4
$\begingroup$

I'm studying for an exam an I came across a problem that I am having difficultly solving.

Let $\mathcal{F}$ is a family of analytic functions on the closed unit disc, $D$.

Suppose

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

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

Can I conclude that $\mathcal{F}$ is locally bounded?

$\endgroup$
  • $\begingroup$ Consider the family {(2(u-z))^(-1/2) | u > 1, u Real} near the point z = 1. (If I have computed correctly, all integrals are less than 1.) I hope I'm not misleading you, but the intuition behind this suggestion is that these approach a function (when u = 1) which (a) blows up at a boundary point yet (b) whose L2 norm is finite. $\endgroup$ – whuber Oct 6 '10 at 3:22
  • $\begingroup$ See also math.stackexchange.com/questions/33146/…. $\endgroup$ – Jonas Meyer Apr 15 '11 at 20:16
8
$\begingroup$

Yes. Cauchy's formula states that $$f(z) = \frac{1}{2\pi} \int_{\theta} f(z + re^{i \theta}) d\theta$$ which gives an expression for $f(z)$ in terms of the average around a circle. By integration it follows 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 implies that if the square integral of $f$ is bounded, then (by Cauchy-Schwarz) $f$ is locally bounded by a bound depending on the square integral of $f$ and the circle in question. From this your claim follows.

Basically, the point is that $f$ is the average of its values in a neighborhood. Note that this implies that the space of holomorphic functions in an open set $U$ such that $\int_{U} |f|^2$ is finite is actually a Hilbert space (i.e., complete) under the usual inner product.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The space is complete - it is called the Bergman space, easier to define the norm compared to the similar Hardy space but less rich with respect to factorization theory. $\endgroup$ – AD. Oct 6 '10 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.