I saw this problem here (7)
Suppose that $f : [0, 1] \to \mathbb{R}$ is continuous, that $f(0) = f(1) = 0$, and that for every $x \in (0, 1)$ there exists $0 < \delta < \inf\{x, 1 − x\}$ with $f(x) = \frac{f(x − \delta) + f(x + \delta)}2$. Show that $f(x) = 0$ for all $x$.
I tried to prove this statement by setting $M :=\sup\{f(x)\} , \ m := \inf\{f(x)\}$ and proving that $f(a)= \frac{f(a-\delta_1)+f(a+\delta_1)}{2}=\frac{f(a-\delta_1 -\delta_2)+f(a+\delta_1-\delta_2)+f(a-\delta_1 +\delta_2)+f(a+\delta_1+\delta_2)}{4}=\dots$ will converge. The problem was I couldn't handle the sequences of deltas. There must be some trick in "$0 < \delta < \inf\{x, 1 − x\}$" but I couldn't found it. I tried to prove that eventually that sequence of delta will make $a+\sum_{k=0}^n \varepsilon_k \delta_k$ will converge to $0$ or $1$ where $\varepsilon_k \in \{-1,1\}$ for all $k$ but this didn't work.
Suppose $f$ is not identically zero. WLOG we can assume it is positive for some $x$. Then its maximum $M$, which is achieved, is positive. Let $f(x_0)=M$. Let $x_1= \sup\{x: f(x)=M\}$. By continuity, $f(x_1)=M$ and $x_1\in(0,1)$. Then we must have $$ M=f(x_1)=[f(x_1-\delta)+f(x_1+\delta)]/2 $$ Which is clearly impossible because $f(x_1-\delta)\le M$ and $f(x_1+\delta)<M$ by definition of $x_1$.