# Inequality is exactly the one Rudin uses

suppose that $g:[0,1] \rightarrow \re$ is continuous, $g(0)=g(1)=0$ and for every $c \in (0,1)$, there is a $k > 0$ such that $$0 < c-k < c < c+k < 1$$ and $$g(c)=\frac(1}{2}$$$$(g(c+k)+g(c-k))$$.
Prove that $g(x) = 0$ for all $x \in [0,1]$ Hint: Consider sup{$x \in [0,1] | f(x)=M$} where M is maximum of $f$ on [0,1].

I see that $c=\frac{1}{2}((c+k)+(c-k))$. I also see that the inequality is exactly the one Rudin uses to prove that the derivative of a local maximum is 0. I dont really understand what the hint is. There is no supremum, right?
What I tried to do and decided I couldnt make it work was to take $\delta > 0$ and take $x_0, x_1$ such that $d(x_1,1)=d(x_0,0)<\delta$ and let $k=d(x_1,1)=(x_0,0)$ so that now $g(c)=0$ when $c=\frac{1}{2}$ and I was going to show that $g(c+k),g(c-k)$ would always have to equal 0, but then I was thinking that what if the function oscillated and intersected the x-axis at 0,1/2, and 1 so that $g(c-k)=-g(c+k)$. Seems like it would hold for my proof, also I didnt use the hint. HELP!! This problem seems easy but I can't seem to wrap my head around it.

Last edited: