# proof of weak maximum principle for real domains

First, we show that, if $\Delta f>0$ (where $\Delta$ denotes the Laplacian  on $\mathbb{R}^{d}$) on $K$, then $f$ cannot attain a maximum on the interior of $K$. Assume, to the contrary, that $f$ did attain a maximum at a point $p$ located on the interior of $K$. By the second derivative test  , the matrix of second partial derivatives  of $f$ at $p$ would have to be negative semi-definite. This would imply that the trace of the matrix is negative. But the trace of this matrix is the Laplacian, which was assumed to be strictly positive on $K$, so it is impossible for $f$ to attain a maximum on the interior of $K$.

Next, suppose that $\Delta f=0$ on $K$ but that $f$ does not attain its maximum on the boundary of $K$. Since $K$ is compact, $f$ must attain its maximum somewhere, and hence there exists a point $p$ located in the interior of $K$ at which $f$ does attain its maximum. Since $K$ is compact, the boundary of $K$ is also compact, and hence the image of the boundary of $K$ under $f$ is also compact. Since every element of this image is strictly smaller than $f(p)$, there must exist a constant $C$ such that $f(x) whenever $x$ lies on the boundary of $K$. Furthermore Since $K$ is a compact subset of $\mathbb{R}^{d}$, it is bounded. Hence, there exists a constant $R>0$ so that $|x-p| for all $x\in K$.

Consider the function  $g$ defined as

 $g(x)=f(x)+(f(p)-C){|x-p|^{2}\over R^{2}}$

At any point $x\in K$,

 $g(x)

In particular, if $x$ lies on the boundary of $K$, this implies that

 $g(x)

Since $g(p)=f(p)$ this inequality  implies that $g$ cannot attain a maximum on the boundary of $K$.

This leads to a contradiction. Note that, since $\Delta f=0$ on $K$,

 $\Delta g={d(f(p)-C)\over R^{2}}>0$

which implies that $g$ cannot attain a maximum on the interior of $K$. However, since $K$ is compact, $g$ must attain a maximum somewhere on $K$. Since we have ruled out both the possibility that this maximum occurs in the interior and the possibility that it occurs on the boundary, we have a contradiction. The only way out of this contradiction is to conclude that $f$ does attain its maximum on the boundary of $K$.

Title proof of weak maximum principle for real domains ProofOfWeakMaximumPrincipleForRealDomains 2013-03-22 14:35:21 2013-03-22 14:35:21 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Proof msc 30F15 msc 31B05 msc 31A05 msc 30C80