Let $f:D\subseteq\mathbb{C}\to\mathbb{C}$ be an analytic function (except perhaps at the origin) defined over a disk $D$ centered at the origin. Its Laurent series on $D$ is $$ f(z) = \sum_{n=-\infty}^{\infty}c_n z^n $$ The question is if this series can be interpreted as an orthonormal series expansion of a Hilbert space (of analytic functions on $D\setminus\{0\}$) similarly as in a Fourier series. Of course, we would have to define the inner product in a different way as the usual to make the base $\{z^n\}_{n=-\infty}^{\infty}$ orthonormal.

My question is motivated by the similarity of how you obtain the coeficients $c_n$ for this series, and how you would obtain it for generic orthonormal series expansions (as in a Fourier series). Take the integral $\frac{1}{2\pi i}\oint_{\gamma} \left(z^n\right)\left(z^{-m}\right)\frac{1}{z}dz$ where $\gamma$ is the contour for the unit circle $|z|=1$. Thus, with $z = \exp(i\theta)$ and $\theta\in[0,2\pi)$: $$ \begin{aligned} \frac{1}{2\pi i}\oint_{\gamma} \left(z^n\right)\left(z^{-m}\right)\frac{1}{z}dz &=\frac{1}{2\pi i} \oint_{\gamma} z^{n-m-1}dz = \frac{1}{2\pi i}\int_{0}^{2\pi}\exp(i\theta(n-m-1))i\exp(i\theta)d\theta \\&=\frac{1}{2\pi}\int_{0}^{2\pi}\exp(i\theta(n-m))\\ &= \left\{ \begin{array}{ll} \frac{1}{2\pi}\int_0^{2\pi}d\theta = 1 & \text{if }n=m \\ \frac{1}{2\pi(n-m)}(\exp(2\pi(n-m)) - 1) = 0 & \text{if }n\neq m \end{array} \right. \\ & = \delta_{nm} \end{aligned} $$ where $\delta_{nm}$ is the Kronecker delta. Hence, we can multiply the series of $f(z)$ from both sides by $\frac{1}{2\pi i}z^{-m}\frac{1}{z}$ and integrate over $\gamma$: $$ \begin{aligned} \frac{1}{2\pi i}\oint_\gamma f(z)(z^{-m})\frac{1}{z}dz &= \frac{1}{2\pi i}\oint_\gamma\left(\sum_{n=-\infty}^{\infty}c_nz^n \right)z^{-m}\frac{1}{z}dz \\ &=\sum_{n=-\infty}^{\infty}c_n \frac{1}{2\pi i}\oint_\gamma z^nz^{-m}\frac{1}{z}dz \\ & =\sum_{n=-\infty}^{\infty}c_n \delta_{nm}\\ &= c_m \end{aligned} $$ Thus, we obtain the clasical result that $$ c_m = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z^{m+1}}dz $$ Note that this could be possible since, $\frac{1}{2\pi i}\oint_{\gamma} \left(z^n\right)\left(z^{-m}\right)\frac{1}{z}dz = \delta_{nm}$, similar to an orthonomal expansion where you have a basis of functions $\{u_n\}_{n=0}^\infty$ which comply $\langle u_n, u_m\rangle = \delta_{nm}$ and perform a similar trick: apply the inner product $\langle\bullet, u_m \rangle$ at both sides of $f(x) = \sum_{n=0}^\infty c_n u_n $ to obtain $c_m = \langle f(z) , u_m\rangle$.

Thus, this is why it may be possible to think of $\{z^n\}_{n=-\infty}^{\infty}$ as an orthonormal basis for functions as $f$, with some sort of inner product which I still don't know how to define. Motivated by the previous integrals, I tried:

  • $\langle f(z), g(z) \rangle = \frac{1}{2\pi i}\oint_\gamma f(z)g(z)^* \frac{1}{z}dz$
  • $\langle f(z), g(z) \rangle = \frac{1}{2\pi i}\oint_\gamma f(z)\frac{1}{g(z)} \frac{1}{z}dz$

where $*$ stands for complex conjugate. However, non of these have the properties required to be an inner product. In particular the first proposal does not satisfy $\langle f(z), g(z) \rangle = \langle g(z), f(z) \rangle^*$ and the second one it linear in the second argument.

What are your thoughts on this? Do you think it is possible to find definition of the inner product? Do you known any reference which discuss something similar?


The Laurent series of a function analytic in an annulus containing the unit circle is the Fourier series of its restriction to the unit circle. Your first inner product works.

  • $\begingroup$ Thanks for your answer! You are right about the Fourier series, it was kinda obvious and I didn't see that until you mentioned it. Apart from that, I don't see how my first inner product works. I can't show that $\langle f(z), g(z) \rangle = \langle g(z), f(z) \rangle^*$ due to the $\frac{1}{z}$ factor. Maybe I am missing something simple. $\endgroup$ – FeedbackLooper Aug 28 '20 at 8:24
  • $\begingroup$ Compute both sides with the substitution $z = e^{i \theta}$ you used earlier. $\endgroup$ – Qiaochu Yuan Aug 28 '20 at 8:25
  • $\begingroup$ Right... Before marking this as the answer. So we agree that $\{z^n\}_{n=-\infty}^\infty$ is an orthonormal basis with such inner product? Do you know any reference which makes a treatment in this sense? $\endgroup$ – FeedbackLooper Aug 28 '20 at 8:30
  • $\begingroup$ Yes, that's right, as it would have to be for this to be an orthonormal basis expansion. $\endgroup$ – Qiaochu Yuan Aug 28 '20 at 8:31
  • $\begingroup$ Googling "Laurent series Fourier series" gives, for example, this: math.lsa.umich.edu/~rauch/555/laurentfourier.pdf $\endgroup$ – Qiaochu Yuan Aug 28 '20 at 8:31

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