Let $p(z,t)=\frac{1}{2\pi}.\frac{1-|z|^2}{|z-t|^2}$ be the Poisson kernel on the open unit disk $\mathbb{D}$, fix $0<\alpha<1$ . Let $a\in \partial\mathbb{D}=S^1$ be fixed. Then my question is :

what is the limit of $\frac{\int_{S^1}|t-a|^{1 + \alpha}.p(z,t)|dt|}{|z-a|}$ as $z \to a, z\in \mathbb{D}$. I am tending to believe that the limit is zero, because of the following reason :

Take any $0<\alpha' < \alpha$, then $t \mapsto |t-a|^{1 + \alpha'}\in C^{1,\alpha'}(S^1)$.Therfore by Kellog's (or by Kellog-Warschawski's) theorem, its harmonic extension,extended by $H(z)= \int_{S^1}|t-a|^{1 + \alpha'}.p(z,t)|dt|$on $\mathbb{D}$ is $C^{1,\alpha'}(\mathbb{D})$, therefore the harmonic extension is Holder continuous, that is :

$\frac{\int_{S^1}|t-a|^{1 + \alpha'}.p(z,t)|dt|}{|z-a|} \leq M \equiv M(\alpha')$ [Note that, at $a\in \partial \mathbb{D}, H(a)=0$].

But then there should be the 'effect' of this "extra" $\alpha - \alpha'$ in the integration, which, heuristically, should make the limit go to zero. But I am not sure how to prove that ? Is it right at least ? Any help will be highly appreciated, thank you !