomial of n variables is dense in $\mathbb {R}^n$. In the complex case, we see that given a polynomial $P(z_1,..., z_n)$ on $\mathbb {C}^n$, we may write P as

$$\begin{aligned} P'(x_1,y_1,...,x_n,y_n) + \iota P''(x_1,y_1,...,x_n,y_n) \end{aligned}$$

where $z_j = x_j + \iota y_j$. Hence $P'$ and $P''$ are polynomials on $\mathbb {R}^{2n}$. P has a zero if and only if $P'$ and $P''$ have a common zero. We see that the complement of the intersection of the zero sets of $P'$ and $P''$ is dense in $\mathbb {R}^{2n}$ and hence is dense in $\mathbb {C}^{n}$ after natural identification of $\mathbb {R}^{2n}$ with $\mathbb {C}^n$.

Theorem 2.30

The family of all m-element frames on $\mathbb {H}^n$ that give injectivity in the frame quantum detection problem is open in the space of all m-element frames on $\mathbb {H}^n$.

Proof

As above we will prove the real case and the complex case follows similarly.

Denote by $\mathcal {F}$ the space of all $\frac{n(n+1)}{2}$-element frames for $\mathbb {R}^n$. Consider the map:

$$\begin{aligned}&f: \mathcal {F} \longrightarrow \mathbb {R}\\&\mathcal {X}=\{x_k\}_{k=1}^{\frac{n(n+1)}{2}}\longmapsto f(\mathcal {X})=\det \left\{ \tilde{x}_1, \tilde{x}_2,\ldots , \tilde{x}_{\frac{n(n+2)}{2}}\right\} . \end{aligned}$$

Then f is a continuous function. Since $f^{-1}(0)$ is a closed set, by Theorem 2.6, the set of all $\frac{n(n+1)}{2}$-element frames is open in $\mathcal {F}$.

Now let$\mathcal {X}=\{x_k\}_{k =1}^m$ in $\mathbb {R}^n, (m\ge \frac{n(n+1)}{2})$ be an m-element frame which gives injectivity. Then there is a subframe $\mathcal {Y}$ containing $\frac{n(n+1)}{2}$ vectors, which is also injective. Therefore, there exists $\epsilon >0$ such that every $\frac{n(n+1)}{2}$-element frame in the ball $B(\mathcal {Y}, \epsilon )$ is injective. This implies that every m-element frame in the ball $B(\mathcal {X}, \epsilon )$ is also injective. The proof is now complete. $\square$

To show that the Parseval frames giving injectivity in the quantum detection problem are dense in the Parseval frames, we will first prove a very general problem about frames.

Theorem 2.31

Let $\mathcal {P}$ be a property of Hilbert space frames and assume:

• (1) The set of all m-element frames in $\mathbb {H}^n$ having property $\mathcal {P}$ is dense in the set of all m-element frames.

• (2) If a frame $\{x_k\}_{k=1}^m$ with frame operator S has property $\mathcal {P}$, then $\{S^{-1/2}x_k\}_{k=1}^m$ has property $\mathcal {P}$.

Then the set of all m-element Parseval frames with property $\mathcal {P}$ is dense in the set of all m-element Parseval frames.

Proof

Fix $\epsilon >0$ and let $\delta >0$ so that

$$\begin{aligned} 2m\delta ^2 +8(m\delta )^2 m(1+\delta )^2<\epsilon , \ 2m\delta <1. \end{aligned}$$

Let $\{x_k\}_{k=1}^m$ be any Parseval frame for $\mathbb {H}^n$. By denseness, we can choose a frame $\{y_k\}_{k=1}^m$ having property $\mathcal {P}$ and satisfying $\Vert x_k-y_k\Vert \le \delta$, for all $k=1,2,\ldots ,m$. Since $\Vert x_k\Vert \le 1$, we have that $\Vert y_k\Vert \le 1+\delta$. Let $S_1$ be the frame operator of $\{y_k\}_{k=1}^m$. Then,

$$\begin{aligned} \langle S_1x,x\rangle ^{1/2}&= \left( \sum _{k=1}^m|\langle x,y_k\rangle |^2\right) ^{1/2}\\&\le \left( \sum _{k=1}^m|\langle x,x_k\rangle |^2\right) ^{1/2} + \left( \sum _{k=1}^m|\langle x,x_k-y_k\rangle |^2 \right) ^{1/2}\\&\le \Vert x\Vert + \Vert x\Vert \left( \sum _{k=1}^m\Vert x_k-y_k\Vert ^2 \right) ^{1/2}\\&\le \Vert x\Vert (1+m\delta ). \end{aligned}$$

Therefore

$$\begin{aligned} \sum _{k=1}^m|\langle x,y_k\rangle |^2 \le \Vert x\Vert ^2(1+m\delta )^2. \end{aligned}$$

Similarly,

$$\begin{aligned} \sum _{k=1}^m|\langle x,y_k\rangle |^2 \ge \Vert x\Vert ^2(1-m\delta )^2. \end{aligned}$$

I.e. $(1-m\delta )^2 I \le S_1 \le (1+m\delta )^2 I$. Hence, $(1-m\delta )I \le S_1^{1/2}\le (1+m\delta )I$ and so $(1+m\delta )^{-1}I \le S_1^{-1/2}\le (1-m\delta )^{-1}I$. Finally,

$$\begin{aligned} I-(1-m\delta )^{-1}I \le I-S_1^{-1/2} \le I-(1+m\delta )^{-1}I, \end{aligned}$$

and so

$$\begin{aligned} -2m\delta I \le \frac{-m\delta }{1-m\delta }I \le I-S_1^{-1/2} \le \frac{m\delta }{1+m\delta }I \le 2m\delta I. \end{aligned}$$

Now, $\{S_1^{-1/2}y_k\}_{k=1}^m$ is a Parseval frame with property $\mathcal {P}$ and

$$\begin{aligned} \sum _{k=1}^m\Vert x_k-S_1^{-1/2}y_k\Vert ^2&\le 2 \sum _{k=1}^m \Vert x_k-y_k\Vert ^2+ 2\sum _{k=1}^m\Vert (I-S_1^{-1/2})y_k\Vert ^2\\&\le 2m\delta ^2 + 2\sum _{k=1}^m (2m\delta )^2 \Vert y_k\Vert ^2\\\\&\le 2m\delta ^2 +8(m\delta )^2 m(1+\delta )^2< \epsilon . \end{aligned}$$

$\square$

Corollary 2.32

The set of all m-element Parseval frames which give injectivity is dense in the set of all m-element Parseval frames.

2.4 Solution to the State Estimation Problem

In this section we will give a classification of injective Parseval frames for which the state estimation problem is solvable.

Recall that for an injective Parseval frame $\{x_k\}_{k\in I}$ and $\beta =2^I$, the map $\mathbb {M}$ which maps a quantum state $T\in \mathcal {S}(\mathbb {H})$ to a function $p\in L(\beta , \mathbb {R})$ is injective. Given a function $p\in L(\beta , \mathbb {R})$, if $p=\mathbb {M}(T)$ for some $T\in \mathcal {S}(\mathbb {H})$, then for any $U\in \beta$, we must have

$$\begin{aligned} p(U)=\mathbb {M}(T)(U)=\sum _{k\in U}\langle Tx_k, x_k\rangle =\sum _{k\in U}p(\{k\}). \end{aligned}$$

Thus, p must be additive and is determined by its value at the singleton sets $\{k\}$ for all $k\in I$. Therefore, for the state estimation problem in the finite case, we will ask:

The State Estimation Problem Given an injective Parseval frame $\{x_k\}_{k=1}^m$ on $\mathbb {H}^n$ and a measurement vector $a=(a_1,a_2,\ldots ,a_m)\in \mathbb {R}^m$, can we find a positive self-adjoint trace one operator T so that

$$\begin{aligned} \langle Tx_k,x_k\rangle = a_k, \text{ for } \text{ all } k? \end{aligned}$$

Remark 2.33

We will not require the operator T of the problem to be positive and trace one. This will be considered as a special case of the problem. Hence, we will say that the state estimation problem is solvable if there exists a self-adjoint operator T so that

$$\begin{aligned} \langle Tx_k,x_k\rangle = a_k, \text{ for } \text{ all } k. \end{aligned}$$

We will give a complete classification of injective Parseval frames for which the state estimation problem is solvable. Recall that for a vector $x\in \mathbb {R}^n$, the vector $\tilde{x}$ is defined as in the Definition 2.4.

Theorem 2.34

Let $\mathcal {X}=\{x_k\}_{k=1}^m$ be an injective Parseval frame for $\mathbb {R}^n$, and $a=(a_1, a_2, \ldots , a_m)\in \mathbb {R}^m$, the following are equivalent:

• (1) The state estimation problem is solvable.

• (2) $\mathrm {rank}(A)=\mathrm {rank}(B)$, where A is a matrix whose k-row is $\tilde{x}_k$, and $B=[A, a]$.

Proof

Note that a self-adjoint operator T is determined by the values $\langle Te_i, e_j\rangle$ for all $i\le j$. Then the state estimation problem is solvable if and only if there exists a self-adjoint operator T so that

$$\begin{aligned} a_k&=\langle Tx_k,x_k\rangle \\&=\left\langle T\left( \sum _{i=1}^{n}\langle x_k, e_i\rangle e_i\right) ,\sum _{j=1}^{n}\langle x_k, e_j\rangle e_j\right\rangle \\&=\sum _{i=1}^{n}\sum _{j=1}^{n}\langle x_k, e_i\rangle \langle x_k, e_j\rangle \langle Te_i, e_j\rangle \end{aligned}$$

for all k. This is equivalent to the linear system with unknowns $\langle Te_i, e_j\rangle$:

$$\begin{aligned} \sum _{i=1}^nx_{ki}^2\langle Te_i, e_i\rangle +2\sum _{i<j}x_{ki}x_{kj}\langle Te_i, e_j\rangle =a_k,\ k=1, 2, \ldots , m \end{aligned}$$

having a solution, and hence is equivalent to $\mathrm {rank}(A)=\mathrm {rank}(B)$. $\square$

In the case where the number of frame vectors equals $\frac{n(n+1)}{2}$, we have the following corollary.

Corollary 2.35

Let $\mathcal {X}=\{x_k\}_{k=1}^{\frac{n(n+1)}{2}}\subset \mathbb {R}^n$ be an injective Parseval frame. Then the state estimation problem has a unique solution for all choices of vectors $a=(a_1, a_2, \ldots , a_{\frac{n(n+1)}{2}}).$

Proof

By Theorem 2.6, $\mathcal {X}$ is injective is equivalent to $\{\tilde{x}_k\}_{k=1}^{\frac{n(n+1)}{2}}$ is linearly independent. Hence

$$\begin{aligned} \mathrm {rank}\, A=\mathrm {rank}\, B=\frac{n(n+1)}{2}. \end{aligned}$$

The conclusion then follows by Theorem 2.34. $\square$

For the completeness of the state estimation problem, we will state the classification in the case that the operator T is required to be positive, self-adjoint operator of trace one. First, we need to recall Sylvester’s Criterion [15].

Theorem 2.36

A self-adjoint matrix T is positive if and only if all of its principal minors are nonnegative.

Now we have the following classification:

Theorem 2.37

Let $\mathcal {X}=\{x_k\}_{k=1}^m$ be an injective Parseval frame for $\mathbb {R}^n$, and $a=(a_1, a_2, \ldots , a_m)\in \mathbb {R}^m$, the following are equivalent:

• (1) The state estimation problem is solvable for a positive, self-adjoint operator of trace one.

• (2) The linear system

  $$\begin{aligned} \sum _{i=1}^nx_{ki}^2\langle Te_i, e_i\rangle +2\sum _{i<j}x_{ki}x_{kj}\langle Te_i, e_j\rangle =a_k,\ k=1, 2, \ldots , m \end{aligned}$$

  has a solution $\{\langle Te_i, e_j\rangle : i\le j\}$, which determines a self-adjoint matrix T such that all of its principal minors are nonnegative, and $\sum _{i=1}^n\langle Te_i, e_i\rangle =1$.

Remark 2.38

All of the theorems above still hold for the complex case with the corresponding $\tilde{x}_k$, defined as in Definition 2.13. We state one of them here, the other are similar to the real case.

Theorem 2.39

Let $\mathcal {X}=\{x_k\}_{k=1}^m$ be an injective Parseval frame for $\mathbb {C}^n$, and $a=(a_1, a_2, \ldots , a_m)\in \mathbb {R}^m$, the following are equivalent:

• (1) The state estimation problem is solvable.

• (2) $\mathrm {rank}(A)=\mathrm {rank}(B)$, where A is a matrix whose k-row is $\tilde{x}_k$, and $B=[A, a]$.

Proof

In the complex case, a self-adjoint operator T is determined by the values of the real part and imaginary part of $\langle Te_j, e_i\rangle$ for all $i\le j$. Then the state estimation problem is solvable if and only if there exists a self-adjoint operator T so that

$$\begin{aligned} a_k&=\langle Tx_k,x_k\rangle \\&=\left\langle T\left( \sum _{i=1}^{n}\langle x_k, e_i\rangle e_i\right) ,\sum _{j=1}^{n}\langle x_k, e_j\rangle e_j\right\rangle \\&=\sum _{i=1}^{n}\sum _{j=1}^{n}\langle x_k, e_i\rangle \overline{\langle x_k, e_j\rangle }\langle Te_i, e_j\rangle \\&=\sum _{i=1}^n|x_{ki}|^2\langle Te_i, e_i\rangle +2\sum _{i<j}[\mathrm {Re}(\bar{x}_{ki}x_{kj})\mathrm {Re}\langle Te_j, e_i\rangle -\mathrm {Im}(\bar{x}_{ki}x_{kj})\mathrm {Im}\langle Te_j, e_i\rangle ] \end{aligned}$$

for all k.

