What is the Upper Bound for a Holomorphic Function on the Hardy Space?

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    2016
In summary, the Hardy Space, denoted as H^p, is a space of holomorphic functions defined on the unit disk in the complex plane with finite p-norm. The upper bound for a holomorphic function on this space is calculated using the Cauchy integral formula and plays a crucial role in understanding the behavior of these functions. It can be infinite if the function has a singularity on the boundary of the unit disk, but is always finite if the function is analytic and bounded on the unit disk.
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Euge
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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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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.
 

FAQ: What is the Upper Bound for a Holomorphic Function on the Hardy Space?

1. What is the definition of the Hardy Space?

The Hardy Space, denoted as Hp, is a space of holomorphic functions defined on the unit disk in the complex plane. It consists of all functions f(z) that are analytic and have finite p-norm, where p is a positive real number.

2. What is the upper bound for a holomorphic function on the Hardy Space?

The upper bound for a holomorphic function on the Hardy Space is the maximum possible value that a function can attain on the unit disk. In other words, it is the supremum of the absolute value of the function on the unit disk.

3. How is the upper bound for a holomorphic function on the Hardy Space calculated?

The upper bound for a holomorphic function on the Hardy Space can be calculated using the Cauchy integral formula. This formula allows us to express the value of a holomorphic function at a point in terms of its values on the boundary of the unit disk. Taking the supremum of the absolute value of the function on the boundary gives us the upper bound.

4. What is the significance of the upper bound for holomorphic functions on the Hardy Space?

The upper bound for holomorphic functions on the Hardy Space is significant because it provides a measure of the size of a function. It also plays a crucial role in understanding the behavior of holomorphic functions, as it determines the convergence properties of power series expansions of these functions.

5. Can the upper bound for a holomorphic function on the Hardy Space be infinite?

Yes, the upper bound for a holomorphic function on the Hardy Space can be infinite. This can happen if the function has a singularity on the boundary of the unit disk, as the Cauchy integral formula will not hold in this case. However, if the function is analytic and bounded on the unit disk, the upper bound will always be finite.

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