Generalized orthogonality equations in finite-dimensional normed spaces

Let X, Y be real normed spaces and let ρ+′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho '_+$$\end{document}, ρ-′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho '_-$$\end{document} be norm derivatives. In this work, we solve a system of functional equations ρ+′(f(x),f(y))=g(x)ρ+′(x,y),ρ-′(f(x),f(y))=g(x)ρ-′(x,y),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} {\left\{ \begin{array}{ll}\rho '_+(f(x),f(y))=g(x)\rho '_+(x,y),\\ \rho '_-(f(x),f(y))=g(x)\rho '_-(x,y), \end{array}\right. } \end{aligned}$$\end{document}with unknown functions f:X→Y\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f:X\!\rightarrow \!Y$$\end{document}, g:X→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$g:X\rightarrow \mathbb {R}$$\end{document}. Moreover, we give partial answer to open problem posed in 2010.

are useful generalizations of inner product and the classical relation of orthogonality ⊥.

Theorem 1.1 If X is a finite-dimensional normed linear space, then N sm (X ) is dense in X .
It is also worthwhile, before proceeding, explicitly to reformulate the properties (ρ2) and (ρ3). Namely, let us define the function s : R → {+, −} by the formula Combining (ρ2) with (ρ3) we can assert that for all x, y ∈ X and for all α ∈ R ρ + (x, αy) = αρ s(α) (x, y) and ρ − (x, αy) = αρ −s(α) (x, y), (1. 2) The notation −s(α) will not lead to any confusion because we will write: For a ∈ X \ {0}, we put J (a) := {x * ∈ X * : x * = 1, x * (a) = a }, i.e., the symbol J (a) denotes the set of supporting functionals at the point a. By the Hahn-Banach Theorem, we get J (a) = ∅. After some calculations, explicit formulas for ρ ± can be derived (e.g., see [3]): In the great paper [6], another geometrical notion was introduced. We say that a Banach space X is alternatively convex or smooth (briefly, (acs)), if for all x, y ∈ X \{0} the implication holds. It is not difficult to check that strictly convex spaces (or smooth spaces) are (acs). Example of a sphere in (acs) space that is neither smooth nor strictly convex. The segment BC is contained in S X while A is not a smooth point Moreover, it is worth mentioned that the following result gives full characterization of (acs) spaces in the terms of the Daugavet equation.
Theorem 1.2 [6,10] Let X be a finitely dimensional normed space and let L(X ) denote the set of all linear mappings going from X into X . Then, the following conditions are equivalent: (a) X is (acs); (b) for any T ∈ L(X ) we have that I d + T = 1 + T if and only if a number T is an eigenvalue of T .
The authors of [6,10] proved a far more general theorems. We quote only the special case above and do not generalize their original version. We refer to the classical paper [4] and more recent by the second author [10] for motivations, history, various aspects and problems connected with the notion of (acs).

Motivation
One of the dominant tools in the study of the functional equations is the paper [12], where another open problem was solved. Alsina et al. obtained the following result; see also [2].
for a function g : X → R satisfying the condition (g(x) = 0 ⇒ x = 0) if and only if f is a similarity (scalar multiple of a linear isometry).
The aim of this paper is to fortify Theorem 2.1. More precisely, we will consider finitedimensional space (instead of dim X = 2) and we will show that the assumption of the continuity of f is superfluous.
The second author solved this open problem; see [11] and [12]. In particular, the result [ It is worth recording that the main tool in proving the following Theorem 3.2 is the result (2.3).
It is easy to see that if X is a normed smooth space and a function g : X → R satisfies g(·) = 1, then we have ρ + = ρ − , and therefore, the system (2.1) is reduced to a functional equation ρ + ( f (x), f (y)) = ρ + (x, y). However, there are interesting problems concerning this functional equation on a smooth Banach space (see, e.g., [5] and [14]).
Moreover, similar investigations have been carried for two unknown functions f and g, instead of one, cf. [7]. It is also worth mentioning that the very interesting paper [8] involved a phase version of the above equation, i.e., ρ

Main results
Before launching the main section, we prove the following theorem. Theorem 3.1 Let X be a finite-dimensional normed spaces, and let y, z ∈ X . Then the following statements are true: Proof The property (1.1) and (ρ3)−(ρ6) say that we only need to prove (a). Therefore, assume that ρ + (a, y) = 0 for all a ∈ N sm (X ). The density of N sm (X ) implies that there is a sequence {a n : n = 1, 2, . . .} in N sm (X )\{0} such that a n → y. Now, we have = ρ + (a n , −a n + y) (ρ7) ≤ a n |·|a n − y|, and dividing by a n , we obtain a n ≤ a n − y . Since we know that a n → y, it follows that y ≤ 0 and y = 0.
Let us prove the first main result of this paper.