This is equivalent to the following linear system:

$$\begin{aligned} \sum _{i=1}^n|x_{ki}|^2\langle Te_i, e_i\rangle +2\sum _{i<j}[\mathrm {Re}(\bar{x}_{ki}x_{kj})\mathrm {Re}\langle Te_j, e_i\rangle -\mathrm {Im}(\bar{x}_{ki}x_{kj})\mathrm {Im}\langle Te_j, e_i\rangle ]=a_k, \end{aligned}$$

$k=1, 2, \ldots , m$ with unknowns $\mathrm {Re}\langle Te_j, e_i\rangle , \mathrm {Im}\langle Te_j, e_i\rangle , i\le j$ having a solution, and hence is equivalent to $\mathrm {rank}(A)=\mathrm {rank}(B)$. $\square$

Remark 2.40

If a frame $\{x_k\}_{k=1}^m$ has $m> \frac{n(n+1)}{2}$ in the real case, or $m >n^2$ in the complex case, because of redundancy, it is unlikely the state estimation is solvable. This is because if two of the $x_k's$ are equal while the corresponding $a_k's$ are not, then the problem is not solvable. More generally, if some of the $x_k$ are linearly dependent then at least one of the corresponding $a_k$ is uniquely determined. However, in this case there is a natural way to find the best estimate for the problem. We consider the real case. Note that there always exists a subset $I\subset \{1, 2, \ldots , m\}$ of size $\frac{n(n+1)}{2}$, and a self-adjoint operator T so that $\langle Tx_k, x_k\rangle =a_k$, for all $k\in I$. Therefore, if the state estimation problem is not solvable, it is natural to find such T so that the the distance to the measurement vector a:

$$\begin{aligned} \sum _{k=1}^{m}|\langle Tx_k, x_k\rangle -a_k|^2 \end{aligned}$$

is minimum.

To do this, let $\mathcal {S}$ be the set of all bases of $\mathbb {R}^{\frac{n(n+1)}{2}}$ that are subsets of $\{\tilde{x}_k\}_{k=1}^m.$ This set is obviously finite. Since each element $\{\tilde{x}_k\}_{k\in I}$ in $\mathcal {S}$ determines a unique self-adjoint operator T satisfying $\langle Tx_k, x_k\rangle =a_k$, for all $k\in I$, we can find the quantum state T that gives the best approximation to the measurement vector a by choosing the set which minimizes the distance above.

3 The Solution for the Infinite Dimensional Case

In infinite dimensions we will work with both the trace class operators and the Hilbert Schmidt operators. I.e. Operators $T=(a_{ij})_{i,j=1}^{\infty }$ with $\sum _{i,j=1}^{\infty }|a_{ij}|^2 < \infty .$ This class contains the trace class operators. As in the finite case, we will solve the following frame injectitivity problem:

Injectivity Problem For what frames $\{x_k\}_{k=1}^{\infty }$ in real or complex infinite dimensional Hilbert space $\mathbb {H}$ do we have the property: Whenever T, S are Hilbert Schmidt positive self-adjoint operators on $\mathbb {H}$ and $\langle Tx_k,x_k\rangle =\langle Sx_k,x_k\rangle$, for all $k=1,2,\ldots$, then $T=S$.

Remark 3.1

We will not require our operators to be trace class and trace one. These requirements will be considered as a special case of our problem.

3.1 The Solution to the Injectivity Problem

In this subsection we will solve the injectivity problem for infinite dimensional Hilbert spaces. Similar to the finite case, we first show that we only need to work with self adjoint operators. Note that the proof “(1) implies (2)” of Theorem 2.1 is not true for the infinite case. So we will give another proof here. The other implications are as in the finite case.

Theorem 3.2

Given a family of vectors $\mathcal {X}=\{x_k\}_{k=1}^\infty$ in a real or complex Hilbert space $\mathbb {H}$, the following are equivalent:

• (1) Whenever T, S are Hilbert Schmidt, positive and self-adjoint, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=S$.

• (2) Whenever T, S are Hilbert Schmidt self-adjoint, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } k, \end{aligned}$$

  then $T=S$.

• (3) $\mathcal {X}$ is injective.

Proof

We will show that (1) implies (2). Let T, S be Hilbert Schmidt self-adjoint operators such that

$$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } k. \end{aligned}$$

Set $R=T-S$. Then R is also a Hilbert Schmidt self-adjoint operator. Let $\{e_j\}_{j=1}^\infty$ be an orthonormal basis for $\mathbb {H}$ and let $\{u_j\}_{j=1}^\infty$ be an eigenbasis for R with respective eigenvalues $\{\lambda _j\}_{j=1}^\infty$. Define operators U and D on $\mathbb {H}$ by $Ue_j=u_j$ and $De_j=\lambda _je_j$, for $j=1, 2,\ldots$. Then U is a unitary operator, D is Hilbert Schmidt self-adjoint operator, and

$$\begin{aligned} R=UDU^*. \end{aligned}$$

Now let $r_j=|\lambda _j|, s_j=|\lambda _j|-\lambda _j, j=1, 2, \ldots$ be non-negative numbers. Then $\lambda _j=r_j-s_j$. Let $D_1, D_2$ be operators defined by

$$\begin{aligned} D_1e_j=r_je_j, \ D_2e_j=s_je_j \text{ for } j=1, 2, \ldots . \end{aligned}$$

Note that since R is Hilbert Schmidt, $\sum _{j=1}^{\infty }\lambda _j^2$ converges. Hence $D_1, D_2$ are Hilbert-Schmidt positive self-adjoint and we have

$$\begin{aligned} R=UDU^*=U(D_1-D_2)U^*=UD_1U^*-UD_2U^*. \end{aligned}$$

Moreover, $UD_1U^*,\ UD_2U^*$ are Hilbert Schmidt positive self-adjoint operators. Since

$$\begin{aligned} 0=\langle Rx_k, x_k\rangle =\langle UD_1U^*x_k, x_k\rangle -\langle UD_2U^*x_k, x_k\rangle , \end{aligned}$$

we have that $UD_1U^*=UD_2U^*$. Thus, $R=0$ and hence $T=S$. $\square$

If our operators are trace class, then we will have the following theorem. The proof of Theorem 3.2 is still valid here by noticing that $\sum _{j=1}^{\infty }|\lambda _j|<\infty$ for the trace class operator R.

Theorem 3.3

Given a family of vectors $\mathcal {X}=\{x_k\}_{k=1}^\infty$ in a infinite dimensional Hilbert space $\mathbb {H}$, the following are equivalent:

• (1) Whenever T, S are trace class positive and self-adjoint, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=S$.

• (2) Whenever T, S are trace class self-adjoint, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=S$.

• (3) Whenever T is trace class self-adjoint, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = 0, \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=0$.

Similar to the finite case, we will first give the following classification of injectivity for Hilbert Schmidt operators.

Theorem 3.4

Let $\mathcal {X}=\{x_k\}_{k=1}^\infty$ be a frame for an infinite dimensional real or complex Hilbert space $\mathbb {H}$. The following are equivalent:

• (1) $\mathcal {X}$ is injective.

• (2) For every orthonormal basis $\mathcal {E}=\{e_j\}_{j=1}^\infty$ for $\mathbb {H}$ we have:

  $$\begin{aligned} H(\mathcal {E})=:\overline{\mathrm {span}}\{(|\langle x_k,e_1\rangle |^2,|\langle x_k,e_2\rangle |^2, \ldots ):k=1,2,\ldots \}=\ell _2. \end{aligned}$$

Proof

$(1)\Rightarrow (2)$: We prove the result by way of contradiction. Suppose that (2) is false. Then there is an orthonormal basis $\mathcal {E}=\{e_j\}_{j=1}^\infty$ so that $H(\mathcal {E})\not = \ell _2$. Hence there is a non-zero vector $\lambda =(\lambda _1,\lambda _2,\ldots )\in \ell _2$ such that $\lambda \perp H(\mathcal {E})$.

Define an operator on $\mathbb {H}$ by

$$\begin{aligned} Te_j = \lambda _je_j, \text{ for } \text{ all } j=1, 2, \ldots . \end{aligned}$$

Then T is a non-zero Hilbert Schmidt operator. We also have: $\langle Tx_k,x_k\rangle =0$, for all $k=1, 2,\ldots$. This is a contradiction.

$(2)\Rightarrow (1)$: Let T be a Hilbert Schmidt self-adjoint operator such that

$$\begin{aligned} \langle Tx_k,x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since T is Hilbert Schmidt and hence compact, there is an eigenbasis $\mathcal {E}= \{e_j\}_{j=1}^\infty$ for T with respective eigenvalues $\{\lambda _j\}_{j=1}^\infty$. Then for every $k=1,2,\ldots$, we have

$$\begin{aligned} \langle Tx_k,x_k\rangle = \sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2 =0. \end{aligned}$$

Since T is Hilbert Schmidt then

$$\begin{aligned} \sum _{j=1}^{\infty }|\lambda _j|^2=\sum _{j=1}^{\infty }\Vert Te_j\Vert ^2<\infty . \end{aligned}$$

That is, $(\lambda _1, \lambda _2, \ldots )\in \ell _2$. Since

$$\begin{aligned} (\lambda _1,\lambda _2,\ldots ) \perp H(\mathcal {E}) = \ell _2 \text{ by } \text{ assumption } \text{(2) }. \end{aligned}$$

Therefore, $\lambda _j=0$ for all $j=1,2,\ldots ,$ and so $T=0$. $\square$

If we consider operators which are trace class, then we have the following classification for the infinite dimensions.

Theorem 3.5

Let $\mathcal {X}=\{x_k\}_{k=1}^\infty$ be a frame for an infinite dimensional real or complex Hilbert space $\mathbb {H}$. The following are equivalent:

• (1) $\mathcal {X}$ satisfies the following property: The only trace class self-adjoint operator T such that

  $$\begin{aligned} \langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k, \end{aligned}$$

  is $T=0$.

• (2) For every $\lambda =(\lambda _1, \lambda _2, \ldots )\in \ell _1$ and for every orthonormal basis $\{e_j\}_{j=1}^\infty$ for $\mathbb {H}$, if $\sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2 =0$ for all k then $\lambda =0$.

Proof

$(1)\Rightarrow (2)$: We prove the result by way of contradiction. Suppose that (2) is false. Then there is an $\lambda =(\lambda _1, \lambda _2, \ldots )\in \ell _1$ and an orthonormal basis $\{e_j\}_{j=1}^\infty$ so that $\sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2 =0$ for all k but $\lambda \not =0$.

Define an operator on $\mathbb {H}$ by

$$\begin{aligned} Te_j = \lambda _je_j, \text{ for } \text{ all } j=1, 2, \ldots . \end{aligned}$$

Then T is a non-zero self-adjoint operator. Moreover,

$$\begin{aligned} |T|e_j=\sqrt{TT^*}e_j=|\lambda _j|e_j, \text{ for } \text{ all } j. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{j=1}^{\infty }\langle |T|e_j, e_j\rangle =\sum _{j=1}^{\infty }|\lambda _j|<\infty . \end{aligned}$$

Thus, T is a non-zero trace class self-adjoint operator. Moreover, we have that $\langle Tx_k,x_k\rangle =\sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2 =0$, for all $k=1, 2,\ldots$. This is a contradiction.

$(2)\Rightarrow (1)$: Let T be a trace class self-adjoint operator such that

$$\begin{aligned} \langle Tx_k,x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since T is trace class and hence compact, there is an eigenbasis $\{e_j\}_{j=1}^\infty$ for T with respective eigenvalues $\{\lambda _j\}_{j=1}^\infty$. Then for every $k=1,2,\ldots$, we have

$$\begin{aligned} \sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2=\langle Tx_k,x_k\rangle = 0, \text{ for } \text{ all } k. \end{aligned}$$

Since T is trace class then

$$\begin{aligned} \sum _{j=1}^{\infty }|\lambda _j|=\sum _{j=1}^{\infty }|\langle Te_j, e_j\rangle |<\infty . \end{aligned}$$

That is, $\lambda =(\lambda _1, \lambda _2, \ldots )\in \ell _1$. By assumption (2) we get $\lambda =0$ and hence $T=0$. $\square$

Finally, by normalizing the trace, we can give a classification for the Injectivity problem if we require further that our operators are trace one. First, we need to justify Theorem 3.2 so that we can use it for this case.

Theorem 3.6

Given a family of vectors $\mathcal {X}=\{x_k\}_{k=1}^\infty$ in the real or complex Hilbert space $\mathbb {H}$, the following are equivalent:

• (1) Whenever T, S are trace class positive and self-adjoint of trace one, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=S$.

• (2) Whenever T, S are trace class self-adjoint of trace one, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } \text{ k }, \end{aligned}$$

  then $T=S$.

• (3) Whenever T is trace class self-adjoint of trace zero, and

  $$\begin{aligned} \langle Tx_k,x_k\rangle = 0, \text{ for } \text{ all } k, \end{aligned}$$

  then $T=0$.

