0
$\begingroup$

If $f(x)$ is bounded and uniformly continous on $\mathbb{R}$, and $u(x+iy)$ is harmonic function on the upper half plane $ {\mathbb{H}}$ defined by the convolution of the possion kernel and f(x), then $u(x+iy)$ is uniformly convergent to $f(x)$ when $y\to 0$.

In Axler, Bourdon and Ramey's book “Harmonic Function Theory”, they said that the converse is also true. The accurate statement is exericise 7.4.

Suppose that $u$ is a harmonic function on the upper half plane ${\mathbb{H}}$, and in the harmoninc Hardy space $h^p(\mathbb{H})$ for $p\in [1,\infty]$. If $u_y(x)=u(x+iy)$ converge uniformly on $\mathbb{R}$ when $y \to 0$, then $u$ can be extended to a bounded and uniformly continous harmonic function on the closed upper half plane $\overline{\mathbb{H}}$.

In the case $p=\infty$, clearly $u$ is extended to a bounded continous harmonic function on $\overline{\mathbb{H}}$.

But I can't prove that $u$ is uniformly continous.

Could you show me some hints or references? Thanks a lot.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.