New characterizations of the Takagi function via functional equations

We provide two new characterizations of the Takagi function as the unique bounded solution of some systems of two functional equations. The results are independent of those obtained by Kairies (Wyż Szkoł Ped Krakow Rocznik Nauk Dydakt Prace Mat 196:73–82, 1998), Kairies (Aequ Math 53:207–241, 1997), Kairies (Aequ Math 58:183–191, 1999) and Kairies et al. (Rad Mat 4:361–374, 1989; Errata, Rad Mat 5:179–180, 1989).


Introduction
The Takagi function T : R → R is defined by where d(y) is a distance from y to the closest integer. Below is the approximation of how the Takagi function looks like on the interval [0, 1]. This is a continuous function that is not differentiable at any point in R (see [4]).
The following fact is known due to H.-H. Kairies. It concerns equations fundamental for this presentation.
then f is the Takagi function.
A trivial calculation shows that the function T [0,1] satisfies all of Eqs. (1)-(4). T also satisfies (5) as well as the symmetry equation for every x ∈ R.

Characterization of the Takagi function on the interval [0,1]
We start with the following auxiliary result.
Proof. Assume that equalities (5) and (6) hold for every x ∈ [0, 1]. Putting 1] . Of course the function g is bounded. We shall verify that it satisfies Eq. (5). Fix a number x and represent it in the form x = n + t, where n ∈ Z and t ∈ [0, 1) . If t ∈ 0, 1 2 , then 2t ∈ [0, 1) and thus, by the periodicity of g, On the other hand, if t ∈ 1 2 , 1 , then 2(1 − t) ∈ [0, 1) and whence, making use of the symmetry condition, we get In both cases we obtain the equality 2g(x) = g(2x) + 2d(x). By Theorem K we infer that g is the Takagi function. In particular, There exist functions that do not satisfy the conditions of Theorem K but satisfy those of Proposition 1.
is bounded and, clearly, equality (6)  Proof. If Eq. (2) holds, then putting x = 0 in (6) and (2) we see that f (1) = f (0) = 0. Similarly, if Eq. (3) is satisfied, then setting x = 0 in (6) and x = 1 in (3) we again come to f (1) = f (0) = 0. Let g : R → R be the 1-periodic extension of f . Of course the function g is bounded. Moreover, for every x ∈ R, taking n ∈ Z and t ∈ [0, 1) with x = n + t and making use of the symmetry condition (6), we have This means that the function g is even.