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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.

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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.

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  • $\begingroup$ Elegant proof! Thank you! $\endgroup$
    – user735816
    May 4, 2020 at 15:34

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