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Let $E \in \mathbb{R}^n$ be compact. For simplicity, let's say $E$ is a closed ball. Let $f$ be a smooth function $f: \mathbb{R}^n \rightarrow \mathbb{R}$. We know that global maximum exists in $E$, and it is either on $\partial E$ or in $E^o$. Then, is the following true?

If global maximum takes place at $p \in E^o$, then $p$ must be a critical point, i.e. $Df(p) = 0$.

I think this is false as I've seen some examples that a function with a unique local extrema (wlog take min) with unique critical point fails to attain global minimum at this point, given that the function is defined in open set. (Unique critical point does not imply global maximum/global minimum) Things get really rough in $\mathbb{R}^n$, and I would appreciate if you all can provide some insight!

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  • $\begingroup$ The statement is true. By $D$, I assume you mean the gradient. $\endgroup$
    – Oliver Jones
    Commented Oct 11, 2019 at 8:25
  • $\begingroup$ @OliverJones Would you mind if you can elaborate your answer, why a global maximum needs to be a local maximum? The idea I have is, can't you do some twisting and bendings on the plane to force the maximum to have non-zero Derivative $\endgroup$
    – James C
    Commented Oct 11, 2019 at 8:28
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    $\begingroup$ You're not being clear. A global maximum must be a local maximum; just use the definition. You're asking why the gradient is zero at local extrema? $\endgroup$
    – Oliver Jones
    Commented Oct 11, 2019 at 8:31
  • $\begingroup$ for some $r>0$, and $\forall x \in B(p,r): f(x) \leq f(p)$ $\endgroup$
    – James C
    Commented Oct 11, 2019 at 8:32
  • $\begingroup$ Oh ok. I got it. Thank you very much :) $\endgroup$
    – James C
    Commented Oct 11, 2019 at 8:35

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A global maximum is also a local maximum and $Df(p)=0$ is true for any local maximum.

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