Assume that $\lim_{(nt) |z|\to 1}|f(z)|(1-|z|)^p=0$, where $f$ is analytic in the unit disk and $p>0$,where $(nt)|z|\to 1$ nontangentially. Does this implies that $\lim_{|z|\to 1}|f(z)|(1-|z|)^p=0$ uniformly?
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$\begingroup$ Why the mysterious $nt$ and why is it the derivative? Are you asking if pointwise convergence implies uniform convergence in this case? $\endgroup$– Matthias LudewigCommented Oct 1, 2013 at 16:03
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$\begingroup$ is it related to bloch space? $\endgroup$– KoushikCommented Oct 1, 2013 at 16:08
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$\begingroup$ No, when I said $|z|\to 1$ uniformly I had in mind that $z\to e^{it}$ for some $t$ and throughout the unit disk. Nontangentialy means that $z$ also tends to $e^{it}$ but inside an fixed angle. $\endgroup$– user36162Commented Oct 1, 2013 at 16:10
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$\begingroup$ &Koushik: It is related to little Bloch space. $\endgroup$– user36162Commented Oct 1, 2013 at 16:11
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$\begingroup$ So $(nt)$ just means "nontangential"? $\endgroup$– András BátkaiCommented Oct 1, 2013 at 19:01
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1 Answer
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No, it does not. There can be a region $D$ in the disc with one boundary point on the circle, say $1$, which approaches this point tangentially. And a function which is bounded outside $D$ but grows arbitrarily fast in $D$. To non-tangential limits of your expression at all points are zero, but there is no uniformity, and the growth is arbitrarily high. Such function can be constructed in many ways, using for example approximation theorems, or Cauchy integrals.
answered Oct 2, 2013 at 2:47