A Continuous Function Having Maximum At Every Point

Question

Prove that if there is a continuous function$f(x)$ such that every point is a local maximum then $f(x)$ is a constant function. $$$$Suppose that there exists some $a \neq b$ such that $$f(a) < f(b)$$. Now consider the closed interval $[a, b]$. In this interval there is some point $a<x_0<b$ such that $f(x_0)=f(a)$ because as $a$ is the point of local maximum so for some $\delta>0$ we have $$f(x) \leq f(a)$$ for all $x$ in the interval $(a, a+\delta)$. So if there is some $x$ in this interval such that $$f(x)=f(a)$$ then we are done and if not then for all $x$ in this interval we have $$f(x)<f(a)$$ and as $f(a)<f(b)$ and as $f(x)$ is a continuous function so by intermediate value theorem there exists some $x$ such that $$f(x)=f(a)$$. Now consider a set $A$ which contains all $a<x<b$ such that $$f(x)=f(a)$$. Now as set $A$ is bounded so it must have some least Upper Bound say $c$, then in every neighbourhood of $c$ there is some $x$ such that $$f(x)=f(a)$$ and as $$\lim_{x \to c}f(x)=f(c)$$ so we have $f(c)=f(a)$. But again as $c$ is also the point of local maximum sothere exists some $\delta_0$ such that for all $x$ in the interval $(c, c+\delta_0)$ we have $$f(x) \leq f(c)=f(a)$$ but as $c$ is the least Upper Bound of the set $A$ so we have for all $x$ in this interval $$f(x)<f(c)=f(a)$$ but again as $f(b)>f(a)$ and $f(x)$ is a continuous function so there exists some $c<x<b$ such that $$f(c)=f(a)$$, so any $a<c<b$ cannot be the least Upper Bound of set $A$ and hence $b$ is the least Upper Bound of set $A$ then every neighbourhood of $b$ contains some $x$ such that $$f(x)=f(a)$$ but again as $$\lim_{x \to b}f(x)=f(b)$$ so we have $$f(a)=f(b)$$. Similarly we can deal with the case when $$f(a)>f(b)$$ Is My Proof Correct??

Answer 1

Suppose $f$ is defined on the reals and every point is a local $\max$.

Pick some $x^*$ and let $R = \sup \{ r | f(x) \le f(x^*) \text{ for all } x \in [x^*-r,x^*+r]\}$.

Suppose $R$ is finite, then continuity shows that $f(x^*+R) \le f(x^*)$ and $f(x^*-R) \le f(x^*)$ and since $x^* \pm R$ are also local maxima, we see that there is a slightly larger interval for which the condition holds, which is a contradiction. Hence $R$ is infinite and $f(x^*) \ge f(x)$ for all $x$.

Since the same reasoning holds for any other $y^*$ we see that $f$ is constant.