Abstract
An open problem posed by G. Maksa and Z. Páles is to find the general solution of the functional equation
where \(f: H \rightarrow K\) is between two complex normed spaces and \({\mathbb {T}}_n:=\{e^{i\frac{2k\pi }{n}}: k=1, \cdots ,n\}\) is the set of the nth roots of unity. With the aid of the celebrated Wigner’s unitary-antiunitary theorem, we show that if \(n\ge 3\) and H and K are complex inner product spaces, then f satisfies the above equation if and only if there exists a phase function \(\sigma : H\rightarrow {\mathbb {T}}_n\) such that \(\sigma \cdot f\) is a linear or anti-linear isometry. Moreover, if the solution f is continuous, then f is a linear or anti-linear isometry.
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1 Introduction
Let \({\mathbb {R}}\) be the set of real numbers, \({\mathbb {C}}\) be the set of complex numbers, and \({\mathbb {T}}\) be the unit circle in \({\mathbb {C}}\). Let n be a fixed positive integer and \({\mathbb {T}}_n\) denote the finite cyclic subgroup of \({\mathbb {T}}\) consisting of the nth roots of unity, i.e., \({\mathbb {T}}_n:=\{e^{i\frac{2k\pi }{n}}: k=1, \cdots ,n\}\).
The celebrated Wigner’s theorem plays a fundamental role in the foundations of quantum mechanics and in representation theory in physics. It states that any quantum mechanical symmetry transformation can be represented by a unitary or an anti-unitary operator on a complex inner product space. In mathematical language, the result can be reformulated in the following way. For complex inner product spaces H and K, a mapping \(f: H \rightarrow K\) satisfies
if and only if there exist a function (called phase function) \(\sigma : H \rightarrow {\mathbb {T}}\) and a linear or anti-linear (i.e. conjugate linear) isometry \(U:H \rightarrow K\) such that
We then say that f is phase equivalent to a linear or anti-linear isometry. There are several proofs of this result, see [1, 4, 6, 7, 9, 18, 22] to list just some of them. For further generalizations of this fundamental result, we mention the papers [3, 5, 8, 16, 17, 19].
The real version of Wigner’s theorem was also obtained by Rätz [18, Corollary 8 (a)] and by Turnšek [22, Theorem 2.4 (i)] as follows. A mapping \(f: H \rightarrow K\) between real inner product spaces satisfies (1.1) if and only if there exist a function \(\sigma : H \rightarrow \{-1,1\}\) and a real linear isometry \(U:H \rightarrow K\) such that (1.2) holds. We then say that f is phase equivalent to a real linear isometry. By an easy argument, Maksa and Páles [15, Theorem 2] obtained that if H and K are real inner product spaces then \(f: H \rightarrow K\) satisfies the functional equation
if and only if f satisfies (1.1). Combining the two results above, Maksa and Páles [15, Theorem 2] then proved that the general solution of the functional Eq. (1.3) is phase equivalent to a real linear isometry.
Maksa and Páles [15] also at last formulated two open problems as follows.
Problem 1.1
[15] Under what conditions, when H and K are real normed but not necessarily inner product spaces, the solutions of (1.3) are phase equivalent to a real linear isometry?
Problem 1.2
[15] Let H and K be complex normed spaces and n be a fixed positive integer. Under what conditions the solutions \(f: H \rightarrow K\) of the following generalization of (1.3):
have the form (1.2) for some phase function \(\sigma : H \rightarrow {\mathbb {T}}_n\) and some linear or anti-linear isometry \(U: H\rightarrow K\)?
There are several recent papers dealing with Problem 1.1, see [10,11,12,13,14, 20, 21, 23]. Recently, Ilišević, Turnšek, etc. gave a positive answer to Problem 1.1. Namely, they proved that the solutions of Eq. (1.3) for a mapping \(f: H \rightarrow K\), where H and K are real normed spaces, are phase equivalent to a real linear isometry under the condition that either f is surjective [12] or K is strictly convex [13]. Huang and Tan in [10] introduced a weaker version of Eq. (1.3) as follows:
They also proved that every surjective mapping satisfying this equation between two real normed spaces is phase equivalent to a real linear isometry.
