1 Introduction

It is an easy consequence of the polarisation identity that unitary maps between Hilbert spaces, that is, maps preserving the inner product, are automatically linear. Since inner-product spaces are characterised by the inextricable connection between the norm and the inner product, the aforementioned fact does not have a canonical interpretation in the non-Hilbertian setting. Nonetheless, natural approaches to extending Uhlhorn’s version of Wigner’s theorem on symmetry transformations [5] are available in the Banach-space setting, for example, in terms of Birkhoff–James orthogonality [1] or semi-inner products [3]. In the present paper we focus on the latter approach.

Lumer [4] and Giles [2] proved that reminiscences of inner products are available in arbitrary normed spaces as for every normed space X one may find a pairing \([\,\cdot \, |\, \cdot \,]\) thereon (a semi-inner product) that assumes scalar values, is linear in the first variable, anti-homogeneous in the second variable, and the following form of the Cauchy–Schwarz inequality holds:

$$\begin{aligned} \big | [x|y] \big | \leqslant \Vert x\Vert \cdot \Vert y\Vert \quad (x,y\in X) \end{aligned}$$

with \([x|x] = \Vert x\Vert ^2\). In particular, for each \(w\in X\), \([\,\cdot \,|w]\in X^*\).

Semi-inner products are, in general, non-unique: a normed space X has a unique semi-inner product if and only if it is smooth, that is every non-zero vector \(x\in X\) admits a unique norming functional, that is, a norm-one functional \(\varphi _x\in X^*\) such that \(\langle \varphi _x, x \rangle = \Vert x\Vert \). In the case where X is smooth, we have

$$\begin{aligned}{}[x|y] = \Vert y\Vert \cdot \langle \varphi _y, x\rangle \quad (x,y\in X). \end{aligned}$$
(1)

We observed in [6, Theorem 7] that if X is a non-Hilbertian finite-dimensional space with \(\dim X\geqslant 3\) that is smooth, then there exists a space V of dimension \(\dim X-1\) and a non-linear map \(f:V\rightarrow X\) that preserves semi-inner products. The map f may even be discontinuous.

The first result demonstrates that for X and Y having equal finite dimensions, without any additional hypotheses, a semi-inner product preserving function between X and Y must be a linear isometry.

Theorem 1

Let X and Y be normed spaces with fixed semi-inner products \([\,\cdot \,|\,\cdot \,]_X\) and \([\,\cdot \,|\,\cdot \,]_Y\), respectively. Suppose that \(f:X\rightarrow Y\) is a function such that

$$\begin{aligned} \left[ f(x) | f(y)\right] _Y = [x | y]_X\quad (x, y\in X). \end{aligned}$$
(2)

If either

  1. (a)

    X and Y have the same finite dimension, or

  2. (b)

    X has a Schauder basis \((e_i)\) and \((f(e_i))\) is a Schauder basis of Y,

then f is a linear isometry.

The proof will be presented in the subsequent section. We highlight Theorem 1 as clause (a) appears to be optimal in the case where no further assumptions on f are imposed in the light of the following blatant counterexample of an analogous statement in infinite dimensions.

Theorem 2

There exists a uniformly smooth renorming X of the Hilbert space \(\ell _2\) and a non-linear injection \(f:X\rightarrow X\) such that

$$\begin{aligned} \left[ f(x) | f(y)\right] = [x | y]\quad (x, y\in X). \end{aligned}$$
(3)

Moreover, f may be chosen to be either continuous or discontinuous.

Since X is smooth, the choice of the semi-inner product is unambiguous. Regrettably, Theorem 2 refutes a side result from a recent paper by Ilišević and Turnšek [3, Proposition 2.4(ii)], where it was claimed that if X is a smooth Banach space and \(f:X\rightarrow X\) is a (possibly non-surjective) map satisfying (3), then f is necessarily a linear isometry. Their proof contains a flaw as explained in Remark 6. However, the main results in [3] are dealing with surjective functions between smooth normed spaces which satisfy the Wigner equation. It can be easily verified (see [3, Proposition 2.4(i)]) that, even in arbitrary normed spaces, the additional assumption of surjectivity forces semi-inner product preserving functions to be linear.

