Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $u\in C^2(D)$, $D$ is the closed unit disk in $\mathbf{R}^2$. Assume that $\Delta u>0$. Show that $u$ cannot have a maximum point in $D\setminus\partial D$.

This statement is in a calculus book, after the discussion of extremal values of multivariable functions. So my guess is that I should use the Hessian of $u$ somehow. I started to proof indirectly. Assume that $(x_0,y_0)\in D\setminus\partial D$ is a maximum point. Then $\frac{\partial u}{\partial x}(x_0,y_0),\,\frac{\partial u}{\partial y}(x_0,y_0)=0$. Now I want to investigate the positive/negative definiteness of Hessian and deduce contradiction, but I got stuck.

share|cite|improve this question
If this is a maximum, what can you say about definiteness of the Hessean matrix? And starting at the other hand, how is $\Delta u$ computed from the Hesse matrix? – Hans Engler Nov 28 '12 at 18:06
@HansEngler $\Delta u=tr(Hesse)=\lambda_1+\lambda_2$ (sum of eigvalues). If it is a maximum, then Hesse is negative (semi?)definite, it implies $\lambda_1<0,\,\lambda_2<0$ (or $\leq 0$ ?). So $\Delta u\leq 0$, a contradiction, if I'm correct. Thanks. – vesszabo Nov 28 '12 at 19:35
up vote 2 down vote accepted

To not leave this look unanswered: suppose $a\in \Omega$ is a point of local maximum. (I change the notation to reserve $D$ for derivatives.) Since $u$ is $C^2$ smooth, we have second-order Taylor expansion $$u(x)=u(a)+Du(a) (x-a)+ \frac12 (x-a)^T D^2u(a) (x-a)+o(|x-a|^2)$$ The maximality implies that $Du(a)=0$ and $D^2u(a)$ is negative semidefinite. Therefore, the trace of $D^2u(a)$ is non-positive. This trace is $\Delta u(a)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.