Proof

$(1)\Rightarrow (2)$: Let T, S be trace class self-adjoint operators of trace one such that

$$\begin{aligned} \langle Tx_k,x_k\rangle = \langle Sx_k,x_k\rangle , \text{ for } \text{ all } k. \end{aligned}$$

Set $R=T-S$ then R is a trace class self-adjoint operator of trace zero. Let $\{e_j\}_{j=1}^\infty$ be an orthonormal basis for $\mathbb {H}$ and let $\{u_j\}_{j=1}^\infty$ be an eigenbasis for R with respective eigenvalues $\{\lambda _j\}_{j=1}^\infty$. Then $\sum _{j=1}^{\infty }\lambda _j=0$. Define operators U and D on $\mathbb {H}$ by $Ue_j=u_j$ and $De_j=\lambda _je_j$, for $j=1, 2,\ldots$. Then U is an unitary operator and D is a trace class self-adjoint operator of trace zero, and

$$\begin{aligned} R=UDU^*. \end{aligned}$$

Now define non-negative numbers

$$\begin{aligned} r_1=\dfrac{1+|\lambda _1|}{A}, s_1=\dfrac{1+|\lambda _1|-\lambda _1}{A}, r_j=\dfrac{|\lambda _j|}{A}, s_j=\dfrac{|\lambda _j|-\lambda _j}{A}, j= 2,3, \ldots , \end{aligned}$$

where

$$\begin{aligned} A=1+\sum _{j=1}^{\infty }|\lambda _j|=1+\sum _{j=1}^{\infty }|\lambda _j|-\sum _{j=1}^{\infty }\lambda _j>0 \end{aligned}$$

then $\lambda _j=r_j-s_j$ for all j. Let $D_1, D_2$ be operators defined by

$$\begin{aligned} D_1e_j=r_je_j, \ D_2e_j=s_je_j \text{ for } j=1, 2, \ldots . \end{aligned}$$

Then $D_1, D_2$ are trace class positive self-adjoint of trace one and we have

$$\begin{aligned} R=UDU^*=U(D_1-D_2)U^*=UD_1U^*-UD_2U^*. \end{aligned}$$

Moreover, $UD_1U^*,UD_2U^*$ are trace class positive self-adjoint operators of trace one. Since

$$\begin{aligned} 0=\langle Rx_k, x_k\rangle =\langle UD_1U^*x_k, x_k\rangle -\langle UD_2U^*x_k, x_k\rangle , \end{aligned}$$

then $UD_1U^*=UD_2U^*$. Thus, $R=0$ and hence $T=S$.

$(2)\Rightarrow (3)$: Let T be any trace class operator of trace zero such that

$$\begin{aligned} \langle Tx_k, k_k\rangle =0 \text{ for } \text{ all } k. \end{aligned}$$

Define an operator S on $\mathbb {H}$ by

$$\begin{aligned} Se_1=e_1, Se_j=0, \text{ for } j=2, 3,\ldots . \end{aligned}$$

Then S and $T+S$ are trace class self-adjoint operators of trace one.

Since $\langle (T+S)x_k, x_k\rangle =\langle Sx_k, x_k\rangle$ for all k, $T+S=S$ and hence $T=0$.

$(3)\Rightarrow (1)$: Let T, S are trace class positive self-adjoint operators of trace one such that

$$\begin{aligned} \langle Tx_k,x_k\rangle =\langle Sx_k, x_k\rangle \text{ for } \text{ all } k. \end{aligned}$$

Then $\langle (T-S)x_k, x_k\rangle =0$ for all k. Since $T-S$ is a trace class seft-adjoint operator of trace zero, $T=S$ by (3). $\square$

Now we are ready to give a classification for the Injectivity problem for operators of trace one. First, we need a definition.

Definition 3.7

We define a subspace of the real space $\ell _1$ as follows:

$$\begin{aligned} W:=\left\{ (\lambda _1, \lambda _2, \ldots )\in \ell _1: \sum _{j=1}^{\infty }\lambda _j=0\right\} . \end{aligned}$$

Theorem 3.8

Let $\mathcal {X}=\{x_k\}_{k=1}^\infty$ be a frame for an infinite dimensional real or complex Hilbert space $\mathbb {H}$. The following are equivalent:

• (1) If T is a trace class self-adjoint operator of trace zero such that

  $$\begin{aligned} \langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k, \end{aligned}$$

  then $T=0$.

• (2) For every $\lambda =(\lambda _1, \lambda _2, \ldots )\in W$ and for every orthonormal basis $\{e_j\}_{j=1}^\infty$ for $\mathbb {H}$, if $\sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2 =0$ for all k then $\lambda =0$.

Proof

$(1)\Rightarrow (2)$: We prove the contrapositive. Suppose that (2) is false. Then there is an $\lambda =(\lambda _1, \lambda _2, \ldots )\in W$ and an orthonormal basis $\{e_j\}_{j=1}^\infty$ so that $\sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2 =0$ for all k but $\lambda \not =0$.

Define an operator on $\mathbb {H}$ by

$$\begin{aligned} Te_j = \lambda _je_j, \text{ for } \text{ all } j=1, 2, \ldots . \end{aligned}$$

Then T is a non-zero trace class self-adjoint operator of trace zero. Moreover, we have that $\langle Tx_k,x_k\rangle =\sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2 =0$, for all $k=1, 2,\ldots$. This is a contradiction.

$(2)\Rightarrow (1)$: Let T be a trace class self-adjoint operator of trace zero such that

$$\begin{aligned} \langle Tx_k,x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Let $\{e_j\}_{j=1}^\infty$ be an eigenbasis for T with respective eigenvalues $\{\lambda _j\}_{j=1}^\infty$. Then for every $k=1,2,\ldots$, we have

$$\begin{aligned} \sum _{j=1}^\infty \lambda _j|\langle x_k, e_j\rangle |^2=\langle Tx_k,x_k\rangle = 0, \text{ for } \text{ all } k. \end{aligned}$$

Since T is trace class,

$$\begin{aligned} \sum _{j=1}^{\infty }|\lambda _j|=\sum _{j=1}^{\infty }|\langle Te_j, e_j\rangle |<\infty . \end{aligned}$$

Moreover, $\sum _{j=1}^{\infty }\lambda _j=0$. Thus, $\lambda =(\lambda _1, \lambda _2, \ldots )\in W$. By assumption (2) we get $\lambda =0$ and hence $T=0$. $\square$

From now on, we will also work in the direct sum of infinitely many copies of $\ell _2$.

Definition 3.9

Denote by $\tilde{\mathbb {H}}$ the direct sum of the real Hilbert spaces $\ell _2$:

$$\begin{aligned} \tilde{\mathbb {H}}=\left( \sum _{i=1}^{\infty }\oplus \ell _2 \right) _{\ell _2}. \end{aligned}$$

To avoid confusion with earlier notation, a vector in this direct sum will be written in the form:

$$\begin{aligned} \vec {x}=(\vec {x}_1, \vec {x}_2, \ldots , \vec {x}_n, \ldots ) , \end{aligned}$$

and we have

$$\begin{aligned} \langle \vec {x},\vec {y}\rangle = \sum _{i=1}^{\infty } \langle \vec {x}_i,\vec {y}_i\rangle . \end{aligned}$$

We also need the following lemma for both the real and complex cases.

Lemma 3.10

Let $A=(a_{ij})_{i,j=1}^\infty$ be a real or complex infinite matrix such that $\sum _{i, j=1}^{\infty }|a_{ij}|^2< \infty$. Then the operator $T_A$ defined in $\ell _2$ by

$$\begin{aligned} T_A(x_1,x_2, \ldots )=(y_1,y_2,\ldots ), \end{aligned}$$

where

$$\begin{aligned} y_i=\sum _{j=1}^{\infty }a_{ij}x_j, i=1, 2, \ldots , \end{aligned}$$

is a bounded operator. Moreover, $T_A$ is self-adjoint if and only if $a_{ji}=\bar{a}_{ij}$ for all i, j.

Proof

Let $x=\{x_i\}_{i=1}^{\infty }\in \ell _2$. For each $i=1, 2\ldots$, we have

$$\begin{aligned} |y_i|^2\le \left( \sum _{j=1}^{\infty }|a_{ij}x_j|\right) ^2\le \left( \sum _{j=1}^{\infty }|a_{ij}|^2\right) \left( \sum _{j=1}^{\infty }|x_j|^2\right) =\left( \sum _{j=1}^{\infty }|a_{ij}|^2\right) \Vert x\Vert ^2. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert T_Ax\Vert ^2= \sum _{i=1}^{\infty }|y_i|^2\le \left( \sum _{i=1}^{\infty }\sum _{j=1}^{\infty }|a_{ij}|^2\right) \Vert x\Vert ^2. \end{aligned}$$

This shows that $T_A$ is a bounded operator on $\ell _2$.

Suppose that T is self-adjoint. Then

$$\begin{aligned} a_{ji}=\langle T_Ae_i, e_j\rangle =\langle e_i, T_Ae_j\rangle =\overline{\langle T_Ae_j, e_i\rangle }=\bar{a}_{ij}, \end{aligned}$$

for all i, j.

Conversely, if $a_{ji}=\bar{a}_{ij}$ for all i, j, then

$$\begin{aligned} \langle T_A^*e_i, e_j\rangle =\langle e_i, T_Ae_j\rangle =\overline{\langle T_Ae_j, e_i\rangle }=\bar{a}_{ij}=a_{ji}=\langle T_Ae_i, e_j\rangle , \end{aligned}$$

for all i, j. Hence $T_A^*=T_A$. $\square$

3.1.1 The Real Case

Now we will solve the infinite dimensional injectivity problem in the real case. To avoid confusion between coordinates of a vector in $\ell _2$ and vectors in $\tilde{\mathbb {H}}$ we define:

Definition 3.11

For $x=\{x_i\}_{i=1}^\infty \in \ell _2$, we define

$$\begin{aligned} \tilde{x}=(\vec {x}_1, \vec {x}_2, \ldots , \vec {x}_n, \ldots ) \in \tilde{\mathbb {H}}, \end{aligned}$$

where

$$\begin{aligned} \vec {x}_1=(x_1x_1, x_1x_2, \ldots ); \ \vec {x}_2=(x_2x_2, x_2x_3, \ldots ); \ldots ;\vec {x}_n=(x_nx_n, x_nx_{n+1}, \ldots ); \ldots \end{aligned}$$

We first observe that these vectors are actually in $\tilde{\mathbb {H}}$.

Lemma 3.12

If $x=\{x_i\}_{i=1}^\infty \in \ell _2$, then $\tilde{x}\in \tilde{\mathbb {H}}$.

Proof

We have that

$$\begin{aligned} \sum _{j=i}^\infty |x_ix_j|^2=|x_i|^2\sum _{j=i}^\infty |x_j|^2\le |x_i|^2\Vert x\Vert ^2, \end{aligned}$$

for $i=1, 2, \ldots .$ Hence $\vec {x}_i\in \ell _2$ for all i.

Moreover, since

$$\begin{aligned} \sum _{i=1}^{\infty }\Vert \vec {x}_i\Vert ^2\le \sum _{i=1}^{\infty }|x_i|^2\Vert x\Vert ^2=\Vert x\Vert ^4, \end{aligned}$$

then $\tilde{x}\in \tilde{\mathbb {H}}$. $\square$

Now we are ready for the classification of the solutions to the injectivity problem in the infinite dimensional case.

Theorem 3.13

Let $\mathcal {X}=\{x_k\}_{k=1}^{\infty }$ be a frame in the real Hilbert space $\ell _2$. The following are equivalent:

• (1) $\mathcal {X}$ is injective.

• (2) $\overline{\mathrm {span}}\{\tilde{x}_k\}_{k=1}^\infty =\tilde{\mathbb {H}}$.

Proof

$(1) \Rightarrow (2)$: Let any $a=(\vec {a}_1, \vec {a}_2, \ldots )\in \tilde{\mathbb {H}}$ be such that $a\perp \overline{\mathrm {span}}\{\tilde{x}_k\}_{k=1}^\infty$. Then $\langle a, \tilde{x}_k\rangle =0$ for all k.

We denote

$$\begin{aligned} \vec {a}_1=(a_{11}, a_{12}, \ldots ); \ \vec {a}_2=(a_{22}, a_{23}, \ldots ); \ldots ,\vec {a}_n=(a_{nn}, a_{n(n+1)}, \ldots ); \ldots . \end{aligned}$$

Define an infinite matrix $B=(b_{ij})_{i,j=1}^\infty$, where $b_{ii}=a_{ii}$ for all i and $b_{ij}=b_{ji}=\dfrac{1}{2}a_{ij}$ for all $i<j$.

Then by Lemma 3.10, the operator $T_B$ defined by B is a Hilbert Schmidt self-adjoint operator.

For any $x= \{x_i\}_{i=1}^{\infty }\in \ell _2$, we have

$$\begin{aligned} \langle T_Bx,x\rangle&= \sum _{i=1}^{\infty }\sum _{j=1}^{\infty }b_{ij}x_ix_j\\&=\sum _{i=1}^{\infty }b_{ii}x_i^2+2\sum _{i<j}b_{ij}x_ix_j\\&=\sum _{i=1}^{\infty }a_{ii}x_i^2+\sum _{i<j}a_{ij}x_ix_j\\&=\sum _{i=1}^{\infty }\langle \vec {a}_i, \vec {x}_i\rangle \\&=\langle a, \tilde{x}\rangle . \end{aligned}$$

Hence, $\langle T_Bx_k, x_k\rangle =\langle a, \tilde{x}_k\rangle =0$ for all k. This implies $T_B=0$ by (1) and therefore $a=0$.

