Norm Inequalities Associated with Two Projections

Abstract

Suppose that p and q are projections in a unital $C^*$-algebra $\mathfrak {A}$ such that $\Vert p(1-q)\Vert <1$. It is shown that there exists a unitary u in $\mathfrak {A}$ which is homotopic to the unit of $\mathfrak {A}$, and satisfies $pup=pu^*p$, $u(pqp)u^*=qpq$ and $\Vert 1-u\Vert \le \sqrt{\frac{2\Vert (qp)^\dag \Vert }{1+\Vert (qp)^\dag \Vert }}\cdot \Vert p(1-q)\Vert$, where $(qp)^\dag$ denotes the Moore–Penrose inverse of qp. Under the same restriction of $\Vert p(1-q)\Vert <1$, it is proved that $\Vert p-q\Vert <1$ if and only if there exists a unitary u in $\mathfrak {A}$ such that pup is normal and $q=upu^*$. An example is constructed to show that there exist certain Hilbert space H and projections p and q on H such that $\Vert p-q\Vert =1$ and $q=upu^*$ for some unitary operator u on H.

1 Introduction

Recall that two projections p and q in a (unital) $C^*$-algebra $\mathfrak {A}$ are referred to be homotopic, written $p\sim _h q$, if p and q are connected by a norm-continuous path of projections in $\mathfrak {A}$. It is known that this homotopic equivalence is stronger than the unitary equivalence. Namely, if $p\sim _h q$, then there exists a unitary $u\in \mathfrak {A}$ such that $q=upu^*$. To fulfill a proof of such an assertion, it needs only use the following well-known result:

Lemma 1

([10, Lemma 6.2.1]) Let p, q be projections in a unital $C^*$-algebra $\mathfrak {A}$ and suppose that $\Vert p-q\Vert <1$. Then, there exists a unitary u in $\mathfrak {A}$ such that $q=upu^*$ and

$$\begin{aligned} \Vert 1-u\Vert \le \sqrt{2}\, \Vert p-q\Vert , \end{aligned}$$

(1)

where $u=v|v|^{-1}$, in which $v=1-p-q+2qp$.

Some other characterizations of projections p and q satisfying $\Vert p-q\Vert <1$ can be found in [11, Proposition 2.2.4] and [12, Proposition 5.2.6]. It is notable that the above lemma can be used to clarify the continuity of Moore–Penrose inverses of elements in a $C^*$-algebra [5, 8]. Based on the theory of the Moore–Penrose inverse [4, 5, 13] and Halmos’ two projections theorem [1, 3, 9, 14], in this paper we have managed to provide an analogous result under the condition that $\Vert p(1-q)\Vert < 1$; see Theorem 1 and Remark 3 for the details. Due to $\Vert p(1-q)\Vert =\Vert p(p-q)\Vert \le \Vert p-q\Vert$, or using the following Krein–Krasnoselskii–Milman equality [7, 15]

$$\begin{aligned} \Vert p-q\Vert =\max \big \{\Vert p(1-q)\Vert , \Vert q(1-p)\Vert \big \}, \end{aligned}$$

(2)

we see that Theorem 1 as well as Corollary 1 actually gives a generalization of Lemma 1.

Let p and q be projections in a unital $C^*$-algebra $\mathfrak {A}$. A simple use of (2) gives $\Vert p-q\Vert \le 1$. It is interesting to investigate conditions that ensure $\Vert p-q\Vert <1$ and $\Vert p-q\Vert =1$, respectively. For some recent progress in this direction for Hilbert $C^*$-module operators, the reader is referred to [15]. Our aim here is to give some new characterizations of $\Vert p-q\Vert <1$ under the precondition that $\Vert p(1-q)\Vert <1$.

The rest of the paper is organized as follows. In Sect. 2, some basic knowledge about the Moore–Penrose inverse and Halmos’ two projections theorem are provided. In Sect. 3, we focus on the generalization of the above Lemma 1. Let p and q be projections in a unital $C^*$-algebra $\mathfrak {A}$. Some results analogous to that of Lemma 1 are obtained in Theorem  1 under the weaker condition that $\Vert p(1-q)\Vert <1$. When $\Vert p-q\Vert <1$ (as is considered in Lemma 1), it is shown in Corollary 1 that the number $\sqrt{2}$ in (1) can in fact be replaced with $\sqrt{\frac{2m}{1+m}}$ by choosing a unitary element u in $\mathfrak {A}$ which satisfies $q=upu^*$, where m is defined by (27). Furthermore, as is shown in Theorem 1, this unitary element u is homotopic to the unit of $\mathfrak {A}$. In Sect. 4, we turn to study the case of $\Vert p-q\Vert <1$ under the precondition that $\Vert p(1-q)\Vert <1$. Based on the canonical $2\times 2$ matrix representations for two projections on a Hilbert space, some necessary and sufficient conditions are provided in Theorem 2. As a supplement to Theorem 2, we have managed to construct two projections P and Q on certain Hilbert space H such that $\Vert P-Q\Vert =1$ and $Q=UPU^*$ for some unitary operator $U\in \mathbb {B}(H)$ (see Example 2). This shows the independence of the two conditions stated in (34).

2 Preliminaries

Throughout the rest of this paper, $\mathbb {C}$ is the complex field, H is a Hilbert space, $\mathbb {B}(H)$ is the set of all bounded linear operators on H, $\mathfrak {A}$ is a nonzero unital $C^*$-algebra, whose unit is denoted simply by 1. We use letters in lower case to denote elements in $\mathfrak {A}$, while all operators in $\mathbb {B}(H)$ will be capitalized. The notation $I_H$ will be reserved for the identity operator on H. Let $\mathbb {B}(H)_{+}$ be the subset of $\mathbb {B}(H)$ consisting of all positive elements in $\mathbb {B}(H)$. The notation $A\ge 0$ is also used to indicate that A is an element of $\mathbb {B}(H)_+$. An operator A in $\mathbb {B}(H)$ is said to be positive definite, written $A>0$, if $A\ge 0$ and A is invertible in $\mathbb {B}(H)$. By a projection, we mean that it is idempotent and self-adjoint. A projection $P\in \mathbb {B}(H)$ is said to be non-trivial if $P\ne 0$ and $P\ne I_H$. Given $a\in \mathfrak {A}$, its Moore–Penrose inverse (briefly MP-inverse) is denoted by $a^\dag$, which is the unique element in $\mathfrak {A}$ satisfying

$$\begin{aligned} axa=a, \quad xax=x,\quad (ax)^*=ax,\quad (xa)^*=xa. \end{aligned}$$

From [4], we know that a is MP-invertible (that is, $a^\dag$ is existent) if and only if there exists $b\in \mathfrak {A}$ such that $aba=a$. In the special case that $\mathfrak {A}=\mathbb {B}(H)$, it is well-known (see, e.g., [13, Theorem 2.2]) that an operator $T\in \mathbb {B}(H)$ is MP-invertible if and only if $\mathcal {R}(T)$ is closed in H, where $\mathcal {R}(T)$ and $\mathcal {N}(T)$ denote the range and the null space of T, respectively. Since the closedness of any one of

$$\begin{aligned} \mathcal {R}(T),\quad \mathcal {R}(T^*),\quad \mathcal {R}(TT^*),\quad \mathcal {R}(T^*T) \end{aligned}$$

implies the closedness of the remaining three sets [9, Lemma 5.7], T is MP-invertible if and only if any one of $T^*, TT^*$ and $T^*T$ is. In such case, we have $(T^*T)^\dag T^*=T^\dag$. These characterizations of the MP-invertibility will be used without specified.

