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For a multivariate smooth function $f: [0,1]^m\rightarrow \mathbb{R}$, does the existence of only one point where the Jacobian $=0$, plus the fact that the Hessian is PSD at that point (local convexity) imply that it is a global minimum? Suppose the minimum is not attained at boundaries (In other words, f evaluated at the boundaries will be larger than the local minimum).

My thoughts: I saw some counterexamples in multivariate case but not sure this would fail in general with the additional boundary assumption. The proof in the bounded univariate case seems well extendable in this case, as long as we take care of the "boundaries". Intuitively, what could possibly go wrong if boundary is not the issue?

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    $\begingroup$ Yes it is correct. $\endgroup$
    – Kroki
    May 18 at 18:46

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The answer is Yes! In fact, if $\Omega$ is an open in $\mathbb R^m$ and $f: \Omega \to \mathbb R$ a $C^2$ function, then $x$ is a local minimum of $f$ implies the two conditions: $\nabla f(x) = 0$ and $\nabla^2 f(x)\succeq 0$.

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  • $\begingroup$ Thanks again for your confirmation, Youem! I was only a little worried about extending the local minimum to a global minimum in the multivariate case because although the proof used in the univariate case seems extendable, maybe there's something I'm missing when dimension gets higher... The counterexamples given in this example seems relavant only when ''boundaries'' are in fact smaller. math.stackexchange.com/q/596625/391841 $\endgroup$
    – Kryvtsov
    May 18 at 19:05

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