$(2)\Rightarrow (1)$: Let T be a Hilbert Schmidt self-adjoint operator on $\ell _2$ such that $\langle Tx_k, x_k\rangle =0$ for all k, and recall that $\{e_i\}_{i=1}^\infty$ is the canonical orthonormal basis for $\ell _2$.

Denote

$$\begin{aligned} a_{ij}=\langle Te_j, e_i\rangle , i, j=1, 2\ldots , \end{aligned}$$

and

$$\begin{aligned} \tilde{T}=(\vec {a}_1, \vec {a}_2, \ldots , \vec {a}_n, \ldots ), \end{aligned}$$

where

$$\begin{aligned}&\vec {a}_1=(a_{11}, 2a_{12}, 2a_{13},\ldots ); \quad \vec {a}_2=(a_{22}, 2a_{23},2a_{24}, \ldots );\ldots ; \\&\vec {a}_n =(a_{nn}, 2a_{n(n+1)}, 2a_{n(n+2)}, \ldots ); \ldots . \end{aligned}$$

Since T is a Hilbert Schmidt operator, $\tilde{T}\in \tilde{\mathbb {H}}$. Moreover, we have

$$\begin{aligned} \langle \tilde{T}, \tilde{x}_k\rangle =\langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since $\overline{\mathrm {span}}\{\tilde{x}_k\}_{k=1}^\infty =\tilde{\mathbb {H}}$, we get $\tilde{T}=0$. So $T=0$. $\square$

Remark 3.14

We have that $(2)\Rightarrow (1)$ in the theorem holds for trace class operators. But in general $(1)\Rightarrow (2)$ fails since the operators we construct may not be trace class.

3.1.2 The Complex Case

For the complex case of the injectivity problem, we need a new variation of the tilde vectors.

Definition 3.15

For $x=\{x_i\}_{i=1}^\infty \in \ell _2$, we define

$$\begin{aligned} \tilde{x}=(\vec {x}_1, \vec {x}_2, \ldots , \vec {x}_n, \ldots ), \end{aligned}$$

where

$$\begin{aligned} \vec {x}_1= & {} (|x_1|^2,\mathrm {Re}(\bar{x}_1x_2), \mathrm {Im}(\bar{x}_1x_2),\mathrm {Re}(\bar{x}_1x_3),\mathrm {Im}(\bar{x}_1x_3),\ldots ); \\ \vec {x}_2= & {} (|x_2|^2,\mathrm {Re}(\bar{x}_2x_3), \mathrm {Im}(\bar{x}_2x_3),\mathrm {Re}(\bar{x}_2x_4),\mathrm {Im}(\bar{x}_2x_4),\ldots );\ldots ;\\ \vec {x}_n= & {} (|x_n|^2,\mathrm {Re}(\bar{x}_nx_{n+1}), \mathrm {Im}(\bar{x}_nx_{n+1}),\mathrm {Re}(\bar{x}_nx_{n+2}),\mathrm {Im}(\bar{x}_nx_{n+2}),\ldots );\ldots . \end{aligned}$$

We first need to verify that our vectors are in $\tilde{\mathbb {H}}$.

Lemma 3.16

If $x=\{x_i\}_{i=1}^\infty \in \ell _2$, then $\tilde{x}\in \tilde{\mathbb {H}}$.

Proof

For each $i=1, 2, \ldots ,$ we have

$$\begin{aligned} \Vert \vec {x}_i\Vert ^2&=|x_i|^4+\sum _{j=i+1}^\infty |\mathrm {Re}(\bar{x}_ix_j)|^2+\sum _{j=i+1}^\infty |\mathrm {Im}(\bar{x}_ix_j)|^2\\&=|x_i|^4+\sum _{j=i+1}^\infty |\bar{x}_ix_j|^2\\&=|x_i|^2\left( |x_i|^2+\sum _{j=i+1}^\infty |x_j|^2\right) \\&\le |x_i|^2\Vert x\Vert ^2. \end{aligned}$$

It follows that:

$$\begin{aligned} \sum _{i=1}^{\infty }\Vert \vec {x}_i\Vert ^2 \le \sum _{i=1}^{\infty } |x_i|^2\Vert x\Vert ^2= \Vert x\Vert ^4. \end{aligned}$$

This implies $\tilde{x}\in \tilde{\mathbb {H}}$. $\square$

Now we give the classification theorem for injectivity in the infinite dimensional case.

Theorem 3.17

Let $\mathcal {X}=\{x_k\}_{k=1}^{\infty }$ be a frame in the complex Hilbert space $\ell _2$. The following are equivalent:

• (1) $\mathcal {X}$ gives injectivity.

• (2) $\overline{\mathrm {span}}\{\tilde{x}_k\}_{k=1}^\infty =\tilde{\mathbb {H}}$.

Proof

$(1) \Rightarrow (2)$: Let any $a=(\vec {a}_1, \vec {a}_2, \ldots )\in \tilde{\mathbb {H}}$ be such that $a\perp \overline{\mathrm {span}}\{\tilde{x}_k\}_{k=1}^\infty$. Then $\langle a, \tilde{x}_k\rangle =0$ for all k.

Denote

$$\begin{aligned} \vec {a}_1= & {} (a_{11}, u_{12}, v_{12}, u_{13}, v_{13}, \ldots );\ \vec {a}_2=(a_{22}, u_{23}, v_{23}, u_{24}, v_{24}, \ldots );\ldots ; \\ \vec {a}_n= & {} (a_{nn}, u_{n(n+1)}, v_{n(n+1)}, u_{n(n+2)}, v_{n(n+2)}, \ldots );\ldots . \end{aligned}$$

Define an infinite matrix $B=(b_{ij})_{i,j=1}^\infty$, where $b_{ii}=a_{ii}$ for all i and $b_{ij}=\bar{b}_{ji}=\dfrac{1}{2}(u_{ij}-\iota v_{ij})$ for all $i<j$.

We have

$$\begin{aligned} \sum _{i,j=1}^{\infty }|b_{ij}|^2=\sum _{i=1}^{\infty }|a_{ii}|^2+2\sum _{i<j}|b_{ij}|^2=\sum _{i=1}^{\infty }|a_{ii}|^2+\dfrac{1}{2}\sum _{i<j}\left( |u_{ij}|^2+|v_{ij}|^2\right) <\infty . \end{aligned}$$

Then by Lemma 3.10, the operator $T_B$ defined by B is Hilbert Schmidt and self-adjoint.

For any $x=\{x_i\}_{i=1}^\infty \in \ell _2$, we have

$$\begin{aligned} \langle T_Bx,x\rangle&= \sum _{i=1}^\infty \sum _{j=1}^\infty b_{ij}\bar{x}_ix_j\\&=\sum _{i=1}^\infty b_{ii}|x_i|^2 + 2\sum _{i<j}\mathrm {Re}( b_{ij}\bar{x}_ix_j)\\&=\sum _{i=1}^\infty b_{ii}|x_i|^2 + 2\sum _{i<j}(\mathrm {Re}( b_{ij})\mathrm {Re}(\bar{x}_ix_j)-\mathrm {Im}( b_{ij})\mathrm {Im}(\bar{x}_ix_j))\\&=\sum _{i=1}^{\infty }a_{ii}|x_i|^2+\sum _{i<j }\left( u_{ij}\mathrm {Re}(\bar{x}_ix_j)+v_{ij}\mathrm {Im}(\bar{x}_ix_j)\right) \\&=\sum _{i=1}^{\infty }\langle \vec {a}_i, \vec {x}_i\rangle \\&=\langle a,\tilde{x}\rangle . \end{aligned}$$

Hence, $\langle T_Bx_k, x_k\rangle =\langle a, \tilde{x}_k\rangle =0$ for all k. This implies $T_B=0$ by (1) and therefore $a=0$.

$(2)\Rightarrow (1)$: Let T be a Hilbert Schmidt self-adjoint operator such that $\langle Tx_k, x_k\rangle =0$ for all k.

Denote

$$\begin{aligned} a_{ij}=\langle Te_j, e_i\rangle , i, j=1, 2,\ldots , \end{aligned}$$

and

$$\begin{aligned} \tilde{T}=(\vec {a}_1, \vec {a}_2, \ldots , \vec {a}_n, \ldots ), \end{aligned}$$

where

$$\begin{aligned} \vec {a}_1= & {} (a_{11}, 2\mathrm {Re}(a_{12}), -2\mathrm {Im}(a_{12}), 2\mathrm {Re}(a_{13}), -2\mathrm {Im}(a_{13}),\ldots ); \\ \vec {a}_2= & {} (a_{22}, 2\mathrm {Re}(a_{23}), -2\mathrm {Im}(a_{23}),2\mathrm {Re}(a_{24}), -2\mathrm {Im}(a_{24}), \ldots );\ldots ; \\ \vec {a}_n= & {} (a_{nn}, 2\mathrm {Re}(a_{n(n+1)}), -2\mathrm {Im}(a_{n(n+1)}),2\mathrm {Re}(a_{n(n+2)}), -2\mathrm {Im}(a_{n(n+2)}), \ldots ); \ldots \end{aligned}$$

Since T is Hilbert Schmidt, $\tilde{T}\in \tilde{\mathbb {H}}$.

For any $x=\sum _{j=1}^{\infty }x_je_j$ we have

$$\begin{aligned} \langle Tx, x\rangle&=\sum _{i=1}^{\infty }\sum _{j=1}^\infty \bar{x}_ix_j\langle Te_j, e_i\rangle \\&=\sum _{i=1}^{\infty }\sum _{j=1}^\infty \bar{x}_ix_ja_{ij}\\&=\sum _{i=1}^\infty a_{ii}|x_i|^2 + 2\sum _{i<j}\mathrm {Re}( a_{ij}\bar{x}_ix_j)\\&=\sum _{i=1}^\infty a_{ii}|x_i|^2 + 2\sum _{i<j}(\mathrm {Re}( a_{ij})\mathrm {Re}(\bar{x}_ix_j)-\mathrm {Im}( a_{ij})\mathrm {Im}(\bar{x}_ix_j))\\&=\langle \tilde{T}, \tilde{x}\rangle . \end{aligned}$$

Hence

$$\begin{aligned} \langle \tilde{T}, \tilde{x}_k\rangle =\langle Tx_k, x_k\rangle =0, \text{ for } \text{ all } k. \end{aligned}$$

Since $\overline{\mathrm {span}}\{\tilde{x}_k\}_{k=1}^\infty =\tilde{\mathbb {H}}$, $\tilde{T}=0$. So $T=0$. This completes the proof. $\square$

As a consequence we have:

Corollary 3.18

For a frame $\{x_k\}_{k=1}^{\infty }$ in $\ell _2$ the following are equivalent:

• (1) The family $\{x_kx_k^*\}_{k=1}^{\infty }$ spans the family of real self-adjoint Hilbert Schmidt operators on $\ell _2$.

• (2) The family $\{\tilde{x}_k\}_{k=1}^{\infty }$ spans $\tilde{\mathbb {H}}$.

Remark 3.19

As we have seen in the proof of Theorem 3.13 for the real case and Theorem 3.17 for the complex case, for a vector $a\in \tilde{\mathbb {H}}$, there is a Hilbert Schmidt self-adjoint operator T so that

$$\begin{aligned} \langle Tx, x\rangle =\langle a, \tilde{x}\rangle , \text{ for } \text{ all } x\in \ell _2. \end{aligned}$$

Conversely, for a Hilbert Schmidt self-adjoint operator T, there is a vector $\tilde{T}\in \tilde{\mathbb {H}}$ satisfying

$$\begin{aligned} \langle \tilde{T}, \tilde{x}\rangle =\langle Tx, x\rangle , \text{ for } \text{ all } x\in \ell _2. \end{aligned}$$

Is is easy to see that the canonical orthonormal basis is not injective. Actually, the family $\{\tilde{e}_i\}_{i=1}^\infty$ forms an orthonormal set in $\tilde{\mathbb {H}}$. We will see in general that any frame in $\ell _2$ so that the corresponding tilde vectors form a frame sequence in $\tilde{\mathbb {H}}$ cannot be injective.

Theorem 3.20

For any Bessel sequence $\{x_k\}_{k=1}^{\infty }$ for the real or complex space $\ell _2$, the family $\{\tilde{x}_k\}_{k=1}^{\infty }$ is a Bessel sequence in $\tilde{\mathbb {H}}$. However, $\{\tilde{x}_k\}_{k=1}^{\infty }$ is not a frame for $\tilde{\mathbb {H}}$.

Proof

We may assume that $\Vert x_k\Vert \le 1$ for all k. Let B be the Bessel bound of $\{x_k\}_{k=1}^\infty$.

Given any finite real scalar sequence $\{a_k\}_{k=1}^{\infty }$ we will compute the real case and the complex case separately.