Given closed linear subspaces $M, H_1$ and $H_2$ of H, let $P_M$ denote the projection from H onto M, and let $H_1\oplus H_2$ be the Hilbert space defined by

$$\begin{aligned} H_1\oplus H_2=\left\{ \left( {\begin{array}{c}h_1\\ h_2\end{array}}\right) :h_i\in H_i, i=1,2\right\} . \end{aligned}$$

Now, given projections $P,Q\in \mathbb {B}(H)$, let

$$\begin{aligned} H_1= & {} \mathcal {R}(P)\cap \mathcal {R}(Q),\quad \ H_2=\mathcal {R}(P)\cap \mathcal {N}(Q), \end{aligned}$$

(3)

$$\begin{aligned} H_3= & {} \mathcal {N}(P)\cap \mathcal {R}(Q), \quad H_4=\mathcal {N}(P)\cap \mathcal {N}(Q). \end{aligned}$$

(4)

We denote by $P_i$ the projection on $H_i$ for $1\le i\le 4$. As in [9, 14], we put

$$\begin{aligned} P_5&= P-P_1-P_2\quad \text{ and }\quad H_5=\mathcal {R}(P_5), \end{aligned}$$

(5)

$$\begin{aligned} P_6&= (I_H-P)-P_3-P_4\quad \text{ and }\quad H_6=\mathcal {R}(P_6). \end{aligned}$$

(6)

Then, a unitary operator $U_{P,Q}: H\rightarrow \bigoplus \nolimits _{i=1}^6 H_i$ can be induced by setting

$$\begin{aligned} U_{P,Q}(x)=(P_1(x),P_2(x),\ldots ,P_6(x))^T\quad \text{ for } x\in H \end{aligned}$$

(7)

such that

$$\begin{aligned} U_{P,Q}^*((x_1,x_2,\ldots ,x_6)^T)=\sum _{i=1}^6 x_i,\quad \text{ for } x_i\in H_i, i=1,2,\ldots ,6. \end{aligned}$$

With the notations as above, we have $H_5\ne \{0\}$ if and only if $H_6\ne \{0\}$, and in such case Halmos’ two projections theorem (see, e.g., [14, Theorem 3.3]) indicates that $U_{P,Q} P U_{P,Q}^*$ and $U_{P,Q} Q U_{P,Q}^*$ are given as

$$\begin{aligned} U_{P,Q} P U_{P,Q}^*&= I_{H_1}\oplus I_{H_2}\oplus 0\oplus 0\oplus I_{H_5}\oplus 0, \end{aligned}$$

(8)

$$\begin{aligned} U_{P,Q} Q U_{P,Q}^*&= I_{H_1}\oplus 0\oplus I_{H_3}\oplus 0\oplus \widehat{Q}, \end{aligned}$$

(9)

where $\widehat{Q}\in \mathbb {B}(H_5\oplus H_6)$ can be formulated by

$$\begin{aligned} \widehat{Q}=\left( \begin{array}{cc} A &{}\quad A^\frac{1}{2}(I_{H_5}-A)^\frac{1}{2} U_0^* \\ U_0A^\frac{1}{2}(I_{H_5}-A)^\frac{1}{2} &{}\quad U_0(I_{H_5}-A)U_0^*\\ \end{array} \right) ,\end{aligned}$$

(10)

in which $U_0\in \mathbb {B}(H_5,H_6)$ is a unitary, and $P_5QP_5|_{H_5}$ simplified as A, is a positive contraction in $\mathbb {B}(H_5)$ satisfying $\mathcal {N}(A)=\mathcal {N}(I_{H_5}-A)=\{0\}$.

3 Some Generalizations of Lemma 1

Recall that a pair $(\pi , H)$ is said to be a representation of a $C^*$-algebra $\mathfrak {A}$ if H is a Hilbert space and $\pi : \mathfrak {A}\rightarrow \mathbb {B}(H)$ is a $C^*$-morphism. Given a subset E of $\mathfrak {A}$, let $C^*E$ denote the $C^*$-subalgebra of $\mathfrak {A}$ generated by E.

We begin with a known result, whose proof can be in [6].

Lemma 2

Let a be an element in a $C^*$-algebra $\mathfrak {A}$ such that a is MP-invertible in $\mathfrak {A}$. Then, $a^\dag \in C^*\{a\}$.

Now, we state the technical lemma of this paper as follows.

Lemma 3

Let $P,Q\in \mathbb {B}(H)$ be projections such that $\Vert P(I_H-Q)\Vert < 1$. Then, QP is MP-invertible with $(QP)^\dag \in C^*\{P,Q\}$, and there exists a unitary $U\in C^*\{I_H,P,Q\}$ satisfying

$$\begin{aligned} PUP&= PU^*P, \quad U PQPU^*=QPQ, \end{aligned}$$

(11)

$$\begin{aligned} \Vert I_H-U\Vert&\le \sqrt{\frac{2\Vert (QP)^\dag \Vert }{1+\Vert (QP)^\dag \Vert }}\cdot \Vert P(I_H-Q)\Vert . \end{aligned}$$

(12)

If furthermore $\Vert P-Q\Vert <1$, then the second equation in (11) can be replaced with $Q=UPU^*$.

Proof

Let $H_i(1\le i \le 6)$ be defined by (3)–(6), and let $I_H$ and $I_{H_i}$ be simplified as I and $I_i$, respectively.

First, we consider the case that $H_5\ne \{0\}$ (and thus, $H_6\ne \{0\}$). Let $U_{P,Q}$ be defined by (7). Utilizing (8)–(10), we obtain

$$\begin{aligned} U_{P,Q}P(I-Q)P U_{P,Q}^*=0\oplus I_2\oplus 0\oplus 0\oplus \left( \begin{array}{cc} I_5-A &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) ,\end{aligned}$$

(13)

which gives

$$\begin{aligned} \max \{\Vert I_2\Vert , \Vert I_5-A\Vert \}=\Vert P(I-Q)P\Vert =\Vert P(I-Q)\Vert ^2<1. \end{aligned}$$

Hence, $I_2=0$ and $\Vert I_5-A\Vert <1$. This ensures $H_2=\{0\}$ and the invertibility of A, since A is a positive contraction in $\mathbb {B}(H_5)$. Let $U_1\in \mathbb {B}(H)$ be defined by

$$\begin{aligned} U_1=I-P-Q+2QP. \end{aligned}$$

(14)