In this paper, we give a positive answer to Problem 1.2 for complex inner product spaces. Let H and K be two complex inner product spaces and f be a mapping \(f: H \rightarrow K\). We will show in Theorem 2.5 that for \(n\ge 3\), a mapping f satisfies (1.4) if and only if there is a phase function \(\sigma : H \rightarrow {\mathbb {T}}_n\) such that \(\sigma \cdot f\) is a linear or anti-linear isometry. Moreover, we will show in Corollary 2.7 that for \(n\ge 3\), a continuous mapping f satisfies (1.4) if and only if f is a linear or anti-linear isometry.
2 Results
Throughout this section, let H and K be complex inner product spaces, and n be a fixed positive integer.
First we present a simple and useful lemma. For convenience, we introduce two equations
and
Lemma 2.1
For \(n\ge 2\) and a mapping \(f: H \rightarrow K\), the following statements are equivalent:
Moreover, if this is the case, then f is norm preserving, i.e., \(\Vert f(x)\Vert =\Vert x\Vert \) for every \(x\in H\).
Proof
Putting \(y=x\), we deduce from each of Eqs. (1.4), (2.1) and (2.2) that f is norm preserving. For \(x,y\in H\) and \(\beta \in {\mathbb {T}}_n\), we obtain
and
By the norm-preserving property and the fact that \(\beta \in {\mathbb {T}}_n\) if and only if \({\overline{\beta }}\in {\mathbb {T}}_n\), each of Eqs. (1.4) and (2.1) is equivalent to Eq. (2.2). The proof is complete. \(\square \)
Remark 2.2
In the case \(n=1\), Eqs. (1.4), (2.1) and (2.2) turn out to be
and
respectively. By definition, f satisfies Eq. (2.3) if and only if f is an isometry. Since an inner product space is strictly convex, Baker [2] showed that \(f-f(0)\) is a real linear isometry if f is an isometry. However, an easy argument [15, Theorem 1] shows that Eqs. (2.4) and (2.5) are equivalent to each other and each of them is equivalent to f being a real linear isometry. Hence, in general, Lemma 2.1 does not hold for \(n=1\).
In the case \(n=2\), each of Eqs. (1.4) and (2.1) is the same as Eq. (1.3), and Eq. (2.2) becomes
Note that every complex linear space is obviously a real linear space and if \(\langle \cdot ,\cdot \rangle \) is a complex inner product on H, then \(\text{ Re }\langle \cdot ,\cdot \rangle \) is a real inner product on H which induces the same norm. Therefore, from the real version of Wigner’s theorem, the following theorem follows immediately. See [15, Theorem 2] or [22, Theorem 2.4 (i)] for its proof.
Theorem 2.3
([15, 22]) For a mapping \(f: H \rightarrow K\), the following statements are equivalent:
-
(i)
f satisfies Eq. (1.3);
-
(ii)
f satisfies Eq. (2.6);
-
(iii)
there exist a phase function \(\sigma : H \rightarrow \{-1,1\}\) and a real linear isometry \(U:H \rightarrow K\) such that \(f=\sigma \cdot U\).
It may be worth noting that in (iii) of Theorem 2.3, the phrase “a real linear isometry” cannot be replaced by “a linear or anti-linear isometry”. To show this, we construct an example as follows.
Example
Let \(H={\mathbb {C}}\) and \(K={\mathbb {C}}^2\), both equipped with the usual inner products, i.e.,
and
Define \(f:{\mathbb {C}}\rightarrow {\mathbb {C}}^2\) by
An easy calculation shows that f satisfies Eq. (2.6), but for any phase function \(\sigma : H \rightarrow \{-1, 1\}\), \(\sigma \cdot f\) is neither linear nor anti-linear.
However, in the case \(n\ge 3\), in sharp contrast with the cases \(n=1,2\), each of (1.4), (2.1) and (2.2) is equivalent to there being a phase function \(\sigma : H\rightarrow {\mathbb {T}}_n\) such that \(\sigma \cdot f\) is a linear or anti-linear isometry. This is the content of Theorem 2.5. To obtain it, we need an elementary lemma as follows.
Lemma 2.4
For \(n\ge 3\) and \(s,t\in {\mathbb {C}}\), the equation
holds if and only if \(s=\beta _{0}t\) or \(s=\beta _{0}{\overline{t}}\) for some \(\beta _0\in {\mathbb {T}}_n\).