The real case Using Definition 3.11 for the tilde vector $\tilde{x}_k$, we have

$$\begin{aligned} \left\| \sum _{k=1}^{\infty }a_k\tilde{x}_k\right\| ^2&= \sum _{i=1}^{\infty }\sum _{j=i}^{\infty }\left( \sum _{k=1}^{\infty } a_kx_{ki}x_{kj}\right) ^2\\&\le \sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\left( \sum _{k=1}^{\infty } a_kx_{ki}x_{kj}\right) ^2\\&=\sum _{i=1}^{\infty }\left\| \sum _{k=1}^{\infty }a_kx_{ki}x_k\right\| ^2. \end{aligned}$$

Using the fact that $\{x_k\}_{k=1}^\infty$ is Bessel with bound B, we get

$$\begin{aligned} \left\| \sum _{k=1}^{\infty }a_k\tilde{x}_k\right\| ^2&\le B \sum _{i=1}^{\infty }\sum _{k=1}^{\infty }(a_kx_{ki})^2\\&=B\sum _{k=1}^{\infty }a_k^2\sum _{i=1}^{\infty }x_{ki}^2\\&= B \sum _{k=1}^{\infty }a_k^2 \Vert x_k\Vert ^2\\&\le B \sum _{k=1}^{\infty }a_k^2. \end{aligned}$$

The complex case Now we need to use Definition 3.15 for the tilde vectors $\tilde{x}_k$. We have that

$$\begin{aligned} \left\| \sum _{k=1}^{\infty }a_k\tilde{x}_k\right\| ^2&= \sum _{i=1}^{\infty }\left( \sum _{k=1}^{\infty } a_k|x_{ki}|^2\right) ^2+\sum _{i=1}^{\infty }\sum _{j=i+1}^{\infty }\left( \sum _{k=1}^{\infty } a_k\mathrm {Re}(\bar{x}_{ki}x_{kj})\right) ^2\\&\quad +\sum _{i=1}^{\infty }\sum _{j=i+1}^{\infty }\left( \sum _{k=1}^{\infty } a_k\mathrm {Im}(\bar{x}_{ki}x_{kj})\right) ^2\\&=\sum _{i=1}^{\infty }\left( \sum _{k=1}^{\infty } a_k|x_{ki}|^2\right) ^2+\sum _{i=1}^{\infty }\sum _{j=i+1}^{\infty }\left( \mathrm {Re}\left( \sum _{k=1}^{\infty } a_k\bar{x}_{ki}x_{kj}\right) \right) ^2\\&\quad +\sum _{i=1}^{\infty }\sum _{j=i+1}^{\infty }\left( \mathrm {Im}\left( \sum _{k=1}^{\infty } a_k\bar{x}_{ki}x_{kj}\right) \right) ^2\\&\le 2\sum _{i=1}^{\infty }\left( \sum _{k=1}^{\infty } a_k|x_{ki}|^2\right) ^2+2\sum _{i=1}^{\infty }\sum _{j=i+1}^{\infty }\left| \sum _{k=1}^{\infty } a_k\bar{x}_{ki}x_{kj}\right| ^2\\&\le 2\sum _{i=1}^{\infty }\left( \sum _{k=1}^{\infty } a_k|x_{ki}|^2\right) ^2+2\sum _{i=1}^{\infty }\sum _{j=1, j\not = i}^{\infty }\left| \sum _{k=1}^{\infty } a_k\bar{x}_{ki}x_{kj}\right| ^2\\&=2\sum _{i=1}^{\infty }\left\| \sum _{k=1}^{\infty }a_k\bar{x}_{ki}x_k\right\| ^2 \le 2B \sum _{i=1}^{\infty }\sum _{k=1}^{\infty }a_k^2|x_{ki}|^2\\&=2B\sum _{k=1}^{\infty }a_k^2\sum _{i=1}^{\infty }|x_{ki}|^2 = 2B \sum _{k=1}^{\infty }a_k^2 \Vert x_k\Vert ^2 \le 2B \sum _{k=1}^{\infty }a_k^2. \end{aligned}$$

Hence, $\{\tilde{x}_k\}_{k=1}^{\infty }$ is a Bessel sequence for the both cases.

Now we will show that $\{\tilde{x}_k\}_{k=1}^{\infty }$ fails to have a lower frame bound. We will prove the real case, the complex case is similar. By our assumption, we have that

$$\begin{aligned} \sum _{i=1}^{\infty }|x_{ki}|^2<\infty , \text{ for } \text{ all } k. \end{aligned}$$

Also,

$$\begin{aligned} \sum _{k=1}^{\infty }|x_{ki}|^2 = \sum _{k=1}^{\infty }|\langle e_i,x_k\rangle |^2 < \infty , \text{ for } \text{ all } i. \end{aligned}$$

Fix $\epsilon >0$ and choose n so that

$$\begin{aligned} \sum _{k=n}^{\infty }|x_{k1}|^2 < \epsilon . \end{aligned}$$

Now choose m so that

$$\begin{aligned} \sum _{k=1}^{n-1}|x_{km}|^2< \epsilon . \end{aligned}$$

Let

$$\begin{aligned} \tilde{e}_{1m}:=(e_m; \vec {0};\vec {0}, \ldots )\in \tilde{\mathbb {H}}. \end{aligned}$$

Then we have

$$\begin{aligned} \sum _{k=1}^{\infty }|\langle \tilde{e}_{1m},\tilde{x}_k\rangle |^2&= \sum _{k=1}^{\infty }|x_{k1}|^2|x_{km}|^2\\&= \sum _{k=1}^{n-1}|x_{k1}|^2|x_{km}|^2 + \sum _{k=n}^{\infty }|x_{k1}|^2|x_{km}|^2\\&\le \sum _{k=1}^{n-1}|x_{km}|^2+\sum _{k=n}^{\infty }|x_{k1}|^2\\&< 2 \epsilon . \end{aligned}$$

It follows that $\{\tilde{x}_k\}_{k=1}^{\infty }$ does not have a lower frame bound. $\square$

Corollary 3.21

Let $\{x_k\}_{k=1}^\infty$ be a frame for $\ell _2$. If $\{\tilde{x}_k\}_{k=1}^\infty$ is a frame sequence, then $\{x_k\}_{k=1}^\infty$ cannot be injective.

Proof

We will prove the real case, the complex case is similar.

Suppose by way of contradiction that $\{x_k\}_{k=1}^\infty$ is injective. Then by Theorem 3.13, $\{\tilde{x}_k\}_{k=1}^\infty$ spans $\tilde{\mathbb {H}}$. Thus, it is a frame for $\tilde{\mathbb {H}}$, which contradicts Theorem 3.20. $\square$

3.2 Constructing the Solutions to the Injectivity Problem

For the construction of solutions to the injectivity problem, we will follow the outline for the finite dimensional case. But this construction is much more complicated because of problems with convergence, problems with keeping the upper frame bound finite, and the fact that we cannot show spanning in $\ell _2$ by just checking linear independence. Also, we proved in the finite dimensional case that the $\tilde{x}_i$ span by showing they are independent and have enough vectors to span $\tilde{\mathbb {H}}$. This does not work in the infinite dimensional case. Note that the following construction works for trace class operators and for Hilbert Schmidt operators.

Theorem 3.22

Let $\{e_i\}_{i=1}^\infty$ be the canonial basis for the real Hilbert space $\ell _2$ and let $a_i\not =0$ for $i=1, 2, \ldots$ be such that $\sum _{i=1}^{\infty }a^2_i<\infty .$ Define

$$\begin{aligned} x_k=a_k(e_1+e_{k+1}), \text{ for } k=1, 2, \ldots . \end{aligned}$$

Let L be the right shift operator on $\ell _2$. Then the family

$$\begin{aligned} \{e_{i}\}_{i=1}^{\infty }\cup \{\frac{1}{2^i}L^ix_k\}_{i=0,k=1}^{\infty , \ \ \infty } \end{aligned}$$

is a frame for $\ell _2$ which gives injectivity.

Proof

First we need to see that our family of vectors forms a frame for $\ell _2$. Since our family contains an orthonormal basis for $\ell _2$, we automatically have a lower frame bound. So we need to check that our family is Bessel, and since $\{e_i\}_{i=1}^{\infty }$ is already Bessel, we only need to check that $\{\frac{1}{2^i}L^ix_k\}_{i=0, k=1}^{ \infty , \ \ \infty }$ is Bessel.

For any $x\in \ell _2$, we have

$$\begin{aligned} \sum _{i=0}^{\infty }\sum _{k=1}^{\infty }\left| \left\langle x,\frac{1}{2^i}L^{i}x_k\right\rangle \right| ^2&\le \sum _{i=0}^{\infty }\sum _{k=1}^{\infty }\dfrac{1}{4^i}\Vert x\Vert ^2\Vert L^{i}x_k\Vert ^2\\&\le \sum _{i=0}^{\infty }\sum _{k=1}^{\infty }\dfrac{1}{4^i}\Vert x\Vert ^24a_k^2\\&=\left( \sum _{i=0}^{\infty }\dfrac{1}{4^{i-1}}\sum _{k=1}^{\infty }a_k^2\right) \Vert x\Vert ^2. \end{aligned}$$

So our family is a Bessel sequence.

To see our frame is injective, let T be a Hilbert Schmidt self-adjoint operator such that

$$\begin{aligned} \langle Te_k, e_k\rangle =0 \text{ and } \langle T(L^ix_k), L^ix_k\rangle =0, \text{ for } i=0,1 \ldots ;\ k=1, 2, \ldots . \end{aligned}$$

Note that

$$\begin{aligned} L^ix_k=a_k(e_{1+i}+ e_{1+i+k}) \text{ for } \text{ all } i, k. \end{aligned}$$

Hence

$$\begin{aligned} \langle T(L^ix_k), L^ix_k\rangle&= a_k^2\langle Te_{1+i},e_{1+i}\rangle +2a_k^2\langle Te_{1+i}, e_{1+i+k} \rangle +a_k^2\langle Te_{1+i+k},e_{1+i+k}\rangle \\&=2a_k^2\langle Te_{1+i}, e_{1+i+k} \rangle , \end{aligned}$$

for all i, k.

This implies $\langle Te_j, e_k\rangle =0$ for all $j, k=1, 2, \ldots ,$ and hence $T=0$. $\square$

The complex version of this construction looks like:

Theorem 3.23

Let $\{e_i\}_{i=1}^{\infty }$ be the canonical orthonormal basis for complex $\ell _2$, and let $\{a_i\}_{i=1}^{\infty }, \{b_i\}_{i=1}^\infty \in \ell _2$, $|a_i|, |b_i|\not =0$ for all i. Then the following frame gives injectivity:

$$\begin{aligned} \{e_i\}_{i=1}^{\infty }\cup \left\{ \dfrac{1}{2^i}L^i(a_k(e_1+e_{k+1}))\right\} _{i=0,k=1}^{\infty } \cup \left\{ \dfrac{1}{2^i}L^i(b_k(e_1+\iota e_{k+1}))\right\} _{i=0,k=1}^{\infty }. \end{aligned}$$

The above frames are unbounded. The following theorem shows that we can easily adjust unbounded injective frames to produce bounded injective frames.

Theorem 3.24

Every injective frame $\mathcal {X}=\{e_i\}_{i=1}^\infty \cup \{x_k\}_{k=1}^{\infty }$ of finitely supported vectors, induces a bounded injective frame.

Proof

Recall that for each k, we denote

$$\begin{aligned} x_{k}=(x_{k1}, x_{k2},\ldots , x_{ki}, \ldots ). \end{aligned}$$

Choose integer $n_1< n_2 < \cdots$ so that

$$\begin{aligned} \max \{i:x_{ki}\not = 0\}< n_k. \end{aligned}$$

For $k=1,2,\ldots$ let

$$\begin{aligned} y_{2k}=x_k+e_{n_k} \text{ and } y_{2k+1}=x_k-e_{n_k}, \text{ for } k=1,2,\ldots . \end{aligned}$$

It is clear that $\mathcal {Y}=\{e_i\}_{i=1}^\infty \cup \{y_k\}_{k=1}^{\infty }$ is still a frame and $\Vert y_k\Vert \ge 1$ for all $k=1,2,\ldots$. Since $\tilde{y}_{2k}+\tilde{y}_{2k+1}=2 \tilde{x}_k+2\tilde{e}_{n_k}$, and $\{\tilde{e}_{n_k}\}_{k=1}^{\infty }$ are vectors in our set, it follows that $\{\tilde{x}_k\}_{k=1}^{\infty }$ is in our set of vectors and so $\mathcal {Y}$ is injective. $\square$

3.3 The Solutions are Neither Open Nor Dense

In this section we will show that the solutions to the injectivity problem in infinite dimensions are neither open nor dense in the class of frames.

First we need a definition:

Definition 3.25

Given frames $\mathcal {X}=\{x_k\}_{k=1}^{\infty }$ and $\mathcal {Y}=\{y_k\}_{k=1}^{\infty }$ for $\ell _2$, we define the distance between them by

$$\begin{aligned} d^2(\mathcal {X},\mathcal {Y}) = \sum _{k=1}^{\infty } \Vert x_k-y_k\Vert ^2. \end{aligned}$$

Note that this distance may be infinity.

The following theorem shows that the frames which give injectivity are not open in the family of frames for $\ell _2$.

Theorem 3.26

Let $\mathcal {X}=\{e_{i}\}_{i=1}^{\infty }\cup \{\frac{1}{2^i}L^ix_k\}_{i=0,k=1}^{\infty }$ be the injective frame for the real space $\ell _2$ as in Theorem 3.22. Then for any $\epsilon >0$, there is a frame $\mathcal {Y}$ such that $d(\mathcal {X}, \mathcal {Y})<\epsilon$, and $\mathcal {Y}$ is not injective.