Since $I_2=0$, we have $H_2=\{0\}$. So $H=H_1\oplus H_3\oplus H_4\oplus H_5\oplus H_6$. It follows from (8)–(10) that

$$\begin{aligned}&U_{P,Q}U_1U_{P,Q}^*=I_1 \oplus 0 \oplus I_4 \oplus \left( \begin{array}{cc} A &{}\quad -A^\frac{1}{2} (I_5-A)^\frac{1}{2} U_0^* \\ U_0 A^\frac{1}{2} (I_5-A)^\frac{1}{2} &{}\quad U_0 A U_0^* \\ \end{array} \right) ,\\&U_{P,Q}U_1^*U_1U_{P,Q}^*=I_1 \oplus 0 \oplus I_4 \oplus \left( \begin{array}{cc} A &{}\quad 0 \\ 0 &{}\quad U_0 AU_0^* \\ \end{array} \right) . \end{aligned}$$

Therefore,

$$\begin{aligned}&U_{P,Q} |U_1| U_{P,Q}^*=I_1 \oplus 0 \oplus I_4 \oplus \left( \begin{array}{cc} A^\frac{1}{2} &{}\quad 0 \\ 0 &{}\quad U_0 A^\frac{1}{2} U_0^* \\ \end{array} \right) .\end{aligned}$$

Due to the invertibility of A, the equation above indicates that $|U_1|$ is MP-invertible such that

$$\begin{aligned}U_{P,Q} |U_1|^\dag U_{P,Q}^*=I_1 \oplus 0 \oplus I_4 \oplus \left( \begin{array}{cc} A^{-\frac{1}{2}} &{}\quad 0 \\ 0 &{}\quad U_0 A^{-\frac{1}{2}} U_0^* \\ \end{array} \right) . \end{aligned}$$

So, if we put

$$\begin{aligned} U_2=U_1 |U_1|^\dag ,\quad U=U_2+I-U_2U_2^*, \end{aligned}$$

(15)

then

$$\begin{aligned} U_{P,Q} U_2 U_{P,Q}^*&= I_1 \oplus 0 \oplus I_4 \oplus \left( \begin{array}{cc} A^{\frac{1}{2}} &{}\quad -(I_5-A)^\frac{1}{2} U_0^*\\ U_0(I_5-A)^\frac{1}{2} &{}\quad U_0 A^{\frac{1}{2}} U_0^* \\ \end{array} \right) , \end{aligned}$$

(16)

$$\begin{aligned} U_{P,Q} U U_{P,Q}^*&= I_1\oplus I_3 \oplus I_4 \oplus \left( \begin{array}{cc} A^{\frac{1}{2}} &{}\quad -(I_5-A)^\frac{1}{2} U_0^*\\ U_0(I_5-A)^\frac{1}{2} &{}\quad U_0 A^{\frac{1}{2}} U_0^* \\ \end{array} \right) . \end{aligned}$$

(17)

It follows directly from (17) that $U^*U=UU^*=I$, hence U is a unitary. Based on (17) and (8)–(10), it is routine to check the validity of (11). Indeed, the first equation in (11) is clear, and

$$\begin{aligned}U_{P,Q}UPQP U_{P,Q}^*&=I_1\oplus 0 \oplus 0\oplus \left( \begin{array}{cc} A^{\frac{3}{2}} &{}\quad 0 \\ U_0A(I_5-A)^\frac{1}{2} &{}\quad 0 \\ \end{array} \right) \\&=U_{P,Q}QPQUU_{P,Q}^*. \end{aligned}$$

Furthermore, direct computations yield

$$\begin{aligned} U_{P,Q}(I-U)^*(I-U)U_{P,Q}^*=0\oplus 0\oplus 0 \oplus \left( \begin{array}{cc} 2(I_5-A^{\frac{1}{2}}) &{}\quad 0 \\ 0 &{}\quad 2U_0(I_5-A^{\frac{1}{2}})U_0^* \\ \end{array} \right) , \end{aligned}$$

which clearly gives

$$\begin{aligned} \Vert I-U\Vert =\sqrt{ 2(I_5-A^{\frac{1}{2}})}. \end{aligned}$$

(18)

Meanwhile, from (8)–(10) we can get

$$\begin{aligned} U_{P,Q} (QP)^*(QP) U_{P,Q}^*&= U_{P,Q} PQP U_{P,Q}^*=I_1\oplus 0\oplus 0\oplus \left( \begin{array}{cc} A &{}\quad 0\\ 0 &{}\quad 0 \\ \end{array} \right) ,\\ U_{P,Q} QP U_{P,Q}^*&= I_1\oplus 0\oplus 0\oplus \left( \begin{array}{cc} A &{}\quad 0\\ U_0A^\frac{1}{2} (I_5-A)^\frac{1}{2} &{}\quad 0\\ \end{array} \right) . \end{aligned}$$

Hence, QP is MP-invertible such that

$$\begin{aligned} U_{P,Q}(QP)^\dag U_{P,Q}^*&=U_{P,Q}\left[ (QP)^*(QP)\right] ^\dag U_{P,Q}^*\cdot U_{P,Q}(QP)^*U_{P,Q}^*\\&=I_1\oplus 0\oplus 0\oplus \left( \begin{array}{cc} I_5 &{}\quad A^{-\frac{1}{2}}(I_5-A)^{\frac{1}{2}}U_0^* \\ 0 &{}\quad 0 \\ \end{array} \right) . \end{aligned}$$

Let $\sigma (A)$ denote the spectrum of A. Since $\sigma (A)\subseteq (0,1]$, we have $\Vert A^{-1}\Vert \ge 1$. Therefore,

$$\begin{aligned} \Vert (QP)^\dag \Vert =\sqrt{\left\| (QP)^\dag \left[ (QP)^\dag \right] ^*\right\| }=\sqrt{\max \{\Vert I_1\Vert , \Vert A^{-1}\Vert \}}=\sqrt{\Vert A^{-1}\Vert }. \end{aligned}$$

(19)

To simplify the notation, we put

$$\begin{aligned} \lambda =\frac{1}{\Vert A^{-1}\Vert }. \end{aligned}$$

(20)

In virtue of (19) and (20), we have

$$\begin{aligned} \lambda =\frac{1}{\Vert (QP)^\dag \Vert ^2}, \end{aligned}$$

and $I_5=A^{\frac{1}{2}}A^{-1}A^{\frac{1}{2}}\le \Vert A^{-1}\Vert A=\frac{1}{\lambda } A$. Thus, $\sigma (A) \subseteq [\lambda , 1]$.

Let f and g be continuous functions defined on $\big [\lambda , 1\big ]$ by

$$\begin{aligned} f(t)=2(1-\sqrt{t}), \quad g(t)=1-t. \end{aligned}$$

For each $t\in \big [\lambda , 1\big ]$, it is obvious that

$$\begin{aligned} \frac{f(t)}{g(t)}=\frac{2}{1+\sqrt{t}}\le \frac{2}{1+\sqrt{\lambda }} =\frac{2\Vert (QP)^\dag \Vert }{1+\Vert (QP)^\dag \Vert }. \end{aligned}$$

Consequently,

$$\begin{aligned} \big \Vert 2(I_5- A^\frac{1}{2})\big \Vert =\,&\Vert f(A)\Vert \le \frac{2\Vert (QP)^\dag \Vert }{1+\Vert (QP)^\dag \Vert } \cdot \Vert g(A)\Vert \\ =\,&\frac{2\Vert (QP)^\dag \Vert }{1+\Vert (QP)^\dag \Vert }\Vert I_5-A\Vert , \end{aligned}$$

which leads by (18) and (13) with $I_2=0$ therein to (12). From Lemma 2, we deduce that $(QP)^\dag \in C^*\{QP\}\subseteq C^*\{P,Q\}$. In view of (14), (15) and Lemma 2, it can be concluded that $U\in C^*\{I_H,P,Q\}$.

