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 $f(x) < C < f(p)$ 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 $|x - p| < R$ 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 $x \in K$ , $$ g(x) < f(x) + f(p) - $$ In particular, if $x$ lies on the boundary of $K$ , this implies that $$ g(x) < f(p $$ 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 $K$ , $$ \Delta g = {d (f(p) - C) \over R^2} > $$ 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 $K$ .