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

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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,896
Here is this week's POTW:

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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 $\partial \Bbb D$.
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Remember to read the POTW submission guidelines to find out how to submit your answers!
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,034
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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