Suppose now that $\Vert P-Q\Vert <1$. Then, using (2) we get $\Vert Q(I-P)\Vert <1$, which yields $\Vert U_{P,Q}(I-P)Q(I-P)U_{P,Q}^*\Vert <1$. Consequently, we have $H_3=\{0\}$; hence, both $I_2$ and $I_3$ are zero. So, by (8)–(10) we have

$$\begin{aligned} U_{P,Q}QU U_{P,Q}^*=I_1\oplus 0\oplus 0\oplus \left( \begin{array}{cc} A^\frac{1}{2} &{}\quad 0\\ U_0(I_5-A)^\frac{1}{2} &{}\quad 0\\ \end{array} \right) =U_{P,Q}UPU_{P,Q}^*. \end{aligned}$$

Therefore, $Q=UPU^*$ as desired.

Next, we consider the case that $H_5=\{0\}$ and $H_6=\{0\}$. In this case, we have $U_{P,Q}PU_{P,Q}^*\cdot U_{P,Q}QU_{P,Q}^*=U_{P,Q}QU_{P,Q}^*\cdot U_{P,Q}PU_{P,Q}^*$; hence, $PQ=QP$. So, if we set $U=I$, then (11) and (12) are satisfied obviously. $\square$

Lemma 4

Let $P,Q\in \mathbb {B}(H)$ be projections such that $\Vert P(I_H-Q)\Vert < 1$, and let U be given as in the proof of Lemma 3. Then, there exists a norm-continuous path consisting of unitary operators in $C^*\{I_H,P,Q\}$ that starts at $I_H$ and ends at U.

Proof

We follow the notations as in the proof of Lemma 3. Let $W\in C^*\{I,P,Q\}$ be defined by

$$\begin{aligned} W=PUP+(I-P)U(I-P). \end{aligned}$$

(21)

From (8) and (17), we have

$$\begin{aligned} U_{P,Q}WU_{P,Q}^*=I_1\oplus I_3\oplus I_4\oplus \left( \begin{array}{cc} A^{\frac{1}{2}} &{}\quad 0 \\ 0 &{}\quad U_0A^{\frac{1}{2}}U_0^* \\ \end{array} \right) , \end{aligned}$$

(22)

which means that W is a positive definite contraction, since so is A and $U_0$ is a unitary operator. Let

$$\begin{aligned} S_1(t)&= I+t(W-2I),\quad t\in [0,1/2],\nonumber \\ S_2(t')&= (1-t')W+(2t'-1)U, \quad t'\in [1/2,1]. \end{aligned}$$

(23)

Clearly, $S_1(t)$ and $S_2(t')$ are norm-continuous paths consisting of elements in $C^*\{I,P,Q\}$ such that

$$\begin{aligned} S_1(0)=I,\quad S_1(1/2)=S_2(1/2)=1/2\cdot W,\quad S_2(1)=U. \end{aligned}$$

For every $t\in [0,1/2]$, we have

$$\begin{aligned} S_1(t)\ge S_1(1/2) =1/2\cdot W>0. \end{aligned}$$

Therefore, $S_1(t)$ is invertible for all $t\in [0,1/2]$. In addition, by (23), (22) and (17) we know that for each $t'\in [1/2,1]$,

$$\begin{aligned} U_{P,Q}S_2(t')U_{P,Q}^*=t'I_1\oplus t'I_3 \oplus t'I_4\oplus \widetilde{S_2}(t'), \end{aligned}$$

where

$$\begin{aligned} \widetilde{S_2}(t')=\left( \begin{array}{cc} t'A^{\frac{1}{2}} &{}\quad -(2t'-1)(I_5-A)^{\frac{1}{2}}U_0^* \\ (2t'-1)U_0(I_5-A)^{\frac{1}{2}} &{}\quad t'U_0A^{\frac{1}{2}}U_0^* \\ \end{array} \right) . \end{aligned}$$

Direct computations yield

$$\begin{aligned} \widetilde{S_2}(t')\big [\widetilde{S_2}(t')\big ]^*=\big [\widetilde{S_2}(t')\big ]^*\widetilde{S_2}(t')=\left( \begin{array}{cc} R(t') &{}\quad 0 \\ 0 &{}\quad U_0R(t')U_0^* \\ \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} R(t')=t'^2A+(2t'-1)^2(I_5-A)\ge t'^2A>0. \end{aligned}$$

This shows that $S_2(t')$ is invertible for all $t'\in [1/2,1]$. Joining $S_1(\cdot )$ and $S_2(\cdot )$ together, we see that there exists a norm-continuous path $L(t) (t\in [0,1])$ consisting of invertible elements in $C^*\{I,P,Q\}$ that links I with U. The desired path U(t) is obtained by setting $U(t)=L(t)|L(t)|^{-1}$. $\square$

Remark 1

Let $P,Q\in \mathbb {B}(H)$ be projections such that

$$\begin{aligned} \Vert P(I_H-Q)\Vert < 1\quad \text{ and }\quad H_5\ne \{0\}. \end{aligned}$$

(24)

Following the notations as in the proof of the preceding lemma, we have $\sigma (A)\subseteq (0,1]$. Since $\mathcal {N}(I_5-A)=\{0\}$ and $H_5\ne \{0\}$, we see that $\Vert A^{-1}\Vert >1$. So by (19), we have $\Vert (QP)^\dag \Vert >1$. It is interesting to find out projections P and Q satisfying (24) such that $\Vert (QP)^\dag \Vert$ is close to 1. We provide such an example as follows.

Example 1

Let $H=\mathbb {C}^2$, $P=\left( \begin{array}{cc} 1 &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right)$. For each $t\in (0,\frac{\pi }{2})$, let $Q_t$ be the projection defined by $Q_t=\left( \begin{array}{cc} \cos ^2 t &{}\quad \cos t\sin t \\ \cos t\sin t &{}\quad \sin ^2 t\\ \end{array} \right)$. Direct computations yield

$$\begin{aligned}&\Vert P-Q_t\Vert =\sqrt{\Vert (P-Q_t)^2\Vert }=\Vert P(I_H-Q_t)\Vert =\sin t<1,\\&\Vert (Q_tP)^\dag \Vert =\left\| \Big [(Q_t P)^*(Q_tP)\Big ]^\dag (Q_tP)^*\right\| =\left\| \left( \begin{array}{cc} 1 &{}\quad \tan t \\ 0 &{}\quad 0 \\ \end{array} \right) \right\| =\sec t. \end{aligned}$$

Therefore, $\Vert (Q_tP)^\dag \Vert >1$ for every $t\in (0,\frac{\pi }{2})$, and $\Vert (Q_tP)^\dag \Vert \rightarrow 1$ as $t\rightarrow 0^+$.

