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I was looking at an old Berkeley preliminary exam problem (Fall,88) stated below;

Prove that a real-valued $C^3$ function $f$ on $\mathbb{R}^2$ whose Laplacian

$$\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}$$

is everywhere positive cannot have a local maximum.

[End of question]

In the solution, it is mention that "... since $f\in C^3$, for $f$ to have a relative maximum its Hessian must have negative eigenvalues ..."

What I know is that if Hessian has negative eigenvalue, then it is negative definite and hence the critical point is a local maximum, but is the converse used above true? What is its justification?

I hope someone can shade a light on this, thanks.

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2 Answers 2

up vote 1 down vote accepted

Let $f$ have a local maximum in $(a,b)$. Then also $f^a$ resp. $f^b$ has a local maximum in $b$ resp. $a$, where $f^a(s) = f(a,s)$, $f^b(s) = f(s,b)$. Consequently the second derivative of $f^a$ in $b$ is $\leq 0$ and analogeous for $f^b$. It follows that $\partial^2_x f (a,b) + \partial^2_y f (a,b) = \partial^2_s f^b(a) + \partial^2_s f^a(b)\leq 0$ in $(x,y)$.

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But why do we need $C^3$? is that redundant? –  KWO Dec 2 '12 at 20:26
1  
I think it indeed is redundant. $\mathcal C^2$ should suffice –  wspin Dec 2 '12 at 20:59

The point is, in a local maximum, the Hessian has to be negative semidefinite (that is necessary, but not sufficient), that is all eigenvalues have to be $\le 0$. But as the trace of the Hessian (that is, the Laplacian) is positive by assumption, this cannot hold.

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