Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\zeta\in S^1$(unit circle in the complex plane) and $z\in \mathbb{D}$. Fix $0< \alpha < 1$. Then, is the following true ?

(Question 1) Let $p(z,t) = \frac{1}{2\pi}.\frac{1-|z|^2}{|z-t|^2}$ be the Poisson kernel. Then is $\int_{t\in S^1} |t-\zeta|^{\alpha}p(z,t) |dt| \leq K|z-\zeta|^{\alpha}$ for some $K>0$ independent of $z\in \mathbb{D}, \zeta \in S^1$ ?

Actually, by using weak maximum principle on $|z-\zeta|^{\alpha} - H(z,\zeta)$, where $H(z,\zeta)= \int_{t\in S^1} |t-\zeta|^{\alpha}p(z,t) |dt| $ is the complex harmonic extension of $t\to |t-\zeta|^{\alpha}$ at $z$, I am getting $\int_{t\in S^1} |t-\zeta|^{\alpha}p(z,t) |dt| \geq |z-\zeta|^{\alpha}$, which is quite the contrary ! Any help or suggestions ? Thank you !

(Question 2)And also, as a seemingly related question can we say that if $f\in C^{0,\alpha}(S^1)$, then its complex harmonic extension $\int_{t\in S^1} f(t).p(z,t)|dt|$ is $C^{0,\alpha}(\mathbb{D}) ?$ Thanks.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.