Question: Suppose $f$ is continuous on $[0, 1]$ with $f(0)=f(1)=0$. For $\forall x\in (0, 1)$, there $\exists h>0$ with $0\le x-h<x<x+h\le1$ such that $f(x)=\frac{f(x+h)+f(x-h)}{2}$. Show that $\forall x\in(0, 1), f(x)=0$.
I tried to prove that $f$ is differentiable on $(0, 1)$ using the fact that $\frac{f(x+h)-f(x)}{h}=\frac{f(x)-f(x-h)}{h}$, but I realized that not all $h$ holds the equation, but there exists a particular $h$ in every $x$. So, this is not a proper approach.
I also thought about the concavity of $f$. Since $\forall x\in(0, 1),\exists h>0$ with $0\le x-h<x<x+h\le1$ such that $$f(x)=f\left(\frac{x-h}{2}+\frac{x+h}{2}\right)\ge {1\over2}f\left(x-h\right)+{1\over2}f\left(x+h\right)$$ , which implies $f$ is concave downward, and $$f(x)=f\left(\frac{x-h}{2}+\frac{x+h}{2}\right)\le {1\over2}f\left(x-h\right)+{1\over2}f\left(x+h\right)$$ , which implies $f$ is concave upward.
Two facts might imply that $f$ is constant, which in turn $\forall x\in[0, 1], f(x)=0$ since $f(0)=f(1)=0$. Is this approach correct?
I thought it has to be more precise, so I wanted to use the second derivative. But I actually failed to prove that $f$ is differentiable. Could you please give me some ideas about the question? Thanks a lot.
Since $f$ is continuous on a closed, bounded interval, it attains its bounds.
So, let $x^*\in [0,1]$ be a point such that $f(x^*)$ is a maximum on $[0,1].$ However, we know that $f(x^*)$ is the average of two points around it, which (since $f(x^*)$ is a maximum) must also be maxima, i.e.
$$\exists h_1>0\:\colon f(x^*+h_1)=f(x^*).$$
Repeating for $x^*+h_1,$ we form a sequence $x^*, x^* + h_1,x^*+h_1+ h_2,\dotsc.$ This sequence converges, since it is bounded above by 1.
Suppose it converges to some $L< 1.$ But then, since $f$ is continuous, $f(L)$ is also equal to $f(x^*)$ (by sequential continuity). So we can repeat the above process with $f(L)$ to obtain a new point $L+h$ at which $f$ also attains its maximum (a contradiction).
Hence $L=1$ and
$$f(x^*)=f(1)=0.$$
Repeating with the minimum of $f$ gives us that the maximum and minimum values attained by $f$ are both $0,$ and so $f$ is constant.