Our notation and terminology are standard. We consider normed spaces over the field \({\mathbb {K}}\) of real or complex numbers. A normed space X is strictly convex if the unit sphere of X does not contain non-trivial line segments. We denote by \(\langle \cdot , \cdot \rangle \) the duality pairing between a normed space X and its dual \(X^*\). When X is an inner-product space, we denote by \(\langle \cdot | \cdot \rangle \) the underlying inner product. A normed space is uniformly smooth, if for every \(\varepsilon >0\) there exists \(\delta >0\) such that if \(x,y \in X\) are vectors such that \(\Vert x\Vert =1\) and \(\Vert y\Vert \leqslant \delta \) then \(\Vert x+y\Vert +\Vert x-y\Vert \leqslant 2 + \varepsilon \Vert y\Vert .\) Uniformly smooth spaces are, in particular, smooth. For the sake of completeness, we record the following simple property of smooth spaces.

Lemma 3

Let X be a smooth normed space. If \(u,w\in X\) are non-zero vectors and \(\Vert u+w\Vert =\Vert u\Vert +\Vert w\Vert \), then \(\varphi _u=\varphi _w\).

Proof

Let \(u,w\in X\) be vectors with \(\Vert u+w\Vert =\Vert u\Vert +\Vert w\Vert \). It is clear that \(\Vert u+w\Vert =\varphi _{u+w}(u+w)\). So, it follows that

$$\begin{aligned} \Vert u\Vert +\Vert w\Vert= & {} \Vert u+w\Vert =\varphi _{u+w}(u+w)=\varphi _{u+w}(u)+\varphi _{u+w}(w)\\\leqslant & {} |\varphi _{u+w}(u)+\varphi _{u+w}(w)|\leqslant |\varphi _{u+w}(u)|+|\varphi _{u+w}(w)|\\\leqslant & {} \Vert u\Vert +\Vert w\Vert . \end{aligned}$$

Thus \(\varphi _{u+w}(u)+\varphi _{u+w}(w)=\Vert u\Vert +\Vert w\Vert \). Moreover, we know that \(|\varphi _{u+w}(u)|\leqslant \Vert u\Vert \) and \(|\varphi _{u+w}(w)|\leqslant \Vert w\Vert \). This clearly forces \(\varphi _{u+w}(u)=\Vert u\Vert \) and \(\varphi _{u+w}(w)=\Vert w\Vert \). On the other hand, both \(\varphi _{u}(u)=\Vert u\Vert \) and \(\varphi _{w}(w)=\Vert w\Vert \) hold. Since X is smooth, we get \(\varphi _{u+w}=\varphi _{u}\) and \(\varphi _{u+w}=\varphi _{w}\). Hence \(\varphi _u=\varphi _w\). \(\square \)

2 Proofs of Theorems 1 and 2

We start by proving Theorem 1. For the sake of brevity, we shall use the symbol \(\left[ \,\cdot \,|\,\cdot \,\right] \) for (fixed) semi-inner products both in X and Y, hoping it will not lead to unnecessary confusion. In order to prove the theorem, it suffices to show that f is linear.

Proof of Theorem 1

We will prove clause (b) first. Suppose that \((e_i)\) is a Schauder basis of X and \((f(e_i))\) is a Schauder basis of Y. The proof of the case where X is finite-dimensional (so that a Schauder basis is just an ordinary algebraic basis) is mutatis mutandis the same, so we will keep writing infinite series bearing in mind that the proof works equally well for the finite-dimensional case with \(\infty \) replaced by \(\dim X\)).

We will show that for any scalars \(\beta _1, \beta _2, \ldots \)

$$\begin{aligned} f\left( \sum _{i=1}^\infty \beta _i e_i\right) = \sum _{i=1}^\infty \beta _i f(e_i) \end{aligned}$$

as long as the series \(\sum _{i=1}^\infty \beta _i e_i\) converges in X.

