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}).

  • $\begingroup$ Perhaps I'm confused, but isn't Forster assuming (1) converges in $L^{2}$ (and deducing (2) as a conclusion)? $\endgroup$ – Andrew D. Hwang Apr 23 '16 at 13:05
  • $\begingroup$ @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 $f(a)=c_0$... $\endgroup$ – Arnaud D. Apr 23 '16 at 13:16
  • $\begingroup$ 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)...? $\endgroup$ – Andrew D. Hwang Apr 23 '16 at 13:27

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.

  • $\begingroup$ Can you please explain the third equation? Thank you in advance $\endgroup$ – Shirly Geffen Dec 16 '16 at 10:56
  • $\begingroup$ That's follows from the orthogonality of the exponenetials. $\endgroup$ – zhw. Dec 16 '16 at 16:19
  • $\begingroup$ 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... $\endgroup$ – Shirly Geffen Dec 16 '16 at 16:39
  • $\begingroup$ The Taylor series converges uniformly on $[0,2\pi]$ for each $r,0\le r <1.$ That certainly gives convergence in $L^2[0,2\pi].$ $\endgroup$ – zhw. Dec 16 '16 at 16:49
  • $\begingroup$ 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 ! $\endgroup$ – Arnaud D. Apr 2 '17 at 10:35

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