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I am reading Otto Forster's book "Lecture on Riemann surfaces" and on pages 109-110, he introduces the space $L^2(D,\mathcal{O})$ of holomorphic square-integrable functions $f:D\to \mathbb{C}$ (where $D\subset\mathbb{C}$ is open). In particular for $D=B(a,r)$ he explains that the monomials $\psi_n(z)=(z-a)^n$ form an orthogonal system with $\|\psi_n\|=\sqrt\frac{\pi}{n+1}r^{n+1}$ and that if $$f(z)=\sum_{n=0}^\infty c_n(z-a)^n\tag{1}\label{taylor}$$ is in $L^2(B(a,r),\mathcal{O})$, then by Pythagoras $$\|f\|^2_{L^2}=\sum_{n=0}^\infty \frac{\pi r^{2n+2}}{n+1}|c_n|^2.\label{pythagoras}\tag{2}$$

My only problem with all this is that it seems to me that we can only apply Parseval if the Taylor series \eqref{taylor} converges for the norm $L^2$, and it doesn't seem obvious that it does.

I know that the Taylor series converges pointwise on $B(a,r)$ and uniformly on every compact subset. I tried to apply the dominated convergence theorem to show that \eqref{taylor} also converges for the $L^2$ norm but I can't get a good integrable function to bound the differences $$\left|f-\sum_{n= 0}^Nc_n(z-a)^n\right|^2=\left|\sum_{n= N+1}^\infty c_n(z-a)^n\right|^2.$$

I also tried to use an approach similar to what is discussed in the comments of this answer, but I got stuck because I don't know that $L^2(D,\mathcal{O})$ is a Hilbert space (the proof in the book relies on \eqref{pythagoras}).

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Arnaud D.
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  • Perhaps I'm confused, but isn't Forster assuming (1) converges in $L^{2}$ (and deducing (2) as a conclusion)? – Andrew D. Hwang Apr 23 '16 at 13:05
  • @AndrewD.Hwang I don't think so, he takes $f\in L^2(B(a,r),\mathcal{O})$ and says explitely that (1) is its Taylor series. He later uses the fact that $f(a)=c_0$... – Arnaud D. Apr 23 '16 at 13:16
  • But (1) says $$f = \sum_{n=0}^{\infty} c_{n}\sqrt{\frac{\pi}{n+1}} r^{n+1}, \frac{\psi_{n}}{|\psi_{n}|}.$$Viewing this as the expression of the Taylor series of $f$ at $a$ in terms of the orthonormal basis $(\psi_{n}/|\psi_{n}|)$, saying "(1) is in $L^{2}$" seems to mean the partial sums converge in $L^{2}$, which is what you need for (2)...? – Andrew D. Hwang Apr 23 '16 at 13:27

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Suppose we're in the unit disc $\mathbb D$ for simplicity. Let $\sum_{n=0}^{\infty}a_nz^n$ be the Taylor series of $f$ in $\mathbb D.$ Using the orthogonality of the exponenetials, we see $$\int_{\mathbb D}|f|^2\, dA = \int_0^1 \int_0^{2\pi} |f(re^{it})|^2\, dt \, r\, dr = \int_0^1\int_0^{2\pi}|\sum_{n=0}^{\infty}a_nr^ne^{int}|^2\, dt\, r\, dr$$ $$ = \int_0^1 (2\pi \sum_{n=0}^{\infty}|a_n|^2r^{2n})\, r\, dr = 2\pi \sum_{n=0}^{\infty}|a_n|^2/(2n+2).$$ Now $|a_n|^2/(2n+2) = (\|a_nz^n\|_{L^2})^2.$ So whether $f\in L^2$ or not, the above always holds.

zhw.
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  • Can you please explain the third equation? Thank you in advance – Shirly Geffen Dec 16 '16 at 10:56
  • That's follows from the orthogonality of the exponenetials. – zhw. Dec 16 '16 at 16:19
  • So actually we use the continuity of the inner product, but we don't know the series converges in the norm the inner product induces... – Shirly Geffen Dec 16 '16 at 16:39
  • The Taylor series converges uniformly on $[0,2\pi]$ for each $r,0\le r <1.$ That certainly gives convergence in $L^2[0,2\pi].$ – zhw. Dec 16 '16 at 16:49
  • I think there was something that I couldn't quite get at the time, but looking at this again, your answer solves the problem, and is better than the way I avoided the problem. Thanks ! – Arnaud D. Apr 02 '17 at 10:35
  • https://math.stackexchange.com/questions/1253504/uniform-convergence-on-compact-sets-implies-interchange-of-summation-and-integra?rq=1 Given this, how come Taylor series uniform convergence on [0,r] for each r <1 is enough for L^2 convergence? Thanks! (@zhw, which you also answered! :) ) – abe.nong May 02 '17 at 17:21
  • @abe.nong No $L^2$ convergence was mentioned in my answer. – zhw. May 02 '17 at 17:24
  • yep! i understand no L^2 was mentioned in the link, but i'm just linking it to show that compact convergence doesn't imply you can switch summation and integration. I'm trying to understand your comment in this thread answering Shirly. Greatly appreciated :) – abe.nong May 02 '17 at 17:26
  • @abe.nong The integral and sum can always be interchanged if the summands are nonnegative. This follows from the monotone convergence theorem. – zhw. May 02 '17 at 17:31
  • I guess my question is regarding this comment: "The Taylor series converges uniformly on [0,r] for each r,0≤r<1. That certainly gives convergence in L2[0,2π]." – abe.nong May 02 '17 at 17:33