1
$\begingroup$

I am working on a question in which $f_n$ a sequence of holomorphic functions converge in the $L_1$ sense on closed disks in a domain to $f$, which is a continuous function. The goal is to show that $f$ is analytic and that $f_n \to f$ uniformly on compact subsets of the region on which $f,f_n$ are defined.

My attempt (which may be inefficient right off the bat, so please let me know if you disagree): $L_1$ convergence implies convergence almost everywhere. That means, we get $|f_n(z) - \frac{1}{\pi r^2} \int_0^r \int_0^{2\pi} f(z+re^{i\theta}) r d\theta dr| \to 0$ as $n \to \infty$. As a result, we can claim that almost everywhere, the volume mean-value property holds for $f: \mathbb{C} \to \mathbb{C}$ a continuous function.

I don't know where to go from here: I thought that by choosing $r$ small enough we can use the continuity of $D(z) = f(z) - \frac{1}{\pi r^2} \int_0^r \int_0^{2\pi} f(z+re^{i\theta}) r d\theta dr$ in $z$ to show that the volume mean-value property holds everywhere for $f$. Consequently, we have reduced to the fact that $f$ satisfies the volume mean-value property. However, from here on, I am stuck: I don't know what the volume mean-value property really implies for $f$ (other than, maybe its real and imaginary parts are harmonic, which is not the same as $f$ being analytic necessarily).

Another approach: trying to use the $L_1$ convergence on closed disks to show that some subsequence of $f_n$ must converge uniformly on compact subsets to an analytic $f'$ and use the fact that $f' = f$ a.e. to claim that $f=f'$ since they are both continuous. However, I am not sure how $L_1$ convergence on closed disks implies uniform boundedness. That also doesn't solve the problem of showing uniform convergence on compact subsets. Montel's theorem only shows uniform convergence on compacta of the subsequence.

This is complex analysis homework question - there should be a simpler way to do this that doesn't use so much measure theory, but I can't think of any other than Montel.

EDIT: I'm realizing above has a mistake: we can only assume that some subsequence of $f_n$ approaches $f$ almost everywhere.

I thought of a way of finding uniform bounds on compacta (let me know if it makes sense): choose a compact subset $K$ and some $z \in K$. By fixing some small $R$ less than half the distance of $K$ and the boundary of the region, we can say

$|f_n(z)| = |\frac{1}{\pi R^2} \int_0^{2\pi} \int_0^R f_n(re^{i\theta}) rd\theta dr| \leq \frac{1}{\pi R^2} \int_0^{2\pi} \int_0^R |f_n(re^{i\theta})-f(re^{i\theta})| rd\theta dr + \frac{1}{\pi R^2} \int_0^{2\pi} \int_0^R |f(re^{i\theta})| rd\theta dr $

Now, we can use some compact exhaustion to find a maximum for both integrals (in one instance, using the continuity of $f$ and in the other, using the assumption from the question and some cover of the exhaustion by closed disks so that we can find a max independent of $n$).

Does that work?

$\endgroup$
0

2 Answers 2

0
$\begingroup$

In addition to Cauchy's formulas for holomorphic functions, which integrate over paths (e.g., circles) to recover the function, there are also Bergmann kernels, which integrate over solid regions to recover the function.

(This possibility seems more widely known in "Several Complex Variables" than in a classical presentation of one complex variable...)

For example, on the unit disk, distinct monomials $z^m$ and $z^n$ are orthogonal with respect to the hermitian pairing $\int_D f\cdot \overline{g}$, where $D$ is the disk, and the norm of $z^n$ is $2\pi/(2n+2)$. So let $K(z,w)=\sum_{n\ge 0} {2n+2\over 2\pi} z^n\,\overline{w}^n$. Then $\int_D K(z,w)f(w)\,|dw|=f(z)$ for holomorphic $z$.

Then various hypotheses about integrability on a sequence $f_n$ assure that the Bergmann representations converge, and (necessarily) that presentation gives a holomorphic function.

$\endgroup$
1
  • $\begingroup$ This is very helpful to know for the future but I think this question should have a simpler answer. $\endgroup$ Jul 2, 2021 at 21:01
0
$\begingroup$

Answering my own question: you can find a uniform bound on compact sets using $|f_n| \leq |f-f_n| + |f|$, then use Montel's theorem to find a subsequence $f_{n_k}$ converging uniformly on compacta to some analytic $f'$, then use continuity of $f,f'$ and integral of $|f-f'| \leq |f-f_{n_k}| + |f'-f_{n_k}|$ on disks to show $f = f'$. Now you can show uniform convergence on compacta of the general $f_n$ by repeating the reasoning for any subsequence to find a further subsequence converging to $f$, a fixed limit. Therefore, the "uniform" norm and the subsequence principle show that the general sequence $f_n$ must convergence uniformly on compacta.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .