Let $f$ be an anylytic function on the unid disk $|z|<1$. It is well known that $$\left (\int_0^{2\pi}f(e^{i\theta})d \theta \right)^2 \geq 4\pi \iint_{|z|<1} |f(r e^{i\theta})|^2r dr d \theta.$$

I wonder if the constant $4\pi$ cound be improved on an annulus $a<z<1$. More precisely, does the following inequality

$$\left (\int_0^{2\pi}|f(e^{i\theta})|d \theta +\int_0^{2\pi}|f(ae^{i\theta}|)d \theta \right)^2 \geq C \iint_{a<|z|<1} |f(r e^{i\theta})|^2r dr d \theta$$

hold for some constant $C(a)> 4\pi$ independent of $f$? Can $C(a)$ be computed in terms of $a$? This seems to be a classical problem but I could not find a reference, and was not able to prove it after trying for a couple of days.

  • $\begingroup$ I think you mean absolute values in the integrals, right ? Secondly, the inequality you ask is for $f$ analytic in the disc or in the annulus ? $\endgroup$ – an_ordinary_mathematician Apr 6 at 9:15
  • $\begingroup$ Yes, I just edited the question. $\endgroup$ – MathLearner Apr 6 at 13:37

You can probably prove that $$ \Big( \int_\mathbb{A_r} |f(z)|^2 \frac{dxdy}{\pi(1-r^2)} \Big)^{1/2} \leq \int_{ \mathbb{T}} |f(e^{i\theta})| \frac{d\theta}{2\pi}+\int_{ \mathbb{T_r}} |f(re^{i\theta})| \frac{d\theta}{2\pi r}, \,\,\, \forall f\in H^1(\mathbb{A_r}). $$ Where $\mathbb{A_r} $ is the annulus of internal radius $r$ with boundary $\mathbb{T} \cup \mathbb{T_r}$, and $H^1(\mathbb{A_r})$ is the Hardy space on the annulus, by following step by step the proof of Vukotic in "The isoperimetric inequality and a theorem of Hardy and Littlewood, American mathematical monthly". The only tools you need is the Parseval formula and the inner outer factorization for the Hardy space in an annulus which you can find in "Sarason, The Hp spaces of an annulus, Memoirs or the AMS". If you have problems recustructing the proof, maybe I can help.

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  • $\begingroup$ This looks great. Do you think it would be possible to prove a similar inequality in which the right hand side is exactly what I have in the statement of the question, i.e. to remove $1/r$ in the right hand side of the inequality in your answer? Probably with a different weight in the left hand side. $\endgroup$ – MathLearner Apr 6 at 13:46
  • $\begingroup$ Of course it is true if you replace $2\pi$ on the first term on the rhs by $2\pi r $, but in this case the inequality is not sharp. $\endgroup$ – an_ordinary_mathematician Apr 6 at 15:45
  • $\begingroup$ It kind of bothers me that as $r\rightarrow 0$ we do not get the classical isoperimetric inequality on a disk as the limit of the isoperimetric inequality on an annulus. Does this mean the inequality in your answer is not sharp? $\endgroup$ – MathLearner Apr 7 at 0:27

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