Let $f:[0,1]\to[0,1]$ be a continuous function with $f(0)=f(1)=0$ and verifying the following property : for all $x\in(0,1)$ there exists a $\delta>0$ such that $x-\delta, x+\delta\in(0,1)$ and $f(x)= \frac12(f(x-\delta)+f(x+\delta))$. I have to prove $f(x)=0$ for all $x\in[0,1]$
Well, it didn´t seem so difficult at first sight, but I've been some hours trying and nothing useful comes to my mind. This is what I tried, but it probably doesn´t lead anywhere, so I'd be grateful if you could give me a hint:
We take the maximum M of the function of the function $f$ in $[0,1]$. The fucntion is continuous, so it must be reached at an $x^*\in[0,1]$. If $x^*=0$ or $x^*=1$ we have the result trivially, so let's suppose $x^*\in(0,1)$. Then, there exists a $\delta(x^*)>0$ such that $$f(x^*)=\frac12(f(x^*-\delta(x^*)+f(x^*+\delta(x^*)))$$ and the only possibility is $M=f(x^*)=f(x^*-\delta)=f(x^*+\delta)$. We define a sequence such that:$$x_1=x^*$$$$x_{n+1}=x_n-\delta(x_n)$$ It's easily seen we have $f(x_n)=f(x^*)= M$ for all $n\in\Bbb{N}$. Also, the sequence $x_n$ is decreasing and bounded, so it's convergent. I was trying to prove this sequence was converging to $0$ and, because of the continuity of $f$, we would have $0=f(0)=f(x^*)=M$. However, the sequence won't always go to $0$.
Please, I need some help.
Your approach seems to be on the right track.
Let $X$ be the set of points of $[0,1]$ so that $f(x)=M$. Let $y=\sup M$. By the continuity of $f$, $y\in M$ (so the supremum is actually a maximum). Now, if $y\not=1$, then, by the assumption, there exists a $\delta$ so that $$ M=f(y)=\frac{1}{2}(f(y-\delta)+f(y+\delta)). $$ Since $f(y-\delta)\leq M$, it follows that $$ M\leq \frac{1}{2}(M+f(y+\delta)) $$ or that $$ M\leq f(y+\delta). $$ Since $M$ is the maximum value of $f$ on $[0,1]$, $f(y+\delta)=M$, so $y+\delta\in X$, which contradicts the definition of $y$ as being the supremum of $X$.