Fix \(x\in X\). Since \((f(e_i))_{i=1}^\infty \) is a basis, there are uniquely determined scalars \(\beta _1, \beta _2, \ldots \in {\mathbb {K}}\) such that

$$\begin{aligned} f(x)=\sum _{i=1}^\infty \beta _if(e_i). \end{aligned}$$

Let \(x_m = \sum _{i=1}^m\beta _ie_i.\) It is enough to show that \(x_m\rightarrow x\) as \(m\rightarrow \infty \). Let us define the numbers \(\varepsilon _m:=\left\| f(x)-\sum _{i=1}^m\beta _if(e_i)\right\| \). Clearly, \(\varepsilon _m\rightarrow 0\) as \(m\rightarrow \infty \). For every unit vector \(u\in X\), we have \(1= \Vert u\Vert =\Vert f(u)\Vert \). Thus, for every m we have

$$\begin{aligned} \Bigg | \left[ f(x)-\sum \limits _{i=1}^m\beta _if(e_i)\Big |f(u)\right] \Bigg |\leqslant \Big \Vert f(x)-\sum \limits _{i=1}^m\beta _if(e_i)\Big \Vert \cdot \Vert f(u)\Vert \leqslant \varepsilon _m\cdot 1. \end{aligned}$$

Using the linearity of semi-inner products in the first variable, it follows that

$$\begin{aligned} \Bigg | \left[ f(x)|f(u)\right] -\left[ \sum \limits _{i=1}^m\beta _if(e_i)\Big |f(u)\right] \Bigg | = \Bigg | \left[ f(x)|f(u)\right] -\sum \limits _{i=1}^m\beta _i\left[ f(e_i)|f(u)\right] \Bigg |\leqslant \varepsilon _m. \end{aligned}$$

Combining the above inequality with (2) yields

$$\begin{aligned} \Bigg | \left[ x|u\right] -\sum \limits _{i=1}^m\beta _i\left[ e_i|u\right] \Bigg |\leqslant \varepsilon _m. \end{aligned}$$

Consequently,

$$\begin{aligned} \Bigg | \left[ x-\sum \limits _{i=1}^m\beta _ie_i\Big |u\right] \Bigg |\leqslant \varepsilon _m, \end{aligned}$$

which means that \(\big |\left[ x-x_m|u\right] \big |\leqslant \varepsilon _m\). Since \(\big \{\left[ \,\cdot \,|w\right] \in X^*:\Vert w\Vert =1, w\in X\big \}\) is a 1-norming subset in the dual ball of \(X^*\), we can conclude that we have \(\Vert x-x_m\Vert \leqslant \varepsilon _m\), so \(x_m\rightarrow x\). We have thus proved that \(f(\sum _{i=1}^\infty \beta _i e_i) = \sum _{i=1}^\infty \beta _i f(e_i)\). In particular, f is linear, hence also isometric because it preserves the semi-inner products.

Now, in order to prove clause (a), it is enough to show that f maps linearly independent sets to linearly independent sets.

Let \(n = \dim X\). Fix a basis \(\{b_1,\ldots ,b_n\}\) for X. We claim that the set \(\{f(b_1),\ldots ,f(b_n)\}\) is linearly independent in Y. To see this, suppose that \(\sum _{k=1}^n\alpha _kf(b_k)=0\). It follows from (2) that

$$\begin{aligned} \left\| \sum \limits _{k=1}^n\alpha _kb_k\right\| ^2= & {} \left[ \sum \limits _{k=1}^n\alpha _kb_k\Big |\sum \limits _{k=1}^n\alpha _kb_k\right] =\sum \limits _{k=1}^n\alpha _k\left[ b_k\Big |\sum \limits _{k=1}^n\alpha _kb_k\right] \\= & {} \sum \limits _{k=1}^n\alpha _k\left[ f(b_k)\Big |f\left( \sum \limits _{k=1}^n\alpha _kb_k\right) \right] \\= & {} \left[ \sum \limits _{k=1}^n\alpha _kf(b_k)\Big |f\left( \sum \limits _{k=1}^n\alpha _kb_k\right) \right] \\= & {} \left[ 0\Big |f\left( \sum \limits _{k=1}^n\alpha _kb_k\right) \right] =0. \end{aligned}$$

