Taylor expansions of harmonic functions.

Question

Let $D \subset \mathbb{R}^2$ be the unit open disc. Note that any harmonic function on $D$ is real analytic. How can one prove that there exits a constant $C>0$ such that the Taylor expansion of any harmonic function on $D$ converges on the open disc of radius $C$? In other words, there is a uniform radius of convergence.

Answer 1

The story is the same as for holomorphic functions. Given a harmonic function $u$ on a disk, write it as $u=\operatorname{Re}f$ with $f$ holomorphic. The radius of convergence of the Taylor series of $f$ centered at $a$ is equal to the supremum of all $r$ such that $f$ has a holomorphic extension to $|z-a|<r$. Let $R$ be this supremum.

(Proof of the above claim: the Cauchy integral estimate for the coefficients yields $$|c_n|\le \frac{1}{ (R-\epsilon)^n}\max_{|z-a|=R-\epsilon } |f(z)|$$ which tells you that the radius of convergence is at least $R$. Conversely, if the power series converges in some disk, it defines a holomorphic function there. )

The Taylor series of $u$ centered at $a$ is $$u(z)=\sum_{n=0}^\infty \operatorname{Re}\left( c_n (z-a)^n\right)\tag{1}$$ where the $n$th term is a homogeneous polynomial in $x$ and $y$ of degree $n$. The maximum of said polynomial on the circle $|z-a|=r$ is $|c_n|r^n$. Thus, on a disk of radius greater than $R$ the terms fail to converge to zero. It follows that the radius of convergence of (1) is the same ($R$) as for the holomorphic function $f$.

On any compact subset $K$ of the domain of $u$ this radius is bounded from below by the distance from $K$ to the boundary of the domain. In general, it is not greater than that. For example, the function $u(z)=\operatorname{Re}(1-z)^{-1}$ is harmonic on the open unit disk. By the above, its Taylor series at $a=1-\epsilon$ has radius of convergence equal $\epsilon$. Since $\epsilon$ can be arbitrarily small, you do not have a uniform lower bound on the radius of convergence for all centers $a$ with $|a|<1$.