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I want to prove that a function $f(x)$ has its unique minimum at $f(c)$. By that I mean, $f(c) < f(x)$ for all $x \neq c$. The strict inequality is obviously vital.

My question is, does this follow automatically if $f'(x) = 0$ if and only if $x = c$?

I've heard the words "sufficient, necessary" be thrown around when it comes to derivatives. What do they mean?

edit:

I forgot to mention: $f$ is bounded below by zero. It is quadratic in $\beta$.

Masum
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  • No, it does not follow. Just consider the function $f(x)=x^3$. Then $f'(0)=0$, but 0 is not even a local minimum, let alone a global. --- Clearly, you will need stronger conditions on $f$ to make such a claim. – kholli Oct 10 '18 at 14:41
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    If in addition, you can show the function is bounded below, then I think the statement is correct. – user1101010 Oct 10 '18 at 14:43
  • Please see edit! – Masum Oct 10 '18 at 14:44
  • @user9527 Actually, strict inequality still will not be correct. Just consider $f(x)=(x-2)^2(x+2)^2$. Then if will have two minimums with the same value at different points. – kholli Oct 10 '18 at 14:44
  • @kholli But in that case, $f'(x) = 0$ does not have a unique solution, so it's fine right? I am asking that if $f'(x) = 0$ if and only if $x = c$, then $c$ is unique. In your case, that does not apply. – Masum Oct 10 '18 at 14:45
  • @kholli: In the question, it is assumed to have a unique stationary point, i.e., $f'(x)=0$ if and only if $x=c$. – user1101010 Oct 10 '18 at 14:46
  • @Masum: In response to edit: Yes, under those conditions it does hold. (Bounded below gives you the fact that it has a minimum, quadratic tells you that it is a global minimum). – kholli Oct 10 '18 at 14:46
  • @user9527 you still need stronger assumptions. $f(x)=|x|$ has a unique minimum, but the point is not even differentiable. – kholli Oct 10 '18 at 14:48

2 Answers2

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Although I'm not entirely sure I understand what you are asking, I'm going to have to say no: being bounded and having a global local minima is not sufficient to show that the derivative will only be 0 when $x=c$.

For example take the dirty function that I just whipped up in Desmos to show a counterexample

$$ f(x) = \dfrac{x^5-x^4+x^3+x}{20e^{|x|}}+4 $$

enter image description here

This function is bounded both above and below and the local minima you find at $c=-4.756..$ is also equivilent to the global minima, which holds your condition that $f(x) > f(c) \forall x\in \mathbb{R}$. However there is also a local maxima at $c'=5.134..$. By definition the derivative will be zero at both of these points, which runs counter to your original claim that $f'(x)=0$ only when $x=c$

wjmccann
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  • I will remain neutral regarding your answer. The idea of him is totally right! I think he has only a strong intuition, but he does not have a formal education in such a topic. If the function actually achieves a minimum, it does follow automatically that a function has one global minimum if, $f'(x)=0$ if and only if $x=c$. It's a sufficient condition for having a unique global minimizer. The problem is that such a condition does not hold for all problems, as your example showed. – R. W. Prado Sep 02 '22 at 20:16
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Take $f(x)=x-x^3/3$ at $ [-1,1].$

for all $x\in ]-1, 1],$

$$f(-1)<f(x)$$

but $$f'(1)=0$$

hamam_Abdallah
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