Q:

Functions on $\mathbb{R}$ symmetric about a point $x_0 \in \mathbb{R}$

We can say that function $f(x), x \in \mathbb{R}$ is symmetric about a point $x = x_0$ if $f(x_0 - x) = f(x_0 + x)$ for all $x$ for which $f$ is defined. Now let's try to find a function which is symmetric about every point. It is easy to see that functions like $f(x) = const$ are such functions.
Is it possible to prove that there are no other functions that are symmetric about every point? One can prove this for differentiable functions by demostrating that it has to be $f'(x) = 0$ for a function to satisfy the condition. But what about non differentiable functions defined $\forall x \in \mathbb{R}$?

A:

This is very simple: given points $a,b\in\Bbb R$, we know that $f$ is symmetric about their mid-point $x_0=(a+b)/2$, so
$$f(a)=f(x_0-(b-a)/2)=f(x_0+(b-a)/2)=f(b)$$
Hence $f$ is constant.