Proof
Sufficiency. If \(s=\beta _{0}t\) for some \(\beta _0\in {\mathbb {T}}_n\), then
where the second equality follows from the fact that \(\beta _0{\mathbb {T}}_n={\mathbb {T}}_n\). If \(s=\beta _{0}{\overline{t}}\), then
where the fourth equality follows from the fact that \(\beta \in {\mathbb {T}}_n\) if and only if \({\overline{\beta }}\in {\mathbb {T}}_n\).
Necessity. Suppose that Eq. (2.7) holds. We first claim that \(|s|=|t|\). Indeed, taking \(t=|t|e^{i\theta }\) and \(\beta _k=e^{i\frac{2k\pi }{n}}\in {\mathbb {T}}_n\) for \(k=1, \cdots ,n\), we see that
Let us recall the formula of partial sums of trigonometric series
for \(u\in (0,2\pi )\). It follows that
Similarly,
Therefore, Eq. (2.7) implies that \(|s|=|t|\) and the claim is proved. Furthermore, let us take \(s=|t|e^{i\theta _1}\) and \(t=|t|e^{i\theta }\). It follows from Eq. (2.7) that
Choose \(k_0\in \{1,\cdots , n\}\) such that \(\cos \theta _1=\cos (\theta +\frac{2k_0\pi }{n})\), or equivalently,
for some integer m. Therefore, we have \(s=\beta _{k_0} t\) or \(s=\overline{\beta _{k_0}}{\overline{t}}\). \(\square \)
Now we present our main theorem. For convenience, we introduce the equation
Theorem 2.5
For \(n\ge 3\) and a mapping \(f: H \rightarrow K\), the following statements are equivalent:
-
(i)
f satisfies Eq. (1.4);
-
(ii)
f satisfies Eq. (2.1);
-
(iii)
f satisfies Eq. (2.2);
-
(iv)
f satisfies Eq. (2.8);
-
(v)
there exist a phase function \(\sigma : H\rightarrow {\mathbb {T}}_n\) and a linear or anti-linear isometry \(U:H \rightarrow K\) such that \(f=\sigma \cdot U\).
Proof
The equivalences (i) \(\Longleftrightarrow \) (ii) \(\Longleftrightarrow \) (iii) and (iii) \(\Longleftrightarrow \) (iv) are contained in Lemmas 2.1 and 2.4, respectively. We will complete the proof by establishing the implications (v) \(\Longrightarrow \) (i) and (iv) \(\Longrightarrow \) (v).
(v) \(\Longrightarrow \) (i). Suppose that \(f=\sigma \cdot U\) for some phase function \(\sigma : H\rightarrow {\mathbb {T}}_n\) and some linear isometry U (respectively, some anti-linear isometry U). For \(x,y\in H\) and \(\beta \in {\mathbb {T}}_n\),
(respectively,
Note that \(\beta {\mathbb {T}}_n={\mathbb {T}}_n\) and \({\overline{\beta }}\in {\mathbb {T}}_n\) for every \(\beta \in {\mathbb {T}}_n\). It follows that Eq. (1.4) holds.
(iv) \(\Longrightarrow \) (v). Suppose that Eq. (2.8) holds. Therefore Eq. (1.1) holds. By Wigner’s theorem, there exist a phase function \(\sigma _1: H \rightarrow {\mathbb {T}}\) and a linear or anti-linear isometry \(U_1:H \rightarrow K\) such that \(f=\sigma _1\cdot U_1\). Next, we will divide our proof into two cases: \(\dim H=1\) and \(\dim H\ge 2\).
