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

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#### Euge

##### MHB Global Moderator
Staff member
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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#### Jameson

Staff member
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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