Could you tell me how to prove or disprove that if a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ has only one critical point and it is a local minimum, then it is a global minimum?

I know that if $f$ is convex, then its local minimum is global. Is it true in general for real functions with one variable?

I would really appreciate all your help.

Thank you.

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    $\begingroup$ What do you mean by 'general' functions? And your function is derivable twice or not? $\endgroup$
    – Valent
    Dec 10, 2013 at 18:41
  • $\begingroup$ I meant not necessarily convex. The function is supposed to be differentiable (once). If it has to be $C^2$, I am very interested in the result, too. $\endgroup$
    – Hagrid
    Dec 10, 2013 at 18:53
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    $\begingroup$ @Hagrid : it might make an interesting new question if you weaken the differentiability assumption. $\endgroup$ Dec 10, 2013 at 21:31
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    $\begingroup$ @Hagrid : all you really need to assume is that $f$ is continuous (provided you define "critical point" properly). And it is necessary for $f$ to be continuous. Please see my newly-revised answer. $\endgroup$ Dec 10, 2013 at 23:29
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    $\begingroup$ @Hagrid : perhaps a more interesting question is: suppose $f$ is a continuously differentiable function of $x$ and $y$, has a local minimum at $(0,0)$, and no other critical points. Does $f$ necessarily have a global minimum at $(0,0)$? It would not surprise me if this question has been asked and answered in this forum. If the answer is "no", then a counterexample probably appears in most good calculus books (I don't remember what the answer is). $\endgroup$ Dec 11, 2013 at 2:00

1 Answer 1


Without loss of generality, assume the critical point (or critical number) is $0$, and $f(0)=0$. You can do this by sliding the graph of the function around in the plane.

Argue by contradiction and assume $f$ does not have a global minimum at $x=0$. Than $f(b) < 0$ for some $b \neq 0$. For convenience, assume $b>0$. Since $f$ has a local minimum at $x=0$, either $f\equiv 0$ on $[0,b/2]$ or $f(c)>0$ for some $c\in (0,b/2]$. The first alternative is impossible since it implies $f'(b/4)=0$ (and $0$ is the only critical point of $f$). So the second alternative follows. By the Intermediate Value Theorem, there exists $d \in (c, b)$ with $f(d)=0$.

To conclude the proof you essentially need Rolle's Theorem. I looked up Rolle's Theorem and the hypotheses are that $f$ is continuous on $[0,d]$ and "differentiable" on $(0,d)$. I'm not sure if the term "differentiable" requires $f'$ to be continuous. You actually don't need $f'$ to be continuous, you just need $f'(x)$ to exist for all $x \in (0,d)$. So I'll go ahead and finish the proof of this problem using basically the standard proof of Rolle's Theorem.

Since $f$ is continuous, $f$ attains maximum and minimum values on the interval $[0,d]$. If $f$ is constant on $[0,d]$, then $f'(d/2)=0$, which gives you a contradiction. If $f$ has a positive maximum on $[0,d]$, which occurs at some $w \in (0,d)$, then Fermat's Theorem gives you $f'(w)=0$, also a contradiction. If $f$ has a negative minimum on $[0,b]$, which occurs at some $w \in (0,b)$, then Fermat's Theorem gives you $f'(w)=0$, also a contradiction.

Therefore $f$ has a global minimum at $x=0$.

Note that I never assumed that $f'$ was continuous, only that $f'(x)$ exists for all $x\neq 0$ and $f$ is continuous at $x=0$.

If you use the standard Calc I definition of critical point (number): "$c$ is a critical number of $f$ if $f'(c)=0$ or doesn't exist", the conclusion holds if you replace "$f$ is differentiable" with "$f$ is continuous": if $f$ has a nonzero critical point, you're done. Otherwise, $f'(x)$ exists for all $x \neq 0$, and if you also assume $f$ is continuous at $x=0$, just repeat the above proof.

Without assuming $f$ is continuous, the conclusion seems to be false: consider $f:\mathbb{R}\to \mathbb{R}$ defined by $f(0)=0$ and $f(x)=1-|x|$ for $x \neq 0$ (sorry, I forgot how to do \cases). $f$ has a local minimum at $x=0$, but no nonzero critical points. I'm not sure whether $0$ is officially considered a critical point here ($f$ is discontinuous at $x=0$).

  • $\begingroup$ Thank you a lot. I've studied your solution, but I have one question. Why exactly can we assume that there exists $b \neq 0$ such that $f(b)=0$? $\endgroup$
    – Hagrid
    Dec 10, 2013 at 20:04
  • $\begingroup$ Ok, I see. Thank you. $\endgroup$
    – Hagrid
    Dec 10, 2013 at 20:17
  • $\begingroup$ @Hagrid : After I posted, I thought of a gap in my proof, and it appears you might have spotted it too. I fixed my answer. Thanks for improving your question, by the way. $\endgroup$ Dec 10, 2013 at 20:57

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