Remark 2

Let P and $Q_t$ be as in Example 1, and let $H_i(1\le i \le 4)$ be defined by (3)– (4) with Q therein be replaced by $Q_t$. It is easy to show that $H_i=\{0\}(i\le i\le 4)$ for every $t\in (0,\frac{\pi }{2})$. Such a pair of projections is known as in generic position in the literature.

Our next result reads as follows.

Theorem 1

Let p, q be projections in a unital $C^*$-algebra $\mathfrak {A}$ such that $\Vert p(1-q)\Vert < 1$. Then, qp is MP-invertible with $(qp)^\dag \in C^*\{p,q\}$, and there exists a unitary $u\in C^*\{1,p,q\}$ satisfying

$$\begin{aligned} pup&= pu^*p, \quad u pqp u^*=qpq,\nonumber \\ \Vert 1-u\Vert&\le \sqrt{\frac{2\Vert (qp)^\dag \Vert }{1+\Vert (qp)^\dag \Vert }}\cdot \Vert p(1-q)\Vert . \end{aligned}$$

(25)

Moreover, there exists a norm-continuous path consisting of unitary elements in $C^*\{1,p,q\}$ that starts at 1 and ends at u. If furthermore $\Vert p-q\Vert <1$, then the second equation in (25) can be replaced with $q=upu^*$.

Proof

Choose any faithful unital representation $(\pi , H)$ of $\mathfrak {A}$. Let $P=\pi (p)$ and $Q=\pi (q)$. Then, $\Vert P(I_H-Q)\Vert < 1$, so by Lemma 3 there exists a unitary $U\in C^*\{I_H,P,Q\}$ such that (11) and (12) are satisfied. It is easy to verify that

$$\begin{aligned} \pi \left( C^*\{p,q\}\right) =C^*\{P,Q\},\quad \pi \left( C^*\{1,p,q\}\right) =C^*\{I_H,P,Q\}. \end{aligned}$$

(26)

Since $(QP)^\dag \in C^*\{P,Q\}$ and $U\in C^*\{I_H,P,Q\}$, the equations above indicate that $(QP)^\dag =\pi (w)$ and $U=\pi (u)$ for some $w\in C^*\{p,q\}$ and $u\in C^*\{1,p,q\}$. Hence,

$$\begin{aligned} \pi (u)\pi (u^*)=\pi (u^*)\pi (u)=\pi (1), \end{aligned}$$

which leads by the faithfulness of $\pi$ to $uu^*=u^*u=1$, and thus, u is a unitary in $\mathfrak {A}$. Similarly, by the Moore–Penrose equations

$$\begin{aligned}&\pi (qp)\pi (w)\pi (pq)=\pi (pq),\quad \pi (w)\pi (qp)\pi (w)=\pi (w),\\&\left[ \pi (qp)\pi (w)\right] ^*=\pi (qp)\pi (w),\quad \left[ \pi (w)\pi (qp)\right] ^*=\pi (w)\pi (qp), \end{aligned}$$

and the faithfulness of $\pi$, we conclude that qp is MP-invertible such that $(qp)^\dag =w$. Therefore,

$$\begin{aligned} \pi \left[ (qp)^\dag \right] =\pi (w)=(QP)^\dag =\left[ \pi (q)\pi (p)\right] ^\dag =\left[ \pi (qp)\right] ^\dag . \end{aligned}$$

In view of (26), we see that if $U(t) (t\in [0,1])$ is a norm-continuous path consisting of unitary operators in $C^*\{I_H,P,Q\}$ that starts at $I_H$ and ends at U, then $\pi ^{-1}(U(t))$ is a norm-continuous path consisting of unitary elements in $C^*\{1,p,q\}$ that starts at 1 and ends at u. The desired conclusion follows from Lemmas 3 and 4. $\square$

Remark 3

Let p, q be projections in a unital $C^*$-algebra $\mathfrak {A}$ and let u be a unitary in $\mathfrak {A}$ such that $q=upu^*$. Then, it is easy to show that the second equation in (25) is satisfied if and only if pup is normal.

Corollary 1

Let p, q be projections in a unital $C^*$-algebra $\mathfrak {A}$ such that $\Vert p-q\Vert < 1$. Then, there exists a unitary $u\in C^*\{1,p,q\}$ such that $q=upu^*$ and

$$\begin{aligned} \Vert 1-u\Vert \le \sqrt{\frac{2m}{1+m}}\cdot \Vert p-q\Vert , \end{aligned}$$

where

$$\begin{aligned} m=\min \Big \{\Vert (qp)^\dag \Vert , \big \Vert \left[ (1-q)(1-p)\right] ^\dag \big \Vert \Big \}. \end{aligned}$$

(27)

Proof

Observe that $\Vert (1-p)-(1-q)\Vert =\Vert p-q\Vert <1$, and the operator $U_1$ defined by (14) is invariant if P and Q are replaced with $I-P$ and $I-Q$, respectively. The same is true for the operator U defined by (15), so the conclusion follows immediately from Theorem 1. $\square$

4 Some New Characterizations of $\Vert p-q\Vert <1$

To give some new characterizations of norm inequalities, we now turn to use $2\times 2$ block matrix representations for two projections. Suppose that $P, Q\in \mathbb {B}(H)$ are projections such that P is non-trivial. Let $U_P: H\rightarrow \mathcal {R}(P)\dotplus \mathcal {N}(P)$ be defined by

$$\begin{aligned} U_Ph=\left( {\begin{array}{c}Ph\\ (I_H-P)h\end{array}}\right) ,\quad \forall \ h\in H. \end{aligned}$$

(28)

With the orthogonal decomposition $H=\mathcal {R}(P)\dotplus \mathcal {N}(P)$, the canonical $2\times 2$ matrix representations for P and Q are known as follows.

The following lemma and its generalization to the Hilbert $C^*$-module case can be found in [2, 14], respectively.

Lemma 5

[2, Theorem 4.5] Suppose that $P,Q\in \mathbb {B}(H)$ are projections such that P is non-trivial. Let $U_P$ be defined by (28). Then,

$$\begin{aligned} U_PPU_P^*&= \left( \begin{array}{cc} I_{\mathcal {R}(P)} &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) , \end{aligned}$$

(29)

$$\begin{aligned} U_PQU_P^*&= \left( \begin{array}{cc} A &{}\quad A^\frac{1}{2}(I_{\mathcal {R}(P)}-A)^\frac{1}{2} U_0^* \\ U_0A^\frac{1}{2}(I_{\mathcal {R}(P)}-A)^\frac{1}{2} &{}\quad U_0(I_{\mathcal {R}(P)}-A)U_0^*+Q_0\\ \end{array} \right) , \end{aligned}$$

(30)

where $A=PQP|_{\mathcal {R}(P)}\in \mathbb {B}(\mathcal {R}(P))$ is a positive contraction, $U_0\in \mathbb {B}(\mathcal {R}(P),\mathcal {N}(P))$ is a partial isometry and $Q_0\in \mathbb {B}(\mathcal {N}(P))$ is a projection satisfying

$$\begin{aligned} \mathcal {R}(U_0^*)=\overline{\mathcal {R}(A-A^2)}\quad \text{ and }\quad U_0^*Q_0=0. \end{aligned}$$

(31)

Lemma 6

Let $P,Q\in \mathbb {B}(H)$ be projections such that P is non-trivial and $\Vert P(I_H-Q)\Vert <1$. Let $U_P$ and $U_P QU_P^*$ be given by (28) and (30), respectively, such that (31) is satisfied. Then, $\Vert P-Q\Vert <1$ if and only if $Q_0=0$.

