proof of weak maximum principle for real domains

First, we show that, if $\mathrm{\Delta }f>0$ (where $\mathrm{\Delta }$ denotes the Laplacian on ${ℝ}^d$) 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 $\mathrm{\Delta }f=0$ 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 whenever $x$ lies on the boundary of $K$. Furthermore Since $K$ is a compact subset of ${ℝ}^d$, it is bounded. Hence, there exists a constant $R>0$ so that for all $x\in K$.

Consider the function $g$ defined as

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

At any point $x\in K$,

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

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 $\mathrm{\Delta }f=0$ on $K$,

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

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
Canonical name	ProofOfWeakMaximumPrincipleForRealDomains
Date of creation	2013-03-22 14:35:21
Last modified on	2013-03-22 14:35:21
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	6
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 30F15
Classification	msc 31B05
Classification	msc 31A05
Classification	msc 30C80