# Maximum principle for harmonic functions on unbounded domain

I am learning PDE by myself now. I am considering converted the problem to bounded domain to use the strong maximum principle.

### My attempt:

Using the $$\lim u(x)=0$$, then exists $$\epsilon$$ and $$N$$, such that if $$x>N$$, $$|u(x)|<\epsilon$$. Now consider the domain $$1<|x|. By maximum principle, $$u$$ get its maximum on the boundary. if it get its maximum on $$|x|=1$$, then we are done. Otherwise, if $$u$$ get its maximum on $$|x|=N$$, and suppose it is $$p$$. Then exists $$N_0>N$$ such that if $$x>N_0$$, $$|u(x)|<\epsilon/2$$. So considering the domain $$1<|x|, then $$p$$ is a maximum in the domain, so by maximum principle, $$u$$ is constant. So $$u$$ attains its maximum on $$|x|=1$$. So it is proved.

I have been thinking about it for nearly 2 hours and I feel something is wrong with my proof in using the limit. I really need some help about the proof.

Can anyone help me? I will be very grateful and thanks so much!

• Looks OK to me. That said, I think you should comment that if $u$ is constant then the result is trivially true. – Ian Oct 7 '14 at 1:07

Let $M=\max_{\partial\Omega }|u|$. We want to prove that $|u|\le M$ in $\Omega$. Fix $x_0\in\Omega$. Given $\epsilon>0$, let $R$ be such that $R>|x_0|$ and $|u(x)|< \epsilon$ when $|x|\ge R$; such $R$ exists because $u\to 0$ at infinity.
Applying the maximum principle to $u$ (and to $-u$) on $\{x:1<|x|<R\}$, conclude that $|u(x_0)|\le \max(M,\epsilon)\le M+\epsilon$. Since $\epsilon>0$ could be arbitrarily small, $|u(x_0)|\le M$. Done.