# Showing a harmonic function takes its minimum on the boundary

Let $u$ be a harmonic function on the bounded region $A$ and continuous on $Cl(A)$. Then $u$ takes its minimum only on $Bd(A)$ unless $u$ is constant

I am trying to show this. Can we let $v = -u$ and apply the maximum modulus principle?

• Yes, your approach is the standard way to do it. (But it's not called the maximum modulus principle for harmonic functions, just the maximum principle.) – mrf Oct 20 '15 at 7:11

Let $f(z)=u+vi$ and be analytic in $A$. Then $g(z)=e^{f(z)}$ is also analytic and not zero in $A$.
Now since $|g(z)|=e^u$, if $u$ takes minimum inside $A$, then so is $|g(z)|$, which violates minimum modulus principle of analytic function. So $u$ must take minimum on boundary of $A$ unless it is constant.
• You seem to be assuming that $A$ is simply connected (and a subset of $\mathbb{R^2}$ or $\mathbb{C}$ which I guess is reasonable, given the tag). The result is true in more general settings. – mrf Oct 20 '15 at 7:12