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Problem of the Week #200 - Tuesday, March 29, 2016

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MHB Global Moderator
Staff member
Jun 20, 2014
Here is this week's POTW:

Let $\Bbb D$ denote the open unit disc in the complex plane. Given a holomorphic function $f$ on $\Bbb D$, define

$$N_p(f) := \sup_{0 < r < 1} \left[\frac{1}{2\pi}\int_{-\pi}^\pi \lvert f(re^{i\theta})\rvert^p\, d\theta\right]^{1/p},\quad 0 < p < \infty$$

The $p$th Hardy space on $\Bbb D$ consists of all holomorphic functions $f\in \mathcal{O}(\Bbb D)$ for which $N_p(f) < \infty$. Prove the following statement:

If $f\in H^p(\Bbb D)$ and $1\le p < \infty$, then for every $z\in \Bbb D$,

$$\lvert f(z)\rvert \le \rho(z,\partial \Bbb D)^{-2/p}N_p(f)$$

where $\rho(z,\partial \Bbb D)$ is the distance from $z$ to the boundary $\partial \Bbb D$.

Remember to read the POTW submission guidelines to find out how to submit your answers!


Staff member
Jan 26, 2012
No one answered this week's POTW correctly. Here is the solution:

The inequality holds even for $0 < p < \infty$. Fix $z\in \Bbb D$. Since $\lvert f\rvert^p$ is subharmonic on $\Bbb D$,

$$\lvert f \rvert^p \le \frac{1}{2\pi}\int_0^{2\pi} \lvert f(z + re^{i\theta})\,d\theta\quad (0 <r <1)$$

Let $\rho <d(z,\partial \Bbb D)$. Then

$$\frac{1}{2}\rho^2 \lvert f\rvert^p \le \int_0^\rho r\,dr\int_0^{2\pi} |f(z + re^{i\theta})|^p\, d\theta = \frac{1}{2\pi}\iint_{B(z,\rho)} \lvert f(x + yi)\rvert^p\, dx\, dy$$
$$ \le \frac{1}{2\pi}\iint_{\Bbb D} |f(x +yi)\rvert^p\, dx\, dy = \frac{1}{2\pi} \int_0^1 r\, dr \int_0^{2\pi} \lvert f(re^{i\theta})\, d\theta \le \frac{1}{2}[N_p(f)]^p$$

Hence, $\lvert f(z)\rvert \le R^{-2/p}N_p(f)$. Letting $R\to d(z,\Bbb D)^{-}$ we obtain the result.
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