Let $f : [0, 1] → R$ be continuous with $f(0) = f(1) = 0$. Suppose that for every $x ∈ (0, 1)$ there exists $\delta > 0$ such that both $x − \delta$ and $x + \delta$ belong to $(0, 1)$ and $$f(x) = \frac{1}{2}(f(x − \delta) + f(x + \delta))$$ Prove that $f(x) = 0 \ \forall x ∈ [0, 1]$
I have no idea where to start with this problem
$f$ has a global maximum on $[0,1]$, let $M = \max_{x\in[0,1]} f(x)$.
Assume $M > 0$.
$f^{-1}(\{M\})$ is a closed set so let $t_0 = \min f^{-1}(\{M\})$.
If $t_0 > 0$ there exists $\delta > 0$ such that $$M = f(t_0) = \frac{\overbrace{f(t_0+\delta)}^{\le M} + \overbrace{f(t_0-\delta)}^{<M}}2 < M$$
Therefore, $t_0 = 0$ so $M = f(0) = 0$.
An analogous discussion would show that the global minimum has to be $0$.
We conclude $f \equiv 0$.