Locally bounded Family

I'm researching for a test an I found a trouble that I am having difficultly addressing.

Allow $\mathcal{F}$ is a family members of analytic features on the shut device disc, $D$.

Intend

$\int_{D} |f|^{2} dA \le 1$

for all $f \in \mathcal{F}$.

Can I end that $\mathcal{F}$ is in your area bounded?

0
2019-05-18 22:40:22
Source Share
Yes. Cauchy is formula states that $$f(z) = \frac{1}{2\pi} \int_{\theta} f(z + re^{i \theta}) d\theta$$ which offers an expression for $f(z)$ in regards to the ordinary around a circle. By integration it adheres to that $$f(z) = \frac{1}{r_0^2 \pi} \int_{r \leq r_0, \theta} f(z+re^{i \theta}) r dr d\theta$$ which indicates that if the square indispensable of $f$ is bounded, after that (by Cauchy - Schwarz) $f$ is in your area bounded by a bound relying on the square indispensable of $f$ and also the circle concerned. From this your case adheres to.
Primarily, the factor is that $f$ is the standard of its values in an area. Keep in mind that this indicates that the room of holomorphic features in an open set $U$ such that $\int_{U} |f|^2$ is limited is in fact a Hilbert room (i.e., full) under the common internal item.