Proof

Let $I_H$, $I_{\mathcal {R}(P)}$ and $I_{\mathcal {N}(P)}$ be simplified as $I, I_1$ and $I_2$, respectively. Utilizing (29)–(30), we obtain

$$\begin{aligned} U_P\cdot P(I-Q)P\cdot U_P^*= \left( \begin{array}{cc} I_1-A &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) , \end{aligned}$$

which leads to

$$\begin{aligned} \Vert I_1-A\Vert =\Vert P(I-Q)P\Vert =\Vert P(I-Q)\Vert ^2<1. \end{aligned}$$

(32)

This ensures the invertibility of A, since A is a positive contraction in $\mathbb {B}(\mathcal {R}(P))$. Therefore, by [14, Lemma 2.3] we have

$$\begin{aligned} \overline{\mathcal {R}(A-A^2)}=\overline{\mathcal {R}\big [(I_1-A)A\big ]}=\overline{\mathcal {R}(I_1-A)}=\overline{\mathcal {R}\big [(I_1-A)^{\frac{1}{2}}\big ]}, \end{aligned}$$

so from the first equation in (31) we can get

$$\begin{aligned} U_0^*U_0(I_1-A)=I_1-A,\quad U_0^*U_0(I_1-A)^{\frac{1}{2}}=(I_1-A)^{\frac{1}{2}}. \end{aligned}$$

(33)

The equations above, together with (29) –(31), yield

$$\begin{aligned} U_P(P-Q)^2U_P^*=\left( \begin{array}{cc} I_1-A &{}\quad 0 \\ 0 &{}\quad U_0(I_1-A)U_0^*+Q_0 \\ \end{array} \right) . \end{aligned}$$

It is clear that $\Vert U_0(I_1-A)U_0^*\Vert \le \Vert I_1-A\Vert$, and

$$\begin{aligned} \Vert U_0(I_1-A)U_0^*+Q_0\Vert =\max \big \{\Vert U_0(I_1-A)U_0^*\Vert ,\Vert Q_0\Vert \big \}. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert P-Q\Vert ^2&=\max \big \{\Vert I_1-A\Vert ,\Vert U_0(I_1-A)U_0^*+Q_0\Vert \big \} \\&=\max \big \{\Vert I_1-A\Vert ,\Vert Q_0\Vert \big \}. \end{aligned}$$

Since $\Vert I_1-A\Vert <1$ (see(32)), we conclude that $\Vert P-Q\Vert <1$ if and only if $\Vert Q_0\Vert <1$, i.e., $Q_0=0$. $\square$

Theorem 2

Let $P,Q\in \mathbb {B}(H)$ be projections such that $\Vert P(I_H-Q)\Vert <1$. Then, the following statement are equivalent:

• (i) There exists a unitary $U\in C^*\{I_H,P,Q\}$ such that

  $$\begin{aligned} PUP \text{ is } \text{ normal } \quad \text{ and } \quad Q=UPU^*. \end{aligned}$$

  (34)

• (ii) There exists a unitary $U\in \mathbb {B}(H)$ satisfying (34).

• (iii) $\Vert P-Q\Vert <1$.

Proof

The implication (i) $\Longrightarrow$ (ii) is obvious, and (iii) $\Longrightarrow$ (i) is immediate from Lemma 3.

(ii) $\Longrightarrow$ (iii). If P is trivial, then $Q=P$, so $\Vert P-Q\Vert <1$ is obviously satisfied. Now, we suppose that P is non-trivial. In this case, $U_PPU_P^*$ and $U_PQU_P^*$ are given by (29)–(30) such that (31) is satisfied, where $U_P$ is defined by (28). Let $K_1=\mathcal {R}(P)$, $K_2=\mathcal {N}(P)$, and let $I_H,I_{K_1}$ and $I_{K_2}$ be denoted simply by $I,I_1$ and $I_2$, respectively. From the proof of Lemma 6, we see that $\Vert I_1-A\Vert =\Vert P(I-Q)\Vert ^2<1$ and thus, A is invertible in $\mathbb {B}(\mathcal {R}(P))$.

Now, suppose that $U\in \mathbb {B}(H)$ is a unitary which satisfies (34). Let

$$\begin{aligned} U_PUU_P^*=\left( \begin{array}{cc} U_{11} &{}\quad U_{12} \\ U_{21} &{}\quad U_{22} \\ \end{array} \right) , \end{aligned}$$

(35)

where $U_{i,j}\in \mathbb {B}(K_j, K_i)(i,j=1,2)$. In view of (34), we have

$$\begin{aligned} PU^*PUP=PUPU^*P=PQP, \quad UP=QU. \end{aligned}$$

Thus,

$$\begin{aligned}&U_P (PUP)^* U_P^* \cdot U_P(PUP) U_P^* =U_P(PUP)U_P^* \cdot U_P (PUP)^* U_P^* =U_PPQPU_P^*, \\&U_P U U_P^* \cdot U_PPU_P^*=U_PQU_P^*\cdot U_PUU_P^*. \end{aligned}$$

The equations above together with (29) –(30) and (35) yield

$$\begin{aligned}&\left( \begin{array}{cc} U_{11}^* &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) \left( \begin{array}{cc} U_{11} &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) = \left( \begin{array}{cc} U_{11} &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) \left( \begin{array}{cc} U_{11}^* &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) = \left( \begin{array}{cc} A &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) , \\&\left( \begin{array}{cc} U_{11} &{}\quad 0 \\ U_{21} &{}\quad 0 \\ \end{array} \right) = \left( \begin{array}{cc} * &{}\quad S_1 \\ * &{}\quad S_2 \\ \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned}&S_1=AU_{12}+A^{\frac{1}{2}}(I_1-A)^{\frac{1}{2}}U_0^*U_{22}, \\&S_2=U_0(I_1-A)^{\frac{1}{2}}A^{\frac{1}{2}}U_{12}+U_0(I_1-A)U_0^*U_{22}+Q_0U_{22}. \end{aligned}$$

It follows that

$$\begin{aligned} U_{11}^*U_{11}=U_{11}U_{11}^*=A,\quad S_1=0,\quad S_2=0. \end{aligned}$$

(36)

Since A is invertible, from $S_1=0$ we get

$$\begin{aligned} U_{12}=-A^{-\frac{1}{2}}(I_1-A)^{\frac{1}{2}}U_0^*U_{22}. \end{aligned}$$

