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Let $f$ be a continuous function on $\mathbb{R}$ which attains a local maximum at ${{x}_{0}}$. Prove that if $f$ doesn't have any other extremum points, then ${{x}_{0}}$ is the global maximum of $f$ on $\mathbb{R}$.

I know this has something to do with Weierstrass's theorem but am unable to formulate a proof.

BTW, $f$ is not necessarily differentiable!

BigM
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Gad Maor
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  • What do you mean by extremum point, you mean a local min or local max? – Seth Jan 27 '15 at 22:21
  • By extremum point, do you mean points $x$ such that there exists an $\epsilon>0$ such that for all $y$ with $0<|x-y|<\epsilon$ we have $f(x)>f(y)$ (resp $f(x)<f(y)$)? – Matt Samuel Jan 27 '15 at 22:21
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    Suppose $x_0$ is not the global maximum. Then there is $x_1$ with $f(x_1)>f(x_0)$. Show that there is a local minimum between the two. – GEdgar Jan 27 '15 at 22:21
  • @GEdgar : You missed an important part of the question, which is to show that $x_0$ is the global maximum, which is different than showing that $x_0$ is a global maximum. – Patrick Da Silva Jan 27 '15 at 22:39

2 Answers2

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Note that if $x_0$ is a local maximum, there is some $B(x_0,r)$ such that $f(x) \le f(x_0)$ in this ball. In fact, we must have $f(x) < f(x_0)$ for all $x \neq x_0$ in this ball, as otherwise this would contradict the unique extremum assumption.

If there is some point $x_1$ such that $f(x_1) \ge f(x_0)$, then we see that there is a minimizer in $[x_0,x_1]$ with strictly lower value than $f(x_0)$, and this contradicts the unique extremum assumption.

Hence $f(x) < f(x_0)$ for all $x \neq x_0$.

copper.hat
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  • Hi copper.hat and thanks for the swift response! Please clarify the terms "ball" and "minimizer" that you used in your proof. – Gad Maor Jan 27 '15 at 23:02
  • $B(x_0,r) = (x_0-r,x_0+r)$ in the $\mathbb{R}$ case. By minimizer, I mean a point $x'$ such that $f(x') \le f(x)$ for all $x \in [x_0,x_1]$. Since, in this case, $f(x')< f(x_0)$, we see that $x' \in (x_0,x_1)$, so it is a local minimizer (and hence an extremum). – copper.hat Jan 27 '15 at 23:06
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Suppose $x$ is a local maximum for $f$ and let $y \in \mathbb R$ ; assume $f(y) \ge f(x)$. Since $x$ is a local maximum, there exists $\delta > 0$ such that $f : ]x-\delta,x+\delta[ \to \mathbb R$ has its maximum at $x$. The function $f$ restricted to $]x-\delta,x[$ cannot be constant, otherwise it would have other extrema. Therefore there exists $x' \in ]x-\delta,x[$ with $f(x') < f(x)$ (and also in $]x,x+\delta[$, but this case is symmetric).

Now assume furthermore that $y < x-\delta$. Since $y<x'<x$ and $f(y) \ge f(x) > f(x')$, the function $f : [y,x] \to \mathbb R$ is continuous on a compact domain, so it attains its minimum ; it is not attained at $y$ or at $x$ (because the minimum is at most $f(x')$), so the minimum lies in $]y,x[$ ; this minimum is therefore a local minimum , i.e. an extremum point. This point was not supposed to exist, a contradiction.

Therefore $f(y) < f(x)$. Since $\delta$ can always be picked arbitrarily small, this trick works for all $y < x$ ; repeat in a symmetric way for $y > x$. Then $f(y) < f(x)$ for all $y \in \mathbb R \backslash \{x\}$, hence we are done.

Hope that helps,

Patrick Da Silva
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