Let $u$ be a real valued harmonic function on the open unit disc $D_1(0) \subseteq \mathbb{C}$. Show that there exists a sequence of real valued harmonic polynomials that converges uniformly on compact subsets of $D_1(0)$ to $u$.
Now, consider the harmonic function $u = \ln|z|$ on the open set $\Omega = \{ z : 1<|z| < 2\}$. Is it possible to find a sequence of harmonic polynomials which converge uniformly on compact subsets of $\Omega$ to $u$?
I am preparing for my qualifying exam in complex analysis and this question has come up. I will admit that I have not made much progress, but I do have a few thoughts, which I give below. My hope is to receive a few nice hints, and then I will post a solution of my own.
From my complex analysis course, I know that the convexity of $D_1(0)$ allows us to define a harmonic conjugate $v$ for $u$ on $D_1(0)$. Then $f \equiv u + iv$ is analytic on $D_1(0)$. Then maybe I can use a theorem from complex analysis to produce this sequence of harmonic polynomials?
For the second question, I am assuming that the answer is no, likely due to the fact that we cannot define a harmonic conjugate for $ u = \ln|z|$ on the set $\Omega$ under investigation.
Hints are greatly appreciated.
Best Answer
For the first question, following your thoughts, let $f$ be holomorphic(i.e. analytic) on $D_1(0)$, such that ${\rm Re}f=u$, i.e. $u$ is the real part of $f$. Then the Taylor expansion of $f$ $$f(z)=\sum_{k=0}^\infty a_kz^k$$ converges to $f$ uniformly on compact subsets of $D_1(0)$(and hence so do the corresponding real and imaginary parts). For every $n\ge 0$, let $$u_n(z)={\rm Re}\big(\sum_{k=0}^n a_kz^k \big).$$ Since $u_n$ is the real part of a (holomorphic) polynomial of $z=x+iy$, $u_n$ is a real valued harmonic polynomial of $x$ and $y$. Moreover, $u_n$ converges to the real part of $f$, which is just $u$, uniformly on compact subsets of $D_1(0)$.
For the second question, assume that there is a sequence of real valued harmonic polynomials $(u_n)$, such that it converges to $\ln|z|$ uniformly on compact subsets of $\Omega$. Note that $u_n$ is in fact harmonic on $D_2(0)$, so by maximum/minimum principle, it is easy to see that $u_n$ converges uniformly on compact subsets of $D_2(0)$, and let us denote the limit by $u$. Then using Poisson integral formula or otherwise, it is easy to see that $u$ is harmonic on $D_2(0)$. However, since $\ln|z|$ is not harmonic at $0$, it contradicts to the fact that $u=\ln|z|$ on $\Omega$, so such $(u_n)$ does not exist.