On a limit at the boundary of $\mathbb{D}$ related to complex and harmonic analysis - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T13:24:31Z http://mathoverflow.net/feeds/question/108589 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108589/on-a-limit-at-the-boundary-of-mathbbd-related-to-complex-and-harmonic-analys On a limit at the boundary of $\mathbb{D}$ related to complex and harmonic analysis Analysis Now 2012-10-02T01:47:29Z 2012-10-02T03:11:00Z <p>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 $0&lt;\alpha&lt;1$ . Let $a\in \partial\mathbb{D}=S^1$ be fixed. Then my question is :</p> <p>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 :</p> <p>Take any $0&lt;\alpha' &lt; \alpha$, 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 :</p> <p>$\frac{\int_{S^1}|t-a|^{1 + \alpha'}.p(z,t)|dt|}{|z-a|} \leq M \equiv M(\alpha')$ [Note that, at $a\in \partial \mathbb{D}, H(a)=0$].</p> <p>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 !</p> http://mathoverflow.net/questions/108589/on-a-limit-at-the-boundary-of-mathbbd-related-to-complex-and-harmonic-analys/108591#108591 Answer by Alexandre Eremenko for On a limit at the boundary of $\mathbb{D}$ related to complex and harmonic analysis Alexandre Eremenko 2012-10-02T02:56:07Z 2012-10-02T03:11:00Z <p>The limit is not zero; it does not exist.</p> <p>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 $z_k&lt;1$, and<br> $$\frac{u(1)-u(z_k)}{1-z_k}\; \quad\to c>0.$$ 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$, that is very close to the circle. So the limit does not exist. </p> <p>SECOND proof. WLOG$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$0 &lt; w &lt; v$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\$.</p>