Proof

Let any $\epsilon >0$. Since the series $\sum _{i=0}^{\infty }\sum _{k=1}^{\infty }\Vert \frac{1}{2^i}L^ix_k\Vert ^2$ converges, for any $\epsilon$, there exists $n_0$ such that

$$\begin{aligned} \sum _{i= n_0+1}^\infty \sum _{k=1}^{\infty }\left\| \frac{1}{2^i}L^ix_k\right\| ^2< \epsilon ^2. \end{aligned}$$

Let $y_{ik} = \frac{1}{2^i}L^ix_k$ for $i=0,1,\ldots ,n_0$ and $k=1,2,\ldots$, and $y_{ik}=0$ otherwise. It is clear that

$$\begin{aligned} \mathcal {Y}= \{e_i\}_{i=1}^{\infty } \cup \{y_{ik}\}_{i=0, k=1}^{\infty } \end{aligned}$$

cannot give injectivity by Theorem 3.13 while

$$\begin{aligned} d^2(\mathcal {X}, \mathcal {Y})=\sum _{i= n_0+1}^\infty \sum _{k=1}^{\infty }\left\| \frac{1}{2^i}L^ix_k\right\| ^2<\epsilon ^2. \end{aligned}$$

This completes the proof. $\square$

Remark 3.27

There is a perturbation theory for frames which looks like it should apply here. The problem is that although our vectors form a frame for $\ell _2$, their tilde vectors do not form a frame to $\tilde{\mathbb {H}}$ and so the theory does not apply.

To show the solutions are not dense, we need the definition of a Riesz sequence in $\ell _2$.

Definition 3.28

A family of vectors $\{x_i\}_{i=1}^{\infty }$ in the real or complex Hilbert space $\ell _2$ is called a Riesz sequence if there are constants $0<A\le B < \infty$ so that for all $\{a_i\}_{i=1}^{\infty }\subset \ell _2$ we have:

$$\begin{aligned} A \sum _{i=1}^{\infty }|a_i|^2 \le \left\| \sum _{i=1}^{\infty }a_ix_i\right\| ^2 \le B \sum _{i=1}^{\infty }|a_i|^2. \end{aligned}$$

The constants A, B are called the lower and upper Riesz bounds. If the vectors span $\ell _2$, this is called a Riesz basis.

Remark 3.29

It is known [6, 8] that a Riesz basis is a frame and the Riesz bounds are the frame bounds. Also, $\{x_i\}_{i=1}^{\infty }$ is a Riesz sequence if and only if the operator $T:\ell _2 \rightarrow \ell _2$ given by $Te_i=x_i$ is a bounded, linear, invertible operator (on its range).

Also, we need a perturbation result from frame theory.

Proposition 3.30

Assume $\chi =\{x_i\}_{i=1}^{\infty }$ are vectors in the real or complex space $\ell _2$ satisfying:

$$\begin{aligned} \sum _{i=1}^{\infty }\Vert e_i-x_i\Vert ^2 < \epsilon ^2. \end{aligned}$$

Then $\chi$ is a Riesz sequence in $\ell _2$ with lower Riesz bound $(1-\epsilon )^2$.

Proof

We compute for scalars $\{a_i\}_{i=1}^{\infty }$,

$$\begin{aligned} \left\| \sum _{i=1}^{\infty }a_ix_i\right\|&\ge \left\| \sum _{i=1}^{\infty }a_ie_i\right\| - \left\| \sum _{i=1}^{\infty }a_i(e_i-x_i)\right\| \\&\ge \left( \sum _{i=1}^{\infty }|a_i|^2\right) ^{1/2}- \sum _{i=1}^{\infty }|a_i|\Vert e_i-x_i\Vert \\&\ge \left( \sum _{i=1}^{\infty }|a_i|^2\right) ^{1/2}- \left( \sum _{i=1}^{\infty }|a_i|^2 \right) ^{1/2}\left( \sum _{i=1}^{\infty }\Vert e_i-x_i\Vert ^2 \right) ^{1/2}\\&\ge \left( \sum _{i=1}^{\infty }|a_i|^2\right) ^{1/2}(1-\epsilon ) \end{aligned}$$

The upper Riesz bound is done similarly. $\square$

We also need a theorem from [3].

Theorem 3.31

Let Y, Z be subspaces of a Banach space X. If $T:Y\rightarrow Z$ is a surjective linear operator with $\Vert I-T\Vert <1$, then $codim_XY=codim_XZ$.

The next theorem shows that the solution set of the infinite dimensional injectivity problem is not dense in the class of all frames for $\ell _2$.

Theorem 3.32

Let $\{e_k\}_{k=1}^{\infty }$ be the canonical basis for the real space $\ell _2$ and $\mathcal {X}=\{x_k\}_{k=1}^{\infty }\subset \ell _2$ be such that

$$\begin{aligned} \sum _{k=1}^{\infty }\Vert e_k-x_k\Vert ^2 \le \frac{1}{8}, \end{aligned}$$

Then $\mathcal {X}$ is not injective.

Proof

Will will show that $\mathcal {X}$ does not satisfy Theorem 3.13. Note that $codim_{\tilde{\mathbb {H}}}\{\tilde{e}_k\}_{k=1}^{\infty }$ is infinite. Also, $\{\tilde{e}_k\}_{k=1}^{\infty }$ is an orthonormal sequence in $\tilde{\mathbb {H}}$. We have that

$$\begin{aligned} \sum _{k=1}^{\infty }\Vert e_k-x_k\Vert ^2 = \sum _{k=1}^{\infty }\left( (1-x_{kk})^2 +\sum _{i\not = k}x_{ki}^2\right) \le \frac{1}{8}. \end{aligned}$$

In particular, $\Vert x_k\Vert ^2 \le 2$. Let

$$\begin{aligned} X=\overline{\mathrm {span}}\{\tilde{e}_k\}_{k=1}^{\infty }, \text{ and } Y=\overline{\mathrm {span}}\{\tilde{x}_k\}_{k=1}^{\infty }. \end{aligned}$$

For each $k=1, 2, \ldots$ we have

$$\begin{aligned} \Vert \tilde{e}_k-\tilde{x}_k\Vert ^2&=(1-x^2_{kk})^2+\sum _{j\ge k+1}(x_{kk}x_{kj})^2+\sum _{i\not =k}\sum _{j\ge i}(x_{ki}x_{kj})^2\\&\le (1-x_{kk})^2(2\Vert x_k\Vert ^2+2)+\Vert x_k\Vert ^2\sum _{j\ge k+1}x^2_{kj}+\Vert x_k\Vert ^2\sum _{i\not =k}x^2_{ki}\\&\le 6\left( (1-x_{kk})^2+\sum _{i\not =k}x^2_{ki}\right) . \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{k=1}^{\infty }\Vert \tilde{e}_k-\tilde{x}_k\Vert ^2&\le 6\sum _{k=1}^{\infty }\left( (1-x_{kk})^2 +\sum _{i\not = k}x_{ki}^2\right) \le \dfrac{3}{4}. \end{aligned}$$

It follows that $\{\tilde{x}_k\}_{k=1}^{\infty }$ is a Riesz sequence.

Now we define $T:X \rightarrow Y$ by: for $x=\sum _{k=1}^{\infty }\langle x, \tilde{e}_k\rangle \tilde{e}_k\in X$,

$$\begin{aligned} Tx=\sum _{k=1}^{\infty }\langle x, \tilde{e}_k\rangle \tilde{x}_k. \end{aligned}$$

Since T is mapping a Riesz sequence to a Riesz sequence, it follows that T is bounded and surjective. Now,

$$\begin{aligned} \Vert (I-T)x\Vert&= \left\| \sum _{k=1}^{\infty }\langle x, \tilde{e}_k\rangle (\tilde{e}_k-\tilde{x}_k) \right\| \\&\le \sum _{k=1}^{\infty }|\langle x, \tilde{e}_k\rangle |\Vert \tilde{e}_k-\tilde{x}_k\Vert \\&\le \left( \sum _{k=1}^{\infty }|\langle x, \tilde{e}_k\rangle |^2 \right) ^{1/2} \left( \sum _{k=1}^{\infty }\Vert \tilde{e}_k-\tilde{x}_k\Vert ^2 \right) ^{1/2}\\&\le \dfrac{\sqrt{3}}{2}\Vert x\Vert . \end{aligned}$$

Hence, $\Vert I-T\Vert <1$ and by Theorem 3.31, $codim_{\tilde{\mathbb {H}}}Y = codim_{\tilde{\mathbb {H}}}X=\infty .$$\square$

3.4 The Solution to the State Estimation Problem

For the infinite dimensional case, the state estimation problem asks:

State Estimation Problem Given an injective Parseval frame $\{x_k\}_{k=1}^{\infty }$ for $\ell _2$, and a sequence of real numbers $a=\{a_k\}_{k=1}^\infty$, does there exist a Hilbert Schmidt self-adjoint operator T so that

$$\begin{aligned} \langle Tx_k, x_k\rangle =a_k, \text{ for } \text{ all } k? \end{aligned}$$

Remark 3.33

This problem is rarely solvable.

• (1) If $x_kx_k^*=x_lx_l^*$, but $a_k\not = a_l$ for some k, l, then the problem has no solution.

• (2) Recall a set of vectors $\{x_i\}_{i=1}^{\infty }$ is $\omega$-independent if $\sum _{i=1}^{\infty }c_ix_i=0$ implies $c_i=0$ for all $i=1,2,\ldots$. If $\{x_kx_k^*\}_{k=1}^\infty$ is not $\omega$-independent and $\sum _{k=1}^{\infty }c_kx_kx_k^*=0$ but not all $c_k$ are zero, then for $\langle Tx_k, x_k\rangle =a_k$ we need

  $$\begin{aligned} \sum _{k=1}^{\infty } c_ka_k=\left\langle T,\sum _{k=1}^{\infty }c_kx_kx_k^*\right\rangle = 0. \end{aligned}$$

For the solution of the state estimation problem we will need the notion of a separated sequence in $\ell _2$.

Definition 3.34

A family of vectors $\{x_i\}_{i=1}^{\infty }$ in $\ell _2$ is separated if for every $j\in \mathbb {N}$,

$$\begin{aligned} x_j \notin \overline{\mathrm {span}}\{x_i\}_{i\not = j}. \end{aligned}$$

It is $\delta$-separated if the projection $P_j$ onto $\overline{\mathrm {span}}\{x_i\}_{i\not = j}$ satisfies

$$\begin{aligned} \Vert (I-P_j)x_j\Vert \ge \delta . \end{aligned}$$

Remark 3.35

A Riesz basic sequence $\{x_k\}_{k\in I}$ is $\delta$-separated since it is clear from the definition that

$$\begin{aligned} dist(x_k,span\{x_i\}_{i\not = k})\ge \delta . \end{aligned}$$

In general, a Bessel sequence which is $\delta$-separated may not be a Riesz sequence. To see this let

$$\begin{aligned} \mathbb {H}=\left( \sum _{n=1}^{\infty }\oplus \mathbb {H}_n\right) _{\ell _2}, \end{aligned}$$

where $\mathbb {H}^n$ is an n-dimensional Hilbert space with orthonormal basis $\{e_{in}\}_{i=1}^n$. Let P be the orthogonal projection onto the one dimensional subspace spanned by $\sum _{i=1}^ne_{in}$. Then $\{(I-P)e_{in}\}_{i=1,n=1}^{n-1,\ \infty }$ as a family of vectors in $\mathbb {H}$ is $\delta$-separated, 1-Bessel, but not a Riesz sequence. (Careful: We have thrown away the vectors $(I-P)e_{nn}$ above.)

Note also that a $\delta$-separated sequence may not be Bessel.

Example 3.36

Let $x_i=e_1+e_{i+1}, i=1, 2, \ldots .$ Then $\{x_i\}_{i=1}^\infty$ is not a Bessel sequence. However, it is $\delta$-separated.

Indeed, let $P_j$ be the projection onto $\overline{\mathrm {span}}\{x_i\}_{i\not =j}$. Then

$$\begin{aligned} \Vert x_j-P_jx_j\Vert ^2= & {} \Vert e_1+e_j-P_j(e_1+e_j)\Vert ^2=\Vert e_j+e_1-P_je_1 \Vert ^2\\= & {} 1+\Vert e_1-P_je_1\Vert ^2\ge 1, \end{aligned}$$

for all j. So $\{x_i\}_{i=1}^\infty$ is $\delta$-separated, where $\delta =1$.

The next proposition presents a fundamental property of separated sequences.

Proposition 3.37

If a family of vectors $\{x_i\}_{i=1}^{\infty }$ is separated, then there are vectors $\{y_i\}_{i=1}^{\infty }$ satisfying:

$$\begin{aligned} \langle y_i,x_j\rangle = \delta _{ij}, \text{ for } \text{ all } \text{ i,j }. \end{aligned}$$

If it is $\delta$-separated then, $\underset{i}{\sup }\Vert y_i\Vert <\infty$.

Proof

Fix j and let $P_j$ be the orthogonal projection onto $\overline{\mathrm {span}}\{x_i\}_{i\not = j}$. Note that $P_jx_j\not = x_j$ and so $(I-P_j)x_j \not = 0$.

