Harmonic Maximum modulus

Question

So, i am starting to solve some exercises of complex analysis, and i am a little rusty, so if anyone could help me with this exercise. I think that if i just can prove the mean value theorem for harmonic functions, that would be enough...but i am getting some troubles with that

Let $u$ be an harmonic function on $\Omega=\{z:|z|<R\}$ and continuous on the closure of $\Omega$ such that $u\equiv 0$ on $\partial\Omega$. Prove that $u\equiv 0$ on $\Omega$.

Answer 1

You might want to look at Green's identities. You can alternatively recall that harmonic functions are locally the real part of a holomorphic function.

Answer 2

A corollary to maximum modulus theorem, (or to the open mapping theorem) says that a harmonic function attains its maximum and minimum value on the boundary of a domain. Since it is zero on the boundary, the rest is obvious.

Answer 3

The Divergence Theorem says that $$ \int_{B(r,x)}\Delta f(y)\,\mathrm{d}y=\int_{\partial B(r,x)}\nabla f(y)\cdot\vec{n}\,\mathrm{d}\sigma(y)\tag{1} $$ The right side of $(1)$ is the area of $\partial B(r,x)$ times the rate of change of the mean value of $f$ over $\partial B(r,x)$. Since $\Delta f(y)=0$ for harmonic functions, we get that the mean value of $f$ over $\partial B(r,x)$ is constant. Thus, we have the Mean Value Property.

The Mean Value Property implies $$ |f(x)|\le\frac1{|B(r,x)|}\int_{B(r,x)}|f(y)|\,\mathrm{d}y\tag{2} $$ Suppose that $|f|$ attains its maximum at $x_0$ in the interior of $K$ and $B(r,x_0)\subset K$. The existence of a $y\in B(r,x_0)$ so that $|f(y)|\lt|f(x_0)|$ contradicts $(2)$ since $f$ is continuous. Therefore, $|f(y)|=|f(x_0)|$ for all $y\in B(r,x_0)$. By covering a path from $x_0$ to any $x$ in the interior of $K$ with a sequence of balls such that each ball contains the center of the next, we get that $|f(x)|=|f(x_0)|$ for all $x\in K$. Thus, we have the Maximum Modulus Principle.

Therefore, if a harmonic function is $0$ on $\partial K$, it must be $0$ on all of $K$.