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, $D$.


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

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

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

  • $\begingroup$ 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

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.

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  • $\begingroup$ 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

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