"Tooth" function and its Lipschitz property

Question

Suppose that $\varphi:[0,1]\to \mathbb{R}$ is given by $$\varphi(x) = \begin{cases} x, & \text{if } x\in[0,1/2],\\ 1-x, & \text{if } x\in [1/2,1]. \end{cases}$$ Let's continue this function over $\mathbb{R}$ with period $1$ and label the new function as $f:\mathbb{R}\to\mathbb{R}$. This function is used when we want to construct an example of continuous, nowhere differentiable function (van der Waerden function). I am going to prove that $|f(x)-f(y)|\leq C|x-y|$ and probably $C=1$, i.e. $f(x)$ is Lipshitz function.

I have some difficulties to prove it. It is enough to consider the case when $|x-y|<\frac{1}{2}$. Suppose that $x\in [n,n+1)$. Then $x\in[n,n+\frac{1}{2})$ or $x\in [n+\frac{1}{2},n+1)$. So suppose that $x\in[n,n+\frac{1}{2})$ Since $|y-x|<\frac{1}{2}$, then $y\in (n-\frac{1}{2},n+1)$. Then we can consider the following 3 cases:

I. $y\in (n-\frac{1}{2},n)$; II. $y\in [n,n+\frac{1}{2})$; III. $y\in [n+\frac{1}{2},n+1)$.

For example, in the first case we know that: $f(x)=x-n$ and $f(y)=n-y$. Then $|f(x)-f(y)|=|x-n-(n-y)|=|x+y-2n|$.

How to show that $|f(x)-f(y)|\leq |x-y|$ in that case?

I'd be thankful for help!

Answer 1

Here is an interesting proof: Let $g(x)=1$ for $0 \leq x \leq \frac 1 2$ and $g(x)=-1$ for $\frac 1 2 \leq x \leq 1 $. Extend $g$ to a peridic function with period $1$. Then you can check that $f(x)=\int_0^{x} g(t)dt$. Hence, $|f(x)-f(y)| \leq |\int_x^{y} g(t)dt| \leq |x-y|$.