I'm stuck on this problem, I've given it considerable effort and tried using the bounded nature of continuous functions on closed, bounded intervals but I just can't solve it. I think I might need to pick a clever sequence somewhere but I can't see exactly what to do.
Suppose that $f:[0,1] \to\mathbb{R}$ is continuous and $f(0) = f(1) = 0$. Also, suppose further that for all $x \in (0,1)$ there exists a $0 < d < \min\{x,1-x\}$ such that: $f(x) = \frac12\Big(f(x-d) + f(x+d)\Big)\;.$ Prove that $f(x)=0$ everywhere.
Thank You.