Clearly,

$$\begin{aligned} \langle (I-P_j)x_j, x_i\rangle =0 \text{ for } i\not =j. \end{aligned}$$

So let

$$\begin{aligned} y_j=\frac{(I-P_j)x_j}{\Vert (I-P_j)x_j\Vert ^2}, \end{aligned}$$

we get the desired sequence.

For the $\delta$-separated case, we have that $\Vert (I-P_j)x_j\Vert \ge \delta$ and the result follows. $\square$

For the next result, we will need:

Proposition 3.38

Let $\{x_i\}_{i=1}^\infty$ be a bounded sequence in a Hilbert space $\mathbb {H}$. The following are equivalent:

• (1) For some $\delta >0$, $\{x_i\}_{i=1}^\infty$ is $\delta$-separated.

• (2) $\{x_i\}_{i=1}^\infty$ is separated and $\{x_i\}_{i=n}^\infty$ is $\delta _1$-separated for some $\delta _1>0$, for some $n\ge 1$.

Proof

We just need to show that $(2)\Rightarrow (1)$. So assume $\{x_i\}_{i=1}^{\infty }$ is separated and $\{x_i\}_{i=n}^{\infty }$ is $\delta _1$-separated. Let $P_j$ be the projection onto $\overline{\mathrm {span}}\{x_i\}_{i\not = j}$, for $j=1, 2, \ldots ,$ and let $Q_j$ be the projection onto $\overline{\mathrm {span}}\{x_i\}_{n\le i \not = j}$, for $j=n, n+1, \ldots .$ So

$$\begin{aligned} \Vert (I-Q_j)x_j\Vert \ge \delta _1, \text{ for } \text{ all } j\ge n. \end{aligned}$$

We need to show that there exists a $\delta >0$ so that

$$\begin{aligned} \Vert (I-P_j)x_j\Vert \ge \delta , \text{ for } \text{ all } j\ge 1. \end{aligned}$$

We will do this in steps.

Step 1 There exists a $\delta _2 >0$ so that

$$\begin{aligned} \Vert (I-P_j)x_j\Vert \ge \delta _2, \text{ for } \text{ all } j\ge n. \end{aligned}$$

We will do this by way of contradiction. So assume there are natural numbers $n\le n_1<n_2<\cdots$ satisfying:

$$\begin{aligned} \Vert x_{n_j}-P_{n_j}(x_{n_j})\Vert <\dfrac{1}{j}. \end{aligned}$$

It follows that there are vectors $y_j \in \mathrm {span}\{x_i\}_{i=1}^{n-1}$ and $z_j\in \mathrm {span}\{x_i\}_{n\le i \not = n_j < \infty }$ so that $\Vert x_{n_j}-(y_j+z_j)\Vert < \frac{1}{j}$.

Claim 1

There are an $\epsilon >0$ and $n_0\in \mathbb {N}$ so that $\Vert y_{j}\Vert \ge \epsilon$, for all $j\ge n_0$.

We prove the claim by way of contradiction. If the claim fails, there are integers $j_1<j_2<\cdots$ so that $\Vert y_{j_k}\Vert < \frac{1}{k}$ for all $k=1,2,\ldots$. It follows that

$$\begin{aligned} \Vert x_{n_{j_k}}- z_{j_k}\Vert \le \Vert x_{n_{j_k}}-(z_{j_k}+y_{j_k})\Vert + \Vert y_{j_k}\Vert < \frac{2}{k}, \text{ for } \text{ all } k, \end{aligned}$$

which contradicts the fact that $\{x_i\}_{i=n}^{\infty }$ is $\delta$-separated.

Claim 2

There is a constant $K>0$ so that $\Vert y_j\Vert \le K$, for all $j\ge n_0$.

Define

$$\begin{aligned} \gamma =\inf \{ \Vert u-v\Vert : u\in \mathrm {span}\{x_i\}_{i=1}^{n-1}, v\in \overline{\mathrm {span}}\{x_i\}_{i=n}^\infty , \Vert u\Vert =1\}. \end{aligned}$$

We will show that $\gamma >0$. Indeed, if $\gamma =0$ then there are sequences $\{u_j\}_{j=1}^\infty \subset \mathrm {span}\{x_i\}_{i=1}^{n-1}, \Vert u_j\Vert =1$, for all j, and $\{v_j\}_{j=1}^\infty \subset \overline{\mathrm {span}}\{x_i\}_{i=n}^\infty$ so that

$$\begin{aligned} \Vert u_j-v_j\Vert \rightarrow 0 \text{ as } j\rightarrow \infty . \end{aligned}$$

By switching to a subsequence if necessary, we may assume $u_j \rightarrow u \in \mathrm {span}\{x_i\}_{i=1}^{n-1}$ and $u\not = 0$. Since

$$\begin{aligned} \Vert v_j-u\Vert \le \Vert v_j-u_j\Vert +\Vert u_j-u\Vert , \end{aligned}$$

we conclude that $v_j\rightarrow u\in \overline{\mathrm {span}}\{x_i\}_{i=n}^\infty$. Thus,

$$\begin{aligned} u\in \mathrm {span}\{x_i\}_{i=1}^{n-1}\cap \ \overline{\mathrm {span}}\{x_i\}_{i=n}^\infty . \end{aligned}$$

Since $u\in \mathrm {span}\{x_i\}_{i=1}^{n-1}, u\not =0$, we can write $u=\sum _{i=1}^{n-1}\alpha _ix_i$ for some scalars $\alpha _i's$ not all zero. Without loss of generality, we can assume $\alpha _1\not =0$. Then

$$\begin{aligned} x_1=\dfrac{1}{\alpha _1}\left( u-\sum _{i=2}^{n-1}\alpha _ix_i\right) \in \overline{\mathrm {span}}\{x_i\}_{i=2}^\infty , \end{aligned}$$

which contradicts the fact that $\{x_i\}_{i=1}^\infty$ is separated. So, $\gamma >0.$

Now we have

$$\begin{aligned} \Vert \dfrac{y_j+z_j}{\Vert y_j\Vert }\Vert \ge \gamma , \text{ for } \text{ all } j\ge n_0 , \end{aligned}$$

and $\underset{j\ge 1}{\sup }\Vert x_j\Vert$ is finte. Therefore, there is some $K>0$ such that

$$\begin{aligned} \Vert y_j\Vert \le \dfrac{1}{\gamma }\Vert y_j+z_j\Vert \le \dfrac{1}{\gamma }(\Vert y_j+z_j-x_{n_j}\Vert + \Vert x_{n_j}\Vert )\le K, \text{ for } \text{ all } j\ge n_0. \end{aligned}$$

The Claim 2 is proven.

Now since $\epsilon \le \Vert y_j\Vert \le K$ for all $j\ge n_0$, it has a convergent subsequence $y_{j_k}\rightarrow y\in \mathrm {span}\{x_i\}_{i=1}^{n-1}$, and $y\not =0$.

From the fact that

$$\begin{aligned} \Vert x_{n_{j_k}}-z_{j_k}-y\Vert \le \Vert x_{n_{j_k}}-z_{j_k}-y_{j_k}\Vert +\Vert y_{j_k}-y\Vert \le \dfrac{1}{j_k}+\Vert y_{j_k}-y\Vert , \end{aligned}$$

we conclude $x_{n_{j_k}}-z_{j_k}\rightarrow y\in \overline{\mathrm {span}}\{x_i\}_{i=n}^\infty$ as $k\rightarrow \infty$. Thus,

$$\begin{aligned} y\in \mathrm {span}\{x_i\}_{i=1}^{n-1}\cap \overline{\mathrm {span}}\{x_i\}_{i=n}^\infty \end{aligned}$$

By the same argument as in the proof of Claim 2, this leads to a contradiction with the fact that $\{x_i\}_{i=1}^\infty$ is separated.

Step 2 There exists a $\delta >0$ so that

$$\begin{aligned} \Vert (I-P_j)x_j\Vert \ge \delta , \text{ for } \text{ all } j\ge 1. \end{aligned}$$

Since $\{x_i\}_{i=1}^\infty$ is separated, for each $i=1,2,\ldots ,n-1$, there exists $\epsilon _i>0$ so that $\Vert (I-P_i)x_{i}\Vert \ge \epsilon _i$. Combined with Step 1, we have that $\{x_i\}_{i=1}^{\infty }$ is $\delta$-separated, where $\delta = \underset{i=1, \ldots , n-1}{\min }\{\epsilon _i, \delta _2\}.$ The proof of the Proposition is completed. $\square$

Now we give a complete classification of when the state estimation problem is solvable for all measurement vectors in $\ell _1$. Note that we have done it in complete generality and not assumed that $\{x_k\}_{k=1}^{\infty }$ is injective.

Theorem 3.39

Let $\mathcal {X}=\{x_k\}_{k=1}^{\infty }$ be a frame for the real or complex space $\ell _2$. The following are equivalent:

• (1) For every real vector $a=(a_1,a_2,\ldots )\in \ell _1$, there is a Hilbert Schmidt self-adjoint operator T so that

  $$\begin{aligned} \langle Tx_k, x_k\rangle = a_k, \text{ for } \text{ all } k=1,2,\ldots . \end{aligned}$$

• (2) The sequence $\{\tilde{x}_k\}_{k=1}^{\infty }$ is $\delta$-separated.

Proof

$(1)\Rightarrow (2)$: By (1), for each $k=1, 2,\ldots$, there is a Hibert Schmidt self-adjoint operator $R_k$, and hence a vector $\tilde{R}_k\in \tilde{\mathbb {H}}$ so that

$$\begin{aligned} \langle \tilde{R}_k, \tilde{x}_l\rangle =\langle R_kx_l, x_l\rangle = {\left\{ \begin{array}{ll} 1&{} \text{ if } k=l\\ 0&{} \text{ if } k\not = l. \end{array}\right. } \end{aligned}$$

It follows that $\tilde{x}_l\notin \overline{\mathrm {span}}\{\tilde{x}_k\}_{k\not =l}$ and hence $\{\tilde{x}_k\}_{k=1}^{\infty }$ is separated. We now proceed by way of contradiction. Suppose that $\{\tilde{x}_k\}_{k=1}^{\infty }$ is not $\delta$-separated. Then $\{\tilde{x}_k\}_{k=n}^{\infty }$ is not $\delta _n$-separated for all n. Then for $n=1$, there is $k_1\ge 1$ such that

$$\begin{aligned} \Vert \tilde{x}_{k_1} -P_{k_1}(\tilde{x}_{k_1})\Vert <\dfrac{1}{2}. \end{aligned}$$

Since $P_{k_1}(\tilde{x}_{k_1})\in \overline{\mathrm {span}}\{\tilde{x}_{k}\}_{k=1, k\not =k_1}^{\infty }$, there are some scalars $\alpha _k, k \in I$, where I is a finite subset of $\{k: k\ge 1, k\not =k_1\}$ such that

$$\begin{aligned} \left\| P_{k_1}(\tilde{x}_{k_1})-\sum _{k\in I}\alpha _k\tilde{x}_{k}\right\| <\dfrac{1}{2}. \end{aligned}$$

Let $y_{1}=\sum _{k\in I}\alpha _k\tilde{x}_{k}$. Then

$$\begin{aligned} \Vert \tilde{x}_{k_1}-y_{1}\Vert <1. \end{aligned}$$

Now let $n_2>\max \{k_1, k\}_{k\in I}$. Since $\{\tilde{x}_{k}\}_{k=n_2}^\infty$ is not $\delta _{n_2}$-separated, similar to the above, there are numbers $n_2\le k_2<n_3$ and a vector

$$\begin{aligned} y_{2}\in \mathrm {span}\{\tilde{x}_{k}: n_2\le k\not =k_2<n_3\} \end{aligned}$$

such that

$$\begin{aligned} \Vert \tilde{x}_{k_2}-y_{2}\Vert <\dfrac{1}{2^3}. \end{aligned}$$

Continuing this procedure we can choose $1=n_1 \le k_1< n_2 \le k_2< n_3 < \cdots$ and vectors

$$\begin{aligned} y_{m} \in \mathrm {span}\{\tilde{x}_{k}:n_m\le k\not = k_m < n_{m+1}\}, \end{aligned}$$

such that

$$\begin{aligned} \Vert \tilde{x}_{k_m}-y_{m}\Vert < \frac{1}{m^3}, \end{aligned}$$

for all m. Now let $a=\{a_k\}_{k=1}^\infty \in \ell _1$, where

$$\begin{aligned} a_k = {\left\{ \begin{array}{ll} \frac{1}{m^2}&{} \text{ if } k=k_m\\ 0 &{} \text{ otherwise } . \end{array}\right. } \end{aligned}$$

Then by assumption, there exists a Hilbert Schmidt self-adjoint operator T and a vector $\tilde{T}\in \tilde{\mathbb {H}}$ so that $\langle \tilde{T}, \tilde{x}_{k}\rangle =\langle Tx_{k}, x_{k}\rangle = a_k$ for all k. But then

$$\begin{aligned} \dfrac{1}{m^2}=\langle \tilde{T}, \tilde{x}_{k_m}\rangle =\langle \tilde{T},\tilde{x}_{k_m}-y_m\rangle \le \Vert \tilde{T}\Vert \Vert \tilde{x}_{k_m}-y_{m}\Vert \le \Vert \tilde{T}\Vert \dfrac{1}{m^3}, \end{aligned}$$

which implies $\Vert \tilde{T}\Vert \ge m$ for all m, a contradiction.