Hence \(\sum _{k=1}^n\alpha _kb_k=0\). Since the vectors \(b_1,\ldots ,b_n\) are linearly independent, we have \(\alpha _1=\ldots =\alpha _n=0\). This means that the vectors \(f(b_1),\ldots ,f(b_n)\) are linearly independent too. Consequently, \(\{f(b_1),\ldots ,f(b_n)\}\) is a basis for Y. Thus we may apply (b), and the proof is complete. \(\square \)

Before we prove Theorem 2, we need to introduce the main building block that we shall use to construct the sought renorming of \(\ell _2\).

Let \((Z,\Vert \cdot \Vert _o)\) be a two-dimensional normed space that is smooth but not strictly convex. Then there are distinct vectors \(u,w\in Z\) such that the line segment joining u and w lies in the unit sphere \(S_Z\) of Z. Without loss of generality, we may assume that \(Z = {\mathbb {K}}^2\) as a vector space and \(u=(-c,1)\), \(w=(c,1)\) for some real number \(0<c<1\). Thus \((0,1)\in S_{Z}\). Moreover, without loss of generality we may assume that \((1,0)\in S_{Z}\).

Lemma 4

Let \(x_1\in {\mathbb {K}}\) and \(\eta \in (0,c)\). Then \(\Vert (\eta x_1,x_1)\Vert _o=|x_1|\).

Proof

Since \(0<\eta <c\), we have \(\frac{\eta +c}{2c}\in [0,1]\) and

$$\begin{aligned} (\eta ,1)=\left( 1-\frac{\eta +c}{2c}\right) u+\frac{\eta +c}{2c}w\in \mathrm{conv}\{u,w\}\subseteq S_{Z}. \end{aligned}$$

Thus \((\eta ,1)\in S_Z\), i.e., \(\Vert (\eta ,1)\Vert _o=1\). Since \((0,1)\in S_{Z}\), \(\Vert (0,1)\Vert _o=1\). Therefore \(\Vert (\eta x_1,x_1)\Vert _o=|x_1|\cdot \Vert (\eta ,1)\Vert _o=|x_1|\cdot 1=|x_1|\). \(\square \)

Proof of Theorem 2

We shall consider the space \(X = {\mathbb {K}} \oplus _2 \ell _2(Z)\), the \(\ell _2\)-sum of infinitely many copies of Z and the one-dimensional space. The norm in X is thus given by

$$\begin{aligned} \Vert x\Vert :=\sqrt{{|x_1|^2+\sum \nolimits _{k=1}^\infty \big \Vert (x_{2k},x_{2k+1})\big \Vert _o^2}}, \end{aligned}$$
(4)

where \(x=\big (x_1,(x_2,x_3),(x_4,x_5),(x_6,x_7),\ldots \big )\in X\).

The space X is uniformly smooth because Z is smooth (uniformly smooth as it is finite-dimensional) and uniform smoothness passes to \(\ell _2\)-sums of infinitely many copies of a uniformly smooth space [7, Corollary 4.9]. Since Z is isomorphic to the two-dimensional Hilbert space, X is isomorphic to \(\ell _2(\ell _2^2)\), which is isometric to \(\ell _2\).

