# Inequality for harmonic extension : Is $\int_{t\in S^1} |t-\zeta|^{\alpha}p(z,t) |dt| \leq K|z-\zeta|^{\alpha}, 0< \alpha < 1$ for uniform $K$?

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.

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