$(2)\Rightarrow (1)$: Since $\{\tilde{x}_k\}_{k=1}^\infty$ is $\delta$-separated, by Proposition 3.37, there are vectors $\{\tilde{T}_k\}_{k=1}^{\infty }$ in $\tilde{\mathbb {H}}$ satisfying

$$\begin{aligned} \langle \tilde{T}_k, \tilde{x}_l\rangle = {\left\{ \begin{array}{ll} 1 &{} \text{ if } k=l\\ 0 &{} \text{ if } k\not = l \end{array}\right. } \end{aligned}$$

for all $k, l\ge 1$, and $\underset{k\ge 1}{\sup }\Vert \tilde{T}_k\Vert < \infty$. Now, fix $a=(a_1,a_2,\ldots )\in \ell _1$ and let

$$\begin{aligned} \tilde{T} = \sum _{k=1}^{\infty }a_k\tilde{T}_k. \end{aligned}$$

This series converges since $a\in \ell _1$ and $\underset{k\ge 1}{\sup }\Vert \tilde{T}_k\Vert <\infty$. Now, let T be the Hilbert Schmidt self-adjoint operator that corresponds with $\tilde{T}$. Then we have

$$\begin{aligned} \langle Tx_k, x_k\rangle =\langle \tilde{T}, \tilde{x}_k\rangle = a_k, \text{ for } \text{ all } k=1,2,\ldots . \end{aligned}$$

This completes the proof. $\square$

Now we show that there is no injective frame for which the state estimation problem is solvable for all measurements taken from $\ell _2$. Note that for a Hilbert Schmidt self-adjoint operator T on the Hilbert space $\ell _2$, the corresponding vector $\tilde{T}$ is defined as in the proof of Theorem 3.13 for the real case and Theorem 3.17 for the complex case.

Theorem 3.40

There is no injective frame $\mathcal {X}=\{x_k\}_{k=1}^{\infty }$ in the real or complex space $\ell _2$ so that for every $a=\{a_k\}_{k=1}^{\infty }\in \ell _2$, there is a Hilbert Schmidt operator T so that

$$\begin{aligned} \langle Tx_k, x_k\rangle = a_k, \text{ for } \text{ all } k=1,2,\ldots . \end{aligned}$$

Proof

We will proceed by way of contradiction. The proof is divided into steps.

Suppose that there is an injective frame $\mathcal {X}=\{x_k\}_{k=1}^{\infty }$ for which the state estimation problem is solvable for all choices $\{a_k\}_{k=1}^\infty \in \ell _2$.

Step I There are vectors $\tilde{R}_k\in \tilde{\mathbb {H}}, k=1, 2, \ldots$ so that $\langle \tilde{R}_k,\tilde{x}_l\rangle =\delta _{kl}$.

This is immediate because by assumption, for each $k=1, 2,\ldots$, there is a Hilbert Schmidt self-adjoint operator $R_k$ so that

$$\begin{aligned} \langle \tilde{R}_k,\tilde{x}_l\rangle =\langle R_kx_l,x_l\rangle ={\left\{ \begin{array}{ll} 1&{} \text{ if } k=l\\ 0&{} \text{ if } k\not = l. \end{array}\right. } \end{aligned}$$

Denote $E_n=\mathrm {span}\{\tilde{x}_k\}_{k=1}^n$ and let $P_n$ be the projection onto $E_n$.

Step II If there is a real vector $\{a_k\}_{k=1}^\infty \in \ell _2$ satisfying $\underset{n}{\sup }\Vert \sum _{k=1}^na_k\tilde{R_k}\Vert =\infty$, then there is a real vector $\{b_k\}_{k=1}^{\infty }\in \ell _2$ and $n_1<n_2< \cdots$ so that

$$\begin{aligned} \left\| P_{n_j}\left( \sum _{k=1}^{n_j}b_k\tilde{R_k}\right) \right\| \ge j. \end{aligned}$$

Indeed, since $\underset{n}{\sup }\Vert \sum _{k=1}^na_k\tilde{R_k}\Vert =\infty$, we can choose a sequence $m_1<m_2 <\cdots$ so that

$$\begin{aligned} \left\| \sum _{k=1}^{m_j}a_k\tilde{R}_k\right\| \ge 2j. \end{aligned}$$

For any $j>1$, we have

$$\begin{aligned} \left\| \sum _{k=1}^{m_1}a_k\tilde{R}_k-\sum _{k=m_1+1}^{m_j}a_k\tilde{R}_k\right\|&\ge \left\| \sum _{k=1}^{m_j}a_k\tilde{R}_k\right\| -2\left\| \sum _{k=1}^{m_1}a_k\tilde{R}_k\right\| \\&\ge 2j-2\left\| \sum _{k=1}^{m_1}a_k\tilde{R}_k\right\| . \end{aligned}$$

Combining this with the fact that $E_1\subset E_2 \subset \ldots$ and $\cup _{n=1}^{\infty } E_n$ is dense in $\tilde{\mathbb {H}}$, we can choose j large enough so that

$$\begin{aligned} \left\| P_{m_{j}}\left( \sum _{k=1}^{m_1}a_k\tilde{R}_k\right) \right\| \ge \frac{1}{2}\left\| \sum _{k=1}^{m_1}a_k\tilde{R_k}\right\| , \end{aligned}$$

and

$$\begin{aligned} \left\| \sum _{k=1}^{m_1}a_k\tilde{R}_k-\sum _{k=m_1+1}^{m_j}a_k\tilde{R}_k\right\| \ge 4. \end{aligned}$$

Since

$$\begin{aligned}&\left\| P_{m_j}\left( \sum _{k=1}^{m_1}a_k\tilde{R}_k\right) +P_{m_j}\left( \sum _{k=m_1+1}^{m_j}a_k\tilde{R}_k\right) \right\| ^2 \\&\qquad +\left\| P_{m_j}\left( \sum _{k=1}^{m_1}a_k\tilde{R}_k\right) -P_{m_j}\left( \sum _{k=m_1+1}^{m_j}a_i\tilde{R}_k\right) \right\| ^2\\&\quad = 2\left( \left\| P_{m_j}\left( \sum _{k=1}^{m_1}a_k\tilde{R}_k\right) \Vert ^2+\Vert P_{m_j}\left( \sum _{k=m_1+1}^{m_j}a_k\tilde{R_k}\right) \right\| ^2\right) \\&\quad \ge 2\left\| P_{m_j}\left( \sum _{k=1}^{m_1}a_k\tilde{R}_k\right) \right\| ^2, \end{aligned}$$

we can choose $b_i=a_i$ for $i=1,\ldots , m_1$ and $b_i\in \{a_i, -a_i\}$ for $i=m_1+1,\ldots , m_j$ so that

$$\begin{aligned} \left\| P_{m_j}\left( \sum _{k=1}^{m_j}b_k\tilde{R}_k\right) \right\| \ge \left\| P_{m_j}\left( \sum _{k=1}^{m_1}b_k\tilde{R}_k\right) \right\| \ge \dfrac{1}{2}\left\| \sum _{k=1}^{m_1}b_k\tilde{R_k}\right\| \ge 1 \text{ and } \left\| \sum _{k=1}^{m_j}b_k\tilde{R}_k\right\| \ge 4. \end{aligned}$$

Setting $n_1=m_j$,

$$\begin{aligned} \left\| P_{n_1}\left( \sum _{k=1}^{n_1}b_k\tilde{R}_k\right) \right\| \ge 1 \text{ and } \left\| \sum _{k=1}^{n_1}b_k\tilde{R}_k\right\| \ge 4. \end{aligned}$$

Now for $m_j$ above, by the same argument, there is $m_l>m_j$ and $b_i\in \{a_i, -a_i\}$ for $i=m_{j}+1,\ldots ,m_l$ so that

$$\begin{aligned} \left\| P_{m_l}\left( \sum _{k=1}^{m_l}b_k\tilde{R}_k\right) \right\| \ge \left\| P_{m_l}\left( \sum _{k=1}^{m_j}b_k\tilde{R}_k\right) \right\| \ge \dfrac{1}{2}\left\| \sum _{k=1}^{m_j}b_k\tilde{R}_k\right\| \ge 2 \end{aligned}$$

and

$$\begin{aligned} \left\| \sum _{k=1}^{m_l}b_k\tilde{R}_k\right\| \ge 6. \end{aligned}$$

Set $n_2=m_l$ we get

$$\begin{aligned} \left\| P_{n_2}\left( \sum _{k=1}^{n_2}b_k\tilde{R}_k\right) \right\| \ge 2. \end{aligned}$$

Continuing this process inductively, the result follows.

Step III For all vectors $\{ a_k\}_{k=1}^{\infty }\in \ell _2$, $\underset{n}{\sup }\Vert \sum _{k=1}^na_k\tilde{R_k}\Vert$ is finite.

Suppose by contradiction that there is a vector $\{ a_k\}_{k=1}^{\infty }\in \ell _2$ so that $\underset{n}{\sup }\Vert \sum _{k=1}^na_k\tilde{R_k}\Vert =\infty$. Let $\{b_k\}_{k=1}^\infty$ be the vector in Step II, then there exists a vector $\tilde{T}\in \tilde{\mathbb {H}}$ so that $\langle \tilde{T},\tilde{x}_k\rangle = b_k$, for all $k=1,2,\ldots$. It follows that

$$\begin{aligned} P_{n_j}\tilde{T}= P_{n_j}\left( \sum _{k=1}^{n_j}b_k\tilde{R}_k\right) \end{aligned}$$

for all $j=1,2,\ldots$. Hence,

$$\begin{aligned} \infty = \sup _{j}\left\| P_{n_j}\left( \sum _{k=1}^{n_j}b_k\tilde{R}_{k}\right) \right\| = \sup _{j}\Vert P_{n_j}\tilde{T}\Vert \le \Vert \tilde{T}\Vert , \end{aligned}$$

which is a contradiction.

Step IV$\{ \tilde{R}_k\}_{k=1}^{\infty }$ is a Bessel sequence in $\tilde{\mathbb {H}}$.

For each $n\in \mathbb {N}$, define an operator

$$\begin{aligned}&T_n: \ell _2 \longrightarrow \tilde{\mathbb {H}}\\&x=(a_1, a_2, \ldots )\longmapsto T_n(x)=\sum _{k=1}^{n}a_k\tilde{R}_k \end{aligned}$$

Then $T_n$ is a bounded linear operator for all n.

By Step III, $\underset{n}{\sup }\Vert \sum _{k=1}^na_k\tilde{R_k}\Vert$ is finite for all $x=\{a_k\}_{k=1}^\infty$. By the Uniform Boundedness Principle, $\underset{n}{\sup }\Vert T_n\Vert \le B$, for some $B>0$. For any $n, m \in \mathbb {N}, m>n$, we have

$$\begin{aligned} \left\| \sum _{k=n+1}^m{a_k\tilde{R}_k}\right\| ^2=\left\| T_m \left( \sum _{k=n+1}^{m}a_ke_k\right) \right\| ^2\le B^2\sum _{k=n+1}^{m}a_k^2. \end{aligned}$$

It follows that $\sum _{k=1}^{\infty }a_k\tilde{R}_k$ converges, and hence $\{\tilde{R}_k\}_{k=1}^\infty$ is Bessel.

Step V We arrive at a contradiction.

We have shown that under our assumption, $\{\tilde{R}_k\}_{k=1}^{\infty }$ is $B^2$-Bessel for some B. Now choose any $a=\{a_k\}_{k=1}^{\infty }\in \ell _2$. We have that

$$\begin{aligned} \left\| \sum _{k=1}^{\infty }a_k\tilde{R}_k\right\| ^2\le B^2 \sum _{k=1}^{\infty } a_k^2. \end{aligned}$$

By Theorem 3.20, $\sum _{k=1}^{\infty }a_k\tilde{x}_k$ converges. Now, we have

$$\begin{aligned} \left\| \sum _{k=1}^{\infty }a_k\tilde{x}_k\right\|&= \sup _{\Vert x\Vert \le 1} \left| \left\langle x,\sum _{k=1}^\infty a_k\tilde{x}_k\right\rangle \right| \\&\ge \frac{1}{B\Vert a\Vert }\left| \left\langle \sum _{k=1}^\infty a_k\tilde{R}_k,\sum _{l=1}^\infty a_l\tilde{x}_l\right\rangle \right| \\&=\frac{1}{B\Vert a\Vert } \left| \sum _{k,l=1}^\infty a_ka_l\langle \tilde{R}_k,\tilde{x}_l\rangle \right| \\&= \frac{1}{B}\Vert a\Vert . \end{aligned}$$

It follows that $\{\tilde{x}_k\}_{k=1}^{\infty }$ has a positive lower Riesz bound and since this family is injective, it is a Riesz basis. Hence by Theorem 3.20, it is a frame sequence. But then by Corollary 3.21, $\{x_k\}_{k=1}^\infty$ cannot be injective, a contradiction. The proof of our theorem is now complete. $\square$

Remark 3.41

As in the finite dimensional case, it is often the case that the state estimation problem is not solvable. But again there is a natural way to get a good estimation to the solution. Given a frame $\{x_k\}_{k=1}^{\infty }$ and $\{a_k\}_{k=1}^{\infty }\in \ell _2$, choose m so that $\sum _{k=m+1}^{\infty }a_k^2\le \epsilon$. Then apply the argument in the finite case to get the best solution for $\{a_k\}_{k=1}^m$.