# Problem on continuous periodic functions

Problem. Let $$f:\Bbb R\to\Bbb R$$ be a continuous periodic function. Show that for every $$t>0$$, there exists $$x\in\Bbb R$$ with $$f(x)=\frac{f(x+t)+f(x-t)}{2}.$$

My Attempt. First, if $$f$$ is constant, then we are done. Otherwise, we can rewrite the formula as $$f(x+t)-f(x)=f(x)-f(x-t).$$ Denote $$g(x):=f(x)-f(x-t)$$. It then suffices to show that there exists $$x\in\Bbb R$$ such that $$g(x)=g(x+t)$$. In particular, I also know that $$g$$ is periodic with the same period as $$f$$. But I have no idea how to continue.

Here $$t>0$$ is an arbitrary constant, not necessarily the period or integer times the period, so the case looks a bit complicated.

Let $$T$$ be the period of $$f$$.
Let $$h(x) = g(x+t)-g(x)$$. Then $$h$$ is a continuous periodic function with period same as $$g$$ and $$f$$.
Now as $$g$$ is contuinuous and periodic with period $$T$$, we have $$g(\mathbb{R})=g([0,T])$$ so the range of $$g$$ is bounded and compact.
Let $$a,b\in \mathbb{R}$$ such that $$g(a) = \text{sup}\, g(\mathbb{R})$$ and $$g(b) = \text{inf} \,g(\mathbb{R})$$, then $$h(a-t) = g(a)-g(a-t) \ge 0$$ and $$h(b-t) = g(b)-g(b-t) \le 0$$ so by intermediate value theorem there exists a $$c\in \mathbb{R}$$ such that $$h(c)=0$$ implying the result.