(37)

In addition, from $S_2=0$ and $Q_0U_0=0$ we have

$$\begin{aligned} Q_0U_{22}=Q_0S_2=0. \end{aligned}$$

(38)

In virtue of $UU^*=I$ and $U^*U=I$, we obtain

$$\begin{aligned}&U_{11}U_{21}^*+U_{12}U_{22}^*=0, \quad U_{21}U_{21}^*+U_{22}U_{22}^*=I_2, \end{aligned}$$

(39)

$$\begin{aligned}&U_{12}^*U_{12}+U_{22}^*U_{22}=I_2. \end{aligned}$$

(40)

From the first equations in (36) and (39), we have $AU_{21}^*+U_{11}^*U_{12}U_{22}^*=0$, which yields

$$\begin{aligned} U_{21}^*=-A^{-1}U_{11}^*U_{12}U_{22}^*. \end{aligned}$$

Substituting the above equation into the second equation in (39) gives

$$\begin{aligned} U_{22}\big [U_{12}^*U_{11}A^{-2}U_{11}^*U_{12}+I_2\big ]U_{22}^*=I_2. \end{aligned}$$

Meanwhile, by (37) and (40) we can obtain

$$\begin{aligned} U_{22}^*\big [U_0(I_1-A)A^{-1}U_0^*+I_2\big ]U_{22}=I_2. \end{aligned}$$

Therefore, $U_{22}$ is invertible in $\mathbb {B}(K_2)$. Hence, $Q_0=0$ by (38). So according to Lemma 6, we have $\Vert P-Q\Vert <1$. $\square$

Remark 4

Let $P,Q\in \mathbb {B}(H)$ be projections such that $\Vert P(I_H-Q)\Vert <1$ and P is non-trivial. Let $U_1,U_2$ and U be defined by (14) and (15), respectively. Using the Halmos’ two projections theorem, the $6\times 6$ matrix representations for $U_1,U_2$ and U are given in the proof of Lemma 3. Based on (29) and (30), it is interesting to give the $2\times 2$ matrix representation for the operator U, which will be fulfilled by (42).

From (14), it is easy to verify that

$$\begin{aligned} U_PU_1U_P^*= \left( \begin{array}{cc} A &{}\quad -A^\frac{1}{2} (I_1-A)^\frac{1}{2} U_0^* \\ U_0 A^\frac{1}{2} (I_1-A)^\frac{1}{2} &{}\quad I_2-U_0U_0^*-Q_0+U_0 A U_0^* \\ \end{array} \right) , \end{aligned}$$

Let $\varphi$ be an arbitrary continuous function defined on $\sigma (A)$. Since $\mathcal {R}(U_0^*)=\overline{\mathcal {R}(A-A^2)}$, by [14, Lemma 2.4] we have

$$\begin{aligned} \overline{\mathcal {R}\big [\varphi (A)U_0^*\big ]}=\overline{\mathcal {R}\big [\varphi (A)(A-A^2)\big ]}=\overline{\mathcal {R}\big [(A-A^2)\varphi (A)\big ]} \subseteq \overline{\mathcal {R}(A-A^2)}. \end{aligned}$$

Therefore,

$$\begin{aligned} U_0^*U_0\cdot \varphi (A)U_0^*=\varphi (A)U_0^*,\quad U_0\big [\varphi (A)\big ]^*\cdot U_0^*U_0=U_0\big [\varphi (A)\big ]^*. \end{aligned}$$

(41)

It follows that

$$\begin{aligned}&U_PU_1^*U_1U_P^*= \left( \begin{array}{cc} A &{}\quad 0 \\ 0 &{}\quad I_2-U_0U_0^*-Q_0+U_0 AU_0^* \\ \end{array} \right) ,\\&U_P |U_1| U_{P}^*= \left( \begin{array}{cc} A^\frac{1}{2} &{}\quad 0 \\ 0 &{}\quad I_2-U_0U_0^*-Q_0+U_0 A^\frac{1}{2} U_0^* \\ \end{array} \right) . \end{aligned}$$

Due to the invertibility of A, the equation above indicates that $|U_1|$ is MP-invertible such that

$$\begin{aligned}U_P |U_1|^\dag U_P^*= \left( \begin{array}{cc} A^{-\frac{1}{2}} &{}\quad 0 \\ 0 &{}\quad I_2-U_0U_0^*-Q_0+U_0 A^{-\frac{1}{2}} U_0^* \\ \end{array} \right) . \end{aligned}$$

Utilizing (15) and (41) yields

$$\begin{aligned} U_P U_2 U_P^*&=U_PU_1U_P^* \cdot U_P|U_1|^\dag U_P^* \\&=\left( \begin{array}{cc} A^{\frac{1}{2}} &{}\quad -(I_1-A)^\frac{1}{2} U_0^*\\ U_0(I_1-A)^\frac{1}{2} &{}\quad I_2-U_0U_0^*-Q_0+U_0 A^{\frac{1}{2}} U_0^* \\ \end{array} \right) , \end{aligned}$$

which gives

$$\begin{aligned} U_PU_2U_2^*U_P^*=\left( \begin{array}{cc} I_1 &{}\quad 0 \\ 0 &{}\quad I_2-Q_0 \\ \end{array} \right) . \end{aligned}$$

Consequently,

$$\begin{aligned} \nonumber U_PUU_P^*&=U_PU_2U_P^*+ (I_1\oplus I_2)-U_PU_2U_2^*U_P^* \\&=\left( \begin{array}{cc} A^{\frac{1}{2}} &{}\quad -(I_1-A)^\frac{1}{2} U_0^*\\ U_0(I_1-A)^\frac{1}{2} &{}\quad I_2-U_0U_0^*+U_0 A^{\frac{1}{2}} U_0^* \\ \end{array} \right) . \end{aligned}$$

(42)

Remark 5

From (42), it is easy to construct an alternative path of invertible operators in $\mathbb {B}(H)$ which starts at I and ends at U. For instance, if we put

$$\begin{aligned} \widetilde{V}(t)=\left( \begin{array}{cc} A^{\frac{t}{2}} &{}\quad -(I_1-A^{t})^{\frac{1}{2}}U_0^* \\ U_0(I_1-A^t)^{\frac{1}{2}} &{}\quad I_2-U_0U_0^*+U_0A^{\frac{t}{2}}U_0^* \\ \end{array} \right) ,\quad t\in [0,1], \end{aligned}$$

then by (41),

$$\begin{aligned} \widetilde{V}(t)\cdot \big [\widetilde{V}(t)\big ]^*= \big [\widetilde{V}(t)\big ]^* \cdot \widetilde{V}(t)=\left( \begin{array}{cc} A^t+(I_1-A^t)^{\frac{1}{2}}U_0^*U_0(I_1-A^t)^{\frac{1}{2}} &{}\quad 0 \\ 0 &{}\quad I_2 \\ \end{array} \right) , \end{aligned}$$

which means that $\widetilde{V}(t)$ is invertible in $\mathbb {B}\big (\mathcal {R}(P)\oplus \mathcal {N}(P)\big )$, since

$$\begin{aligned} A^t+(I_1-A^t)^{\frac{1}{2}}U_0^*U_0(I_1-A^t)^{\frac{1}{2}}\ge A^t >0. \end{aligned}$$

Hence, if we put $V(t)=U_P^*\widetilde{V}(t)U_P$, then V(t) is a norm-continuous path of invertible operators in $\mathbb {B}(H)$ which starts at I and ends at U. However, these operators $V(t)(t\in [0,1])$ may fail to be in $C^*\{I_H,P,Q\}$.