-
(I)
The case that \(\dim H=1\). In this case, we have that \(f(H)\subset U_1(H)\) and \(\dim U_1(H)=1\). After a suitable identification, the mapping f can be regarded as a mapping \(f:{\mathbb {C}}\rightarrow {\mathbb {C}}\). Set
$$\begin{aligned} A&:=\left\{ x\in {\mathbb {C}}\backslash \{0\}:\frac{x}{{\overline{x}}} \in {\mathbb {T}}_n\right\} ,\\ B&:=\left\{ x\in {\mathbb {C}}\backslash \{0\}:\frac{x}{{\overline{x}}} \notin {\mathbb {T}}_n \text{ and } f(x)=\beta (x,1)f(1)x\right\} , \end{aligned}$$and
$$\begin{aligned} C:=\left\{ x\in {\mathbb {C}}\backslash \{0\}:\frac{x}{{\overline{x}}} \notin {\mathbb {T}}_n \text{ and } f(x)=\beta (x,1)f(1){\overline{x}}\right\} . \end{aligned}$$Obviously, the sets A, B and C are disjoint. By Eq. (2.8), we have that \(A\cup B\cup C={\mathbb {C}}\backslash \{0\}\). We claim that \(B=\emptyset \) or \(C=\emptyset \). Suppose, on contrary that \(x\in B\) and \(y\in C\). Then
$$\begin{aligned} f(x)=\beta (x,1)f(1)x \quad \text{ and }\quad f(y)=\beta (y,1)f(1){\overline{y}}. \end{aligned}$$Substituting these into Eq. (2.8), we obtain that
$$\begin{aligned} \beta (x,1)\overline{\beta (y,1)}xy=\beta (x,y)x{\overline{y}} \quad \text{ or }\quad \beta (x,1)\overline{\beta (y,1)}xy=\beta (x,y){\overline{x}}y, \end{aligned}$$which means that
$$\begin{aligned} \frac{y}{{\overline{y}}}=\frac{\beta (x,y)}{\beta (x,1)\overline{\beta (y,1)}}\in {\mathbb {T}}_n \quad \text{ or }\quad \frac{x}{{\overline{x}}}=\frac{\beta (x,y)}{\beta (x,1)\overline{\beta (y,1)}}\in {\mathbb {T}}_n. \end{aligned}$$However, the fact \(x\in B\) and \(y\in C\) implies that
$$\begin{aligned} \frac{x}{{\overline{x}}}\notin {\mathbb {T}}_n \quad \text{ and }\quad \frac{y}{{\overline{y}}}\notin {\mathbb {T}}_n. \end{aligned}$$This contradiction proves the claim. If \(C=\emptyset \), then we set
$$\begin{aligned} A_1:=\left\{ x\in {\mathbb {C}}\backslash {\mathbb {R}}:\frac{x}{{\overline{x}}}\in {\mathbb {T}}_n \text{ and } f(x)=\beta (x,1)f(1){\overline{x}}\right\} . \end{aligned}$$If \(x\in A_1\), then
$$\begin{aligned} \frac{{\overline{x}}}{x}\in {\mathbb {T}}_n \quad \text{ and }\quad f(x)=\beta (x,1)f(1){\overline{x}}=\beta (x,1)\frac{{\overline{x}}}{x}f(1)x. \end{aligned}$$If \(x\in A\backslash A_1\) or \(x\in B\), then Eq. (2.8) implies that
$$\begin{aligned} f(x)=\beta (x,1)f(1)x. \end{aligned}$$Define
$$\begin{aligned} \sigma (x)={\left\{ \begin{array}{ll} \beta (x,1)\frac{{\overline{x}}}{x}, &{} \quad x\in A_1\\ \beta (x,1), &{} \quad x\in (A\backslash A_1)\cup B\\ 1, &{} \quad x=0 \end{array}\right. } \end{aligned}$$and \(U(x)=f(1)x,\ x\in {\mathbb {C}}\). Then \(\sigma ({\mathbb {C}})\subset {\mathbb {T}}_n\), U is a linear isometry and \(f=\sigma \cdot U\). If \(B=\emptyset \), a similar argument shows that there exist a phase function \(\sigma : {\mathbb {C}}\rightarrow {\mathbb {T}}_n\) and an anti-linear isometry \(U:{\mathbb {C}} \rightarrow {\mathbb {C}}\) such that \(f=\sigma \cdot U\). This completes the proof of the case that \(\dim H=1\).