For a number \(\eta \in (0,c)\), let \(h_{\eta }:X\rightarrow X\) be a linear map given by

$$\begin{aligned} h_{\eta }\left( x_1,\left( x_2,x_3\right) ,\left( x_4,x_5\right) ,\ldots \right) :=\left( 0,\left( \eta x_1,x_1\right) ,\left( x_2,x_3\right) ,\left( x_4,x_5\right) ,\ldots \right) . \end{aligned}$$
(5)

Applying Lemma 4 to (4) we deduce that \(h_\eta \) is a linear isometry. Consequently, as in [3, Corollary 2.5 (ii)\(\Rightarrow \)(i)],

$$\begin{aligned} \left[ h_{\eta }(x)|h_{\eta }(y)\right] =\left[ x|y\right] \quad \left( x,y\!\in \! X,\ \eta \in (0,c)\right) . \end{aligned}$$
(6)

Combining (1) with (6) we may rearrange (6) as

$$\begin{aligned} \big \Vert h_{\eta }(y)\big \Vert _o\cdot \langle \varphi _{h_{\eta }(y)}, h_{\eta }(x)\rangle =\left[ x|y\right] \quad \left( x,y\in X,\ \eta \in (0,c)\right) . \end{aligned}$$
(7)

Moreover, putting y in place of x in (6) we get

$$\begin{aligned} \quad \Vert h_{\eta }(y)\Vert _o=\Vert y\Vert _o\quad \big (\eta \in (0,c)\big ). \end{aligned}$$
(8)

We are now ready to construct the sought non-linear map that preserves semi-inner products.

For this, we fix a function \(\gamma :[0,\infty )\rightarrow [0,\infty )\) with \(\gamma (0)=0\) such that \(0<\gamma (t)<c\) (\(t\in (0,\infty )\)) and \(\gamma \) is not constant on \((0,\infty )\). Next we choose a function \(\eta :X\rightarrow [0,c)\) by

$$\begin{aligned} \eta (x):=\gamma \big (\Vert x\Vert \big )\quad (x\in X). \end{aligned}$$

Then, we define a map \(f:X\rightarrow X\) by the formula

$$\begin{aligned} f\left( x_1,\left( x_2,x_3\right) ,\left( x_4,x_5\right) ,\ldots \right) := \left( 0,\left( \eta (x)x_1,x_1\right) ,\left( x_2,x_3\right) ,\left( x_4,x_5\right) ,\ldots \right) . \end{aligned}$$

Then we may recognise that

$$\begin{aligned} f(x)=h_{\eta (x)}(x)\quad (x\in X). \end{aligned}$$
(9)

Consequently, f fails to be linear. However, in the case where

  • \(\gamma \) is continuous and non-constant on \((0,\infty )\), f is continuous;

  • \(\gamma \) is discontinuous on \((0,\infty )\), f is discontinuous too.

We claim that for all \(x,y\in X\) we have \(\left[ f(x)|f(y)\right] =\left[ x|y\right] \). For this, fix \(x,y\in X\) and consider the associated maps \(h_{\eta (y)},h_{\eta (x)}:X\rightarrow X\). Applying again Lemma 4 to (4) and (5), we conclude that

$$\begin{aligned} \big \Vert h_{\eta (y)}(y)+h_{\eta (x)}(y) \big \Vert _o=\big \Vert h_{\eta (y)}(y)\big \Vert _o+\big \Vert h_{\eta (x)}(y) \big \Vert _o. \end{aligned}$$

It follows from Lemma 3 that , i.e.,

$$\begin{aligned} \langle \varphi _{h_{\eta (y)}(y)}, w \rangle =\langle \varphi _{h_{\eta (x)}(y)},w\rangle \quad (w\in X). \end{aligned}$$
(10)

