Let $f:[0,1]→\Bbb R$ be continuous with $f(0)=f(1)=0$. Suppose that for every $x ∈ (0,1)$ there exists $δ > 0$ such that both $x−δ$ and $x+δ$ belong to $(0,1)$ and $f(x) = \frac{1}{2}(f(x−δ)+f(x+δ))$ . Prove that $f(x) = 0$ for all $x ∈ [0, 1]$.
I have considered several approaches but not had much success with any of them. First, I tried to prove that $f'(x)=0$ for all x, because as we're given $f(0)=f(1)=0$, I think that is sufficient? But I ran into trouble proving that the limit exists for some general $a$ in the domain. So I decided to try and use continuity, specifically around zero: my intuition was that for $x$ sufficiently close to 0, $f(x)$ is approximately zero and I tried to build up that the images of the two perturbations would also be zero but again without much success.
Please could I have some help? Ideally, a discussion of the right kind of intuition as well as a sketch solution. Comments on whether my approaches were at all in the right direction would also be appreciated.
Edit: Please could this be solved with using connectedness and compactness.
Thank you.
Hint: Assume $f$ is not identically zero. Since $[0,1]$ is compact, so is its image under $f$, so there's a numerically largest value of $f$ in $[0,1]$. Furthermore, there must be a smallest $x$ such the $|f(x)|$ has this largest value. Plug this $x$ into the property you're assuming and derive a contradiction.