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 (Wy˙z Szko(cid:2)l Ped Krakow Rocznik Nauk Dydakt 1998), Kairies (Aequ Kairies (Aequ and Kairies al.


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.