Suppose I have a continuous, differentiable (univariate) function $f: \mathbb{R}_{>0}\rightarrow \mathbb{R}$, and I have a local maximum at, say, $x_0$. My intuition now is that in order for the point $x_0$ to not be the global maximum, $f$ should first decrease and then at some point increase again to reach a global maximum somewhere else, implying that there should be at least one local minimum.

How do I formally prove this intuition, that is, that in order for $x_0$ to not be the global maximum, there should be a local minimum? Is there a well-known theorem that implies this result?

Or is my intuition perhaps not correct, in which case, I would love to see a counter example.


1 Answer 1


Let $x_1\in(0,\infty)$ be such that $f(x_1)>f(x_0)$. You can assume without loss of generality that $x_1>x_0$. The restriction of $f$ to $[x_0,x_1]$ has a minimum at some point $a$. Since $f(a)\leqslant f(x_0)<f(x_1)$, $a\ne x_1$. If $a\ne x_0$, then you're done: $a\in(x_0,x_1)$, and therefore $f$ has a local minimum at $a$.

If $a=x_0$ then, since $f$ has a local maximum at $x_0$, there is some $\delta>0$ such that$$x\in[x_0,x_0+\delta)\implies f(x)\leqslant f(x_0).$$Take $\delta$ such that $x_0+\delta<x_1$. You have $a=x_0$ and the minimum of the restriction of $f$ to $[x_0,x_1]$ is reached at $a$. Therefore, $f$ is constant on $[x_0,x_0+\delta)$ and so you can replace $a$ by any number from $(x_0,x_0+\delta)$. So, as above, $f$ has a local minimum at $a$.


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