Theorem 3

Let p and q be projections in a unital $C^*$-algebra $\mathfrak {A}$ such that $\Vert p(I-q)\Vert <1$. Then, the following statements are equivalent:

• (i) $\Vert p-q\Vert <1$.

• (ii) There exists a unitary $u\in C^*\{I, p,q\}$ such that

  $$\begin{aligned} pup \text{ is } \text{ normal } \text{ and } q=upu^*. \end{aligned}$$

  (43)

Proof

Choose any faithful unital representation $(\pi ,H)$ of $\mathfrak {A}$. Let $P=\pi (p)$ and $Q=\pi (q)$. Then, $\Vert P(I_H-Q)\Vert <1$. Hence, the desired conclusion can be derived directly from items (i) and (iii) in Theorem 2, together with (26) and the faithfulness of $\pi$. $\square$

It is remarkable that the two conditions stated in (34) are independent. Actually, there exist two projections P and Q on certain Hilbert space H such that $\Vert P-Q\Vert =1$ and $Q=UPU^*$ for some unitary $U\in \mathbb {B}(H)$. We give such an example as follows.

Example 2

Let $\ell ^2(\mathbb {N})$ be the Hilbert space consisting of all complex-valued sequences $x=(x_1, x_2,\ldots , x_n,\ldots )^T$ such that $\Vert x\Vert ^2=\sum \limits _{i=1}^\infty |x_i|^2< \infty$, and $\{e_n:n\in \mathbb {N}\}$ be the usual orthonormal basis for $\ell ^2(\mathbb {N})$. Let $U_0$ be the isometry in $\mathbb {B}(\ell ^2(\mathbb {N}))$ defined by

$$\begin{aligned} U_0(x)= (0,0,x_1,x_2,\ldots , x_n,\ldots )^T \end{aligned}$$

for $x\in \ell ^2(\mathbb {N})$, which can be characterized as

$$\begin{aligned} U_0(e_n)=e_{n+2}(n\in \mathbb {N}),\ U_0^*(e_1)=U_0^*(e_2)=0, \ U_0^*(e_n)=e_{n-2}(n\ge 3). \end{aligned}$$

(44)

The projections from $\ell ^2(\mathbb {N})$ onto $\text{ span }\{e_1\}$ and $\text{ span }\{e_1,e_2\}$ are denoted by $P_1$ and $P_{1,2}$, respectively. Let I stand for the identity operator on $\ell ^2(\mathbb {N})$. With the notation as above, we have

$$\begin{aligned} U_0^*U_0=I \quad \text{ and }\quad U_0U_0^*=I-P_{1,2}. \end{aligned}$$

Put $H=\ell ^2(\mathbb {N})\oplus \ell ^2(\mathbb {N})$, and choose projections $P,Q\in \mathbb {B}(H)$ to be

$$\begin{aligned} P=\left( \begin{array}{cc} I &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) ,\quad Q=\left( \begin{array}{cc} \frac{1}{4}I &{}\quad \frac{\sqrt{3}}{4}U_0^* \\ \frac{\sqrt{3}}{4}U_0 &{}\quad \frac{3}{4}U_0U_0^*+P_1 \\ \end{array} \right) . \end{aligned}$$

(45)

The well-known Krein–Krasnoselskii–Milman equality (see (2)) indicates that $\Vert P-Q\Vert \le 1$. On the other hand, we have

$$\begin{aligned} \Vert P-Q\Vert \ge \Vert (0\oplus P_1)(P-Q)(0\oplus P_1)\Vert =\Vert 0\oplus P_1\Vert =\Vert P_1\Vert =1. \end{aligned}$$

This shows that $\Vert P-Q\Vert =1$.

Let $U\in \mathbb {B}(H)$ be given by

$$\begin{aligned} U=\left( \begin{array}{cc} \frac{1}{\sqrt{3}}U_0^*U_{21} &{}\quad -\sqrt{3}U_0^*U_{22} \\ U_{21} &{}\quad U_{22} \\ \end{array} \right) , \end{aligned}$$

(46)

where $U_{21}, U_{22}\in \mathbb {B}\big (\ell ^2(\mathbb {N})\big )$ are characterized by

$$\begin{aligned}&U_{21}(e_n)= \left\{ \begin{array}{ll} \frac{\sqrt{3}}{2} e_3, &{} n=1, \\ e_1, &{} n=2, \\ \frac{\sqrt{3}}{2} e_{n+1}, &{} n\ge 3, \end{array} \right. \quad U_{22}(e_n)= \left\{ \begin{array}{ll} e_2, &{} n=1, \\ \frac{1}{2}e_{n+1}, &{} n\ge 2. \end{array} \right. \end{aligned}$$

Then, we have

$$\begin{aligned}&U_{21}^*(e_n)=\left\{ \begin{array}{ll} e_2, &{} n=1, \\ 0, &{} n=2, \\ \frac{\sqrt{3}}{2}e_1, &{} n=3, \\ \frac{\sqrt{3}}{2}e_{n-1}, &{} n\ge 4, \end{array} \right. \quad U_{22}^*(e_n)=\left\{ \begin{array}{ll} 0, &{} n=1, \\ e_1, &{} n=2, \\ \frac{1}{2}e_{n-1}, &{} n\ge 3. \end{array} \right. \end{aligned}$$

It follows that

$$\begin{aligned}&U_{21}^*\big [1/3 \cdot U_0U_0^*+I\big ]U_{21}=I,\quad U_{21}^*\big [I-U_0U_0^*)\big ]U_{22}=0, \end{aligned}$$

(47)

$$\begin{aligned}&U_{22}^*\big [3U_0U_0^*+I\big ]U_{22}=I, \quad U_0^*\big [1/3\cdot U_{21}U_{21}^*+3U_{22}U_{22}^*\big ]U_0=I, \end{aligned}$$

(48)

$$\begin{aligned}&U_0^*\big [1/\sqrt{3}\cdot U_{21}U_{21}^*-\sqrt{3}U_{22}U_{22}^*\big ]=0,\quad U_{21}U_{21}^*+U_{22}U_{22}^*=I, \end{aligned}$$

(49)

$$\begin{aligned}&\big [I-U_0U_0^*-P_1\big ]U_{21}=0,\quad P_1U_{22}=0. \end{aligned}$$

(50)

In view of (46) and (47)–(49), we have $U^*U=UU^*=I$. Therefore, U is a unitary. Furthermore, from (45), (46) and (50) it is easy to verify that $UP=QU$. So, the conclusion is derived.