-
(II)
The case that \(\dim H\ge 2\). In this case, fix a unit vector \(e\in H\). Define
$$\begin{aligned} \sigma (x)={\left\{ \begin{array}{ll} \overline{\sigma _1(e)}\sigma _1(x), &{} \quad x\in H\backslash \{0\}\\ 1, &{} \quad x=0 \end{array}\right. } \end{aligned}$$and \(U=\sigma _1(e)U_1\). Then \(\sigma : H \rightarrow {\mathbb {T}}\) is a phase function, \(U:H \rightarrow K\) is a linear or anti-linear isometry and \(f=\sigma \cdot U\). It suffices to show that \(\sigma (H\backslash \{0\})\subset {\mathbb {T}}_n\). From Eq. (2.8) it follows that if \(\langle x, y \rangle \in {\mathbb {R}}\backslash \{0\}\) then
$$\begin{aligned} \beta (x,y)\langle x, y \rangle =\langle f(x), f(y) \rangle =\sigma (x)\overline{\sigma (y)}\langle x, y \rangle , \end{aligned}$$which implies that \(\sigma (x)\overline{\sigma (y)}=\beta (x,y)\in {\mathbb {T}}_n\). This is equivalent to saying that \(\sigma (x)\in {\mathbb {T}}_n\) if and only if \(\sigma (y)\in {\mathbb {T}}_n\) whenever \(\langle x, y \rangle \in {\mathbb {R}}\backslash \{0\}\). Let \(Z=({\mathbb {C}}e)^{\perp }=\{z\in H: \langle z, e\rangle =0\}\) and choose arbitrarily \(z\in Z\backslash \{0\}\). Since \(\sigma (e)=1\), the equations \(\langle e, e+z \rangle =1\) and \(\langle z, e+z \rangle =\Vert z\Vert ^2\) imply that \(\sigma (e+z)\in {\mathbb {T}}_n\) and hence \(\sigma (z)\in {\mathbb {T}}_n\). Moreover, for every \(t\in {\mathbb {C}}\backslash \{0\}\), the equation \(\langle z, te+z \rangle =\Vert z\Vert ^2\) implies that \(\sigma (te+z)\in {\mathbb {T}}_n\), and then the equation \(\langle te, te+z \rangle =|t|^2\) implies that \(\sigma (te)\in {\mathbb {T}}_n\). This completes the proof of the case that \(\dim H\ge 2\).
The proof is complete. \(\square \)
Remark 2.6
Note that in (v) of Theorem 2.5, the phrase “a linear or anti-linear isometry” cannot be replaced either by “a linear isometry” or by “an anti-linear isometry”. For example, the mappings \(f_1,f_2:{\mathbb {C}}\rightarrow {\mathbb {C}}\) defined by \(f_1(x)={\overline{x}}\) and \(f_2(x)=x\) both satisfy Eq. (2.8). However, it is easy to check that for any phase function \(\sigma : H \rightarrow {\mathbb {T}}_n\) where \(n\ge 3\), \(\sigma \cdot f_1\) cannot be linear and \(\sigma \cdot f_2\) cannot be anti-linear.
The following corollary describes the continuous solutions of (1.4).
Corollary 2.7
For \(n\ge 3\) and a continuous mapping \(f: H \rightarrow K\), each of Eqs. (1.4), (2.1), (2.2) and (2.8) holds if and only if f is a linear or anti-linear isometry.
Proof
By Theorem 2.5, we need only to prove that Eq. (1.4) implies that f is a linear or anti-linear isometry. Assume that f is a continuous mapping satisfying Eq. (1.4). Then there exists a phase function \(\sigma : H\rightarrow {\mathbb {T}}_n\) such that \(\sigma \cdot f\) is a linear or anti-linear isometry. Thus,
or
If \(y\ne 0\), then there exists an open ball \(V_y\) centered at y such that
or
This, by the continuity of f, shows that \(\sigma \) is continuous on \(V_y\). Since the phase function \(\sigma : H\rightarrow {\mathbb {T}}_n\) takes only finite values, we have that \(\sigma \) is constant on \(V_y\). By the fact that \(H\backslash \{0\}\) is connected, \(\sigma \) is constant on \(H\backslash \{0\}\), f must be a linear or anti-linear isometry. \(\square \)
Remark 2.8
The proof of Corollary 2.7 is similar to that of [15, Corollary 3], where the case \(n=2\) was considered as follows. For a continuous mapping \(f: H \rightarrow K\), each of Eqs. (1.3) and (2.6) holds if and only if f is a real linear isometry.
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The first and second authors were supported by the Natural Science Foundation of Tianjin City (Grant No. 22JCYBJC00420) and the National Natural Science Foundation of China (Grant No. 12271402). The third author was supported by the National Natural Science Foundation of China (Grant Nos. 12201459 and 12071358) and is the corresponding author.
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Huang, X., Zhang, L. & Wang, S. A functional equation related to Wigner’s theorem. Aequat. Math. 98, 885–894 (2024). https://doi.org/10.1007/s00010-024-01042-8
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DOI: https://doi.org/10.1007/s00010-024-01042-8