I have recently been trying to understand the theory regarding harmonic extensions in $\mathbb R^n$. I have, however, had some difficulties to find the kind of results I am looking for. For that reason I was hoping to get an answer to the below question here.

Let $g:B(x, 2r) \to \mathbb R$ be real analytic and suppose $u$ is the harmonic function in $B(x, r)$ such that $u=g$ on $\partial B(x, r)$.

Now my question is whether it is possible to extend $u$ to a harmonic function $\tilde u$ in $B(x, (1+\delta)r)$ for some $\delta >0$. In other words, is it possible to find a function $\tilde u$ such that $\tilde u = u$ in $\overline{B(x,r)}$ and $\tilde u$ is harmonic in $B(x, (1+\delta)r)$?

  • $\begingroup$ This can't be true by Hopf's lemma. $\endgroup$
    – k3thomps
    May 15, 2014 at 17:59

1 Answer 1


The answer is No. Consider the function $g(x,y)=\sum_{k=0}^{\infty}r^{k}2^{-k}cos{n_k\varphi}$, where $(r, \varphi)$ are the usial polar coordinates in the plane $(x,y)$. Then $g(x,y)$ is real analytic in the disk $B(0, 2)$. The harmonic extension of $g(x,y)$ from the unit circle is given by $u(x,y)=\sum_{k=0}^{\infty}r^{n_k}2^{-k}cos{n_k\varphi}$. If $u(x,y)$ is harmonic in $B(0, 1+\delta)$ for some $\delta>0$, then the function $f(z)=\sum_{k=0}^{\infty}z^{n_k}2^{-k}$ is an analytic function of $z=x+iy$ in $B(0, 1+\delta)$. But this fails for $n_k=k^k$.

Andrei Pokrovskii, pokrovsk@imath.kiev.ua


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