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Is it true that given a smooth scalar field f on a domain D , if f attains a maximum (minimum) on the interior of D then the hessian of f evaluated at this max (min) is negative (positive) semi-definite?

I have seen this quoted as a fact but usually when i try to find a proof all that comes up is the second derivative test (given hessian is negative definite then we have max)

is this actually just equivalent to the test?

EDIT:

if we consider an approximation via taylor's theorem we have (near $ a $ ) :

$$ f(x) = f(a) + Df(a)^{T} (x-a) + (x-a)^{T} D^2 f(a) (x-a) + o(|x-a|^3)$$

now if $a$ is max then$ f(x) \le f(a) $ so $ D^2 f(a) $ is negative semidef ?

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3 Answers 3

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Yes, at a local maximum the Hessian of a smooth (real) function will be negative semi-definite (and equivalently the Hessian will be positive semi-definite at a local minimum). If this is hard to find, being a weak converse of the second derivative test, it's likely because proving it requires some linear algebra material not yet covered at the time of a 3rd quarter in calculus where multivariable concepts are introduced.

Consider that the Hessian is a symmetric (real) matrix, and thus has a complete basis of eigenvectors. At a local maximum the function will have, on each line passing through the maximum point, a familiar one-dimensional local maximum. If the Hessian were not negative semi-definite, it would have a line (corresponding to an eigenvector of a positive eigenvalue) along which the restricted function would have a concave up appearance. But this would contradict the point being a local maximum.

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  • $\begingroup$ @hardmath Can we say the other way around, that is if $x$ is negative semidefinite then is it local max? $\endgroup$
    – OGC
    Commented Oct 16, 2018 at 3:18
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    $\begingroup$ @OGC: We need more than just the Hessian negative semidefinite; typically one needs the vanishing of the first derivatives and the Hessian negative definite to get sufficient conditions for a local maximum. If you are interested in the semi-definite cases, see this previous Question and this one. If you need further info, it's probably worth asking a Question. $\endgroup$
    – hardmath
    Commented Oct 16, 2018 at 4:03
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I've removed a previous answer that didn't answer the question.

Edit: Yes, this follows from the proof of Theorem 9.6(c) on p. 311 of Calculus, Vol. 2, by Apostol.

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  • $\begingroup$ You're right. I'm going to remove my answer. $\endgroup$
    – user120974
    Commented Jan 13, 2014 at 11:23
  • $\begingroup$ is this text available online? i think i have proven the statement using taylor's theorem, but would be good to see this. thanks! $\endgroup$
    – Derpins
    Commented Jan 13, 2014 at 11:31
  • $\begingroup$ I don't think so, but it is a consequence of Taylor's formula. $\endgroup$
    – user120974
    Commented Jan 13, 2014 at 11:34
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Here is some explanation that may clarify what happens: For every $u\in \mathbb R^2$, if we define $d^2_uf$ to be the second derivative of $f$, then $d^2_uf <0 $. By direct calculations, we obtain $$u^T Hess f u = d^2_uf<0. $$ So by definition $Hess f$ is negative definite at the point maximum happens.

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