Proving that a harmonic function is bounded on a open connected set Let $u:G\to \mathbb{R}$ be harmonic and $K\subset G$ compact where $G$ is open and connected. If $u\leq c$ on $K^c$ then I want to prove that $u\leq c$ on $G$ where $c\in \mathbb{R}$. Here is my attempt.
Since $u$ is continuous on the compact set $K$, it attains a maximum on $K$. By the maximum modulus principle, $u$ (restricted to $K$) attains its maximum on the boundary of $K.$ Similarly, $u$ (restricted to $K^c$) attains its maximum on the boundary of $K^c$. 
 These are all the ideas that I have for now but I just cannot figure out how to bring these together. Any help is appreciated. Thanks

Here's a start: let $v:G\to\mathbb{R}$ be the harmonic conjugate of $u$ so that the function $f:G\to \mathbb{R}$ defined by $f=u+iv$ is analytic. Now set $g(z)=e^{f(z)}$ and observe that $|g(z)|=|e^{u(z)}||e^{iv(z)}|=|e^{u(z)}|$. Since $u$ is bounded on $G\setminus K$, what does that tell you about $g$'s behaviour on the same? 
I was just working on this problem for my qualifying exams, so I'm glad you asked! 
PS: Harmonic does not imply analytic. A function is analytic if it satisfies the Cauchy-Riemann equations. A function $u$ is harmonic if $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$. 
