# Reproducing kernel vs. Riesz kernel.

I'm attending an Analysis course and we are studying the Hardy space $H^2$ in the unit disk, from where the concept of RKHS (Reproducing Kernel Hilbert Space) came out. Acording to my notes:

By the Riesz representation theorem for every evaluation functional $\phi_a:H^2\to\mathbb{C}$, $\phi_af=f(a)$, there is exactly one $k_a\in H^2$ such that $$\phi_af=\langle f,k_a\rangle_{H^2},\ \ f\in H^2.$$ This equation is called the reproducing property, for which $k_a$ is called a reproducing kernel of $H^2$. In fact we can calculate that $k_a(z)=1/(1-\bar az)$ since $$\phi_af=\sum_{n=0}^\infty \hat f(n)a^n=\sum_{n=0}^\infty \hat f(n)\overline{\bar a^n}=\langle f\,, \,\frac{1}{1-\bar az}\rangle_{H^2}.$$ The function $k_a(z)=1/(1-\bar a z)$ receives the name of Riesz kernel.

I have no problem with the math, but with the terminology; why is $k_a$ called both reproducing kernel and Riesz kernel? Is a reproducing kernel exactly the same as a Riesz kernel? If not, how are they different?

• "reproducing" because by taking inner products you "reproduce" the function values, or because by knowing the kernel function you can "reproduce" the space. "Riesz", I'm guessing, in reference to Riesz's lemma that such a kernel exists whenever the evaluation functionals are continuous on a Hilbert space of