Theorem 3.2 Let X and Y be a real normed spaces with
Then, a mapping f : X → Y satisfies the system of functional equations for all x, y ∈ X : if and only if f is a linear similarity and g is constant on X \ {0}.
Proof It is easy to verify that the implication "⇐" is true. Therefore, we want to prove "⇒". First, we show that for all a ∈ N sm (X ), for all x, y ∈ X and for all α, β ∈ R we have In order to prove it, we fix two numbers α, β ∈ R and three vectors x, y ∈ X and a ∈ N sm (X ). It follows from (3.1) and (1.1) that Then, On the other hand, the properties (3.5) and (3.7) are the same. Thus, the string of inequalities becomes a string of equalities and we conclude from (3.4) to (3.6) that (3.2) is true. Now, using induction and (3.3), we also obtain for all a ∈ N sm (X ), for all x 1 , . . . , x k ∈ X and for all α 1 , . . . , α k ∈ R : Now, we prove that f has the following property: Assume that x 1 , . . . , x k are linearly independent and suppose that hence ρ + a, k j=1 α j x j = 0 for all a ∈ N sm (X ). It follows from Theorem 3.1 that k j=1 α j x j = 0. Since x 1 , . . . , x k are linearly independent, we get α 1 = · · · = α k = 0. To summarize, it has been shown that f (x 1 ), . . . , f (x k ) are linearly independent.
Our next goal is to show that f satisfies the functional equation (2.2). The implication z = 0 ⇒ f (z) = 0 holds, which is clear from (3.1) and Theorem 3.1. It remains to prove that f satisfies (2.2). Fix x, y ∈ X . Now, we consider two cases.
Notice that by equations (3.1) and (ρ7), we have and hence, for a non-zero vector x, we have Thus, we can formulate the following corollary about functions preserving norm derivatives.

Corollary 3.3 Let X and Y be a real normed spaces with
Then, a mapping f : X → Y satisfies the system of functional equations (3.1) with g(x) = 1 if and only if f is a linear isometry.
Now, we present that it may be assumed only one equation instead of two-but in some circumstance.

Theorem 3.4 Let X be a normed spaces and let Y be (acs) with
Let g : X → R be a function such that (g(x) = 0 ⇒ x = 0). Then, a mapping f : X → Y satisfies the functional equation

and only if f is a linear similarity and g is constant on X \ {0}.
Proof In light of the preceding theorem, in order to prove Theorem 3.4, it suffices to , . Consider the following equalities: and dividing by f (x) , we obtain 2 , Since Y is (acs), we get x * ∈ J ( f (−x)) by the equality (3.11). Therefore, J (− f (x)) ⊆ J ( f (−x)). Using similar argument we establish that J (− f (x)) ⊇ J ( f (−x)) and thus Combining the above, we obtain which completes the proof.
Every finitely dimensional smooth (or strictly convex) normed space is (acs). Therefore, as an immediate consequence of Theorem 3.4, we have the following. Theorem 3.5 Suppose that X , Y are normed spaces with dim X = dim Y = n. Assume that Y is a smooth (or strictly convex) normed space. Let g : X → R be a function such that (g(x) = 0 ⇒ x = 0). Then, a mapping f : X → Y satisfies the functional equation if and only if f is a linear similarity and g is constant on X \ {0}.
Having the above results, we can solve another functional equation preserving norm derivatives up to something like normalization. Theorem 3.6 Let X and Y be a real normed spaces with dim X = dim Y = n. Suppose that the implication z = 0 ⇒ f (z) = 0 holds for all z ∈ X . Then, a mapping f : X → Y satisfies the system of functional equations x · y and for all x, y ∈ X \{0} such that f (x) = 0 = f (y) if and only if there is a linear isometry L : X → Y and a function λ : We can easily check that if ϕ ∈ C[0, 1] \ {0}, then |ϕ 0 (t)| < ϕ ∞ for t ∈ 1 2 , 1 . On the other hand, since ϕ 0 (t) = ϕ(2t) for t ∈ 0, 1 2 , we get M ϕ 0 ⊆ 0, 1 2 and 1 2 . Thus, applying (4.2), we obtain It is easy to see that f is not linear. Moreover, there is an amazing observation that can be made here. Namely, the reader might think that an isometry satisfying (4.4) would have to be linear. This is untrue as the above example illustrates. Notice once more that the equations (4.4) holds and, furthermore, f is an isometry. Indeed, if 1 2 ≤ t ≤ 1 and ϕ, μ ∈ C[0, 1], then On the other hand, for 0 ≤ t < 1 2 , we have sup |(ϕ 0 − μ 0 )(t)| : t ∈ [0, 1 2 ) = sup {|(ϕ − μ)(t)| : t ∈ [0, 1]} and thus ϕ 0 − μ 0 ∞ = ϕ − μ ∞ , which in turn implies that f is an isometry.
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