On a functional equation characterizing linear similarities

. The aim of this paper is to give an answer to a question posed by Alsina, Sikorska and Tom´as. Namely, we show that, under suitable assumptions, a function f : X → Y from a normed space X into a normed space Y , satisfying the functional equation has to be a linear similarity (scalar multiple of a linear isometry).

(nd1) ∀ x,y∈X ∀ α∈R ρ ± (x, αx + y) = α x 2 + ρ ± (x, y); (nd2) ∀ x,y∈X ∀ α 0 ρ ± (αx, y) = αρ ± (x, y) = ρ ± (x, αy); A normed space X is said to be smooth if for every x ∈ X \{0} there is a unique supporting functional at x, i.e., a unique functional x * ∈ X * such that x * =1 and x * (x)= x . Moreover, we may state this definition in an equivalent form, namely: X is smooth ⇔ ρ + = ρ − ⇔ ∀ x∈X ρ + (x, ·) is linear. If X is smooth, then the following condition holds (see [1]): (nd6) for any two-dimensional subspace P of X and for every x ∈ P , λ ∈ (0, +∞), there exists a y ∈ P such that x⊥ ρ + y and x+y⊥ ρ + λx−y. In a real inner product space (X, ·|· ), given the triangle determined by two linearly independent vectors x, y and the zero vector (i.e., {x, y, 0}), one can compute the height vector from y to the side x and orthogonal to x using the formula h(x, y):=y− x|y x 2 x. Then x⊥h(x, y). The same might be done for normed spaces using the function ρ + as a generalization of an inner product.
In this case we consider the height function h(x, y):=y− ρ + (x,y) Alsina et al. [1] investigated functions f : X → X that transform the height of the triangle with sides x, y, x − y into the corresponding height of the triangle determined by sides which leads to the functional equation f y− ρ + (x,y) In particular, Alsina et al. [1] obtained the following result.

and vanishes only at zero if and only if, f is a linear similarity.
At the end of their book [1, p. 178, Open problem 6] Alsina, Sikorska and Tomás put the following problem.
Open problem Solve the functional equation where f : X → X is injective and f (x) = 0 whenever x = 0. The aim of this paper is to present a partial solution of the above open problem. In particular, we will prove that the assumption of the continuity of f is redundant in some circumstances. Moreover, it is not necessary to assume that f is injective.

Results
Throughout this section we will work with real normed spaces of dimensions not less than 2. We will consider the norm derivatives in various spaces (X and Y ); however, we will use one common symbol ρ + for them. We will prove that f : X → Y is a solution of (1) if and only if it is a linear similarity (scalar multiple of a linear isometry). This assertion, however, can be obtained under the assumption of the smoothness of X. But, unlike Theorem 1, it will not be assumed that a function f is continuous.
Proof. By (1) we get f (0) = f y− ρ + (y,y) Now we prove the first main result of this paper. Theorem 3. Let X, Y be normed spaces and let f : X → Y satisfy (1). Suppose that z = 0 ⇒ f (z) = 0. Then f is additive.
Proof. First we will prove that f preserves the linear independence of two vectors. Suppose that f (y)=αf (x) and x = 0. Then From the assumption (i.e. f (z) = 0 ⇒ z = 0) we have that y − ρ + (x,y) x 2 x = 0, hence the vectors x, y are linearly dependent. So, we have proved that f preserves the linear independence of two vectors.
Fix two linearly independent vectors a, b ∈ X. Then we have It follows from the above equalities that Putting b, a in place of a, b, respectively, in the above equality we get We know that f (a), f(b) are linearly independent. Thus, combining (2) and (3), we immediately get (4) Now let x and y be linearly dependent. We may assume that x = 0 = y. We consider two cases. Assume first that y = γx for some γ ∈ R \ {−1}. There are linearly independent vectors a, b ∈ X such that a + b = x. Then