Consequently,

$$\begin{aligned} \left[ f(x)|f(y)\right]&{\mathop {=}\limits ^{(1)}}&\big \Vert f(y)\big \Vert _o\!\cdot \!\langle \varphi _{f(y)}, f(x)\rangle {\mathop {=}\limits ^{(9)}}\big \Vert h_{\eta (y)}(y)\big \Vert _o\!\cdot \!\langle \varphi _{h_{\eta (y)}(y)}, h_{\eta (x)}(x)\rangle \\&{\mathop {=}\limits ^{(10)}}&\big \Vert h_{\eta (y)}(y)\big \Vert _o\!\cdot \!\langle \varphi _{h_{\eta (x)}(y)}, h_{\eta (x)}(x)\rangle \\&{\mathop {=}\limits ^{(8)}}&\Vert y\Vert _o\!\cdot \!\langle \varphi _{h_{\eta (x)}(y)}, h_{\eta (x)}(x)\rangle \\&{\mathop {=}\limits ^{(8)}}&\big \Vert h_{\eta (x)}(y)\big \Vert _o\!\cdot \!\langle \varphi _{h_{\eta (x)}(y)}, h_{\eta (x)}(x)\rangle {\mathop {=}\limits ^{(7)}}\left[ x|y\right] . \end{aligned}$$

This shows that \(f:X\rightarrow X\) is indeed a non-linear map preserving semi-inner products. \(\square \)

Remark 5

In the above construction one may consider the \(\ell _p\)-sums for \(p\in (1,\infty )\) instead of the \(\ell _2\)-sum. This will lead to a renorming of \(\ell _p\) on which one may find a non-linear injection preserving the (unique) semi-inner products.

Remark 6

In the proof of [3, Proposition 2.4], the authors postulated the following inclusion:

$$\begin{aligned} \left\{ \xi \varphi _{f(z)}\in X^*:z\in X,\, \xi \in {\mathbb {C}}\right\} \supseteq \{\xi \varphi _{f(z)}\circ f\in X^*:z\in X,\, \xi \in {\mathbb {C}}\}, \end{aligned}$$
(11)

where \(\varphi _{f(z)}\circ f=\varphi _z\) (see [3, p. 1265, third line from the bottom]), which fails already in the Hilbert-space setting.

To see this, let us consider the Hilbert space \(\ell _2\). The only semi-inner product on \(\ell _2\) is the inner product \(\left\langle \cdot |\cdot \right\rangle \) itself. We consider the unilateral shift on \(\ell _2\), which is a non-surjective isometry \(f:\ell _2\rightarrow \ell _2\); it is given by the formula

$$\begin{aligned} f(x)=f\left( x_1,x_2,x_3,\ldots \right) =\left( 0,x_1,x_2,x_3,\ldots \right) \quad \left( x=(x_j)_{j=1}^\infty \in \ell _2\right) . \end{aligned}$$

It is easy to see that \(\left\langle f(x)|f(y)\right\rangle =\left\langle x|y\right\rangle \) and for \(z\in \ell _2\setminus \{0\}\) we have

$$\begin{aligned} \xi \left( \varphi _{f(z)}\circ f\right) (\cdot ){\mathop {=}\limits ^{(1)}}\frac{\xi }{\Vert f(z)\Vert }\left\langle f(\cdot )|f(z)\right\rangle =\frac{\xi }{\Vert z\Vert }\left\langle \cdot |z\right\rangle . \end{aligned}$$

Therefore \(\left\{ \xi \varphi _{f(z)}\circ f\in \big (\ell _2\big )^*:z\in \ell _2,\, \xi \in {\mathbb {C}}\right\} =\big (\ell _2\big )^*\). On the other hand,

$$\begin{aligned}&\left\{ \xi \varphi _{f(z)}\in \big (\ell _2\big )^*:z\in \ell _2,\, \xi \in {\mathbb {C}}\right\} \\&\quad =\left\{ \frac{\xi }{\Vert f(z)\Vert }\left\langle \cdot |f(z)\right\rangle \in \big (\ell _2\big )^*:z\in \ell _2,\, \xi \in {\mathbb {C}}\right\} \\&\quad =\left\{ \left\langle \cdot |f(w)\right\rangle \in \big (\ell _2\big )^*:w\in \ell _2\right\} \\&\quad =\left\{ \left\langle \cdot |w\right\rangle \in \big (\ell _2\big )^*:w=(0,w_1,w_2,\ldots )\in \ell _2\right\} \nsupseteq \big (\ell _2\big )^* \end{aligned}$$

which refutes (11).