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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 !

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The numerator is a harmonic function $u$ with $C^{1,\alpha}$ boundary data, so we do expect $C^{1,\alpha}$ regularity from the boundary; in particular, if we subtract the linear part of $u$ at $a$, we expect that the quotient tends to $0$. However, in the expression you give we haven't subtracted the gradient plane, which in general has a normal component (as in the nice examples of Alexandre below). Since $u$ and its tangential derivative at $a$ are $0$, the quotient will tend to $0$ if we travel along the boundary but will be nonzero if we approach from the interior. –  Connor Mooney Oct 3 '12 at 4:43
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up vote 3 down vote accepted

The limit is not zero; it does not exist.

FIRST proof. WLOG let $a=1$. Let $-u(z)$ be your Poisson integral, (in the numerator of your formula) it is a negative harmonic function in the disc, continuous in the closed disc, $u(1)=0$, and negative at every other point of the circle. Let $M(r)=max_{|z|=r}u(z),\; 0\leq r\leq 1$. This is a strictly negative, increasing function on $(0,1)$, and $M(1)=1$. It is known that $M(r)$ is convex with respect to the logarithm, that is $$r\frac{dM(r)}{dr}$$ is increasing. This is called Hadamard's Three Circles Theorem. Thus $M'(1)>0$. Now as $u(re^{i\theta})$ is an even function of $\theta$ decreasing on $(0,\pi)$, we conclude that $M(r)=u(r)$. This means that there exists a sequence $z_k\to 1$,such that $z_k<1$, and
$$\frac{u(1)-u(z_k)}{1-z_k}\; \quad\to c>0.$$ So on this sequence your limit $$\frac{-u(z_k)}{|1-z_k|}\quad =c.$$ On the other hand it is clear that your limit is zero on sequences which tend to $1$ "tangentially$, that is very close to the circle. So the limit does not exist.

SECOND proof. WLOG $a=1$. Let $v$ be your Poisson integral in the numerator. Let $w$ be the Poisson integral of the same function but replaced by $0$ on the right half of the circle. Then $0 < w < v$ in the open disc. But $w(z)=0$ on an arc of a circle near $1$, so by the Symmetry Principle, $\partial w/\partial r \neq 0$ at the point $1$. As $w$ is positive and $w(1)=0$, this derivative is negative. So $v(r)>w(r)>c(1-r)$ for some $c>0$.

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Thanks Prof. Eremenko, that was helpful. I will soon post a related question regarding this which still confuses me. Please don't feel obligated to reply, but if you do, I will appreciate it. –  Analysis Now Oct 23 '12 at 15:03
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