Let be a smooth bounded region (open, connected with smooth boundary). Assume solves
with . Prove that if is not identically then .
Note: Justify all your arguments.
Let a solution of . We can find such that (if it's , we consider ). Since is compact and , we have . Let such that ( can't be in because on ). Since is a critical point, we have and by the Taylor's theorem we have where and is the Hessian matrix ( ). Since , we have . We deduce that for all and each eigenvalue of is . Hence
and we get that . We have to show that we can have , but it's a "well-known result" about harmonic functions.
Here's my approach:
Since we have and so we can extend both to as outside . Take be the standard regularization (convolution with an appropiate test function). For small enough we con find such that . Now, integrating by parts in this ball we get
where for the last inequality we use Cauchy-Schwarz and Young's inequality. Take then all converge uniformly to their respective limits in (and the first and last in ) so we have
Take now a sequence with and and take then and their integrals are bounded by the last inequality so we get that . Now, for any we have
By approximation ( ) we can change for to get
which immediately gives .
It should be noted that in case we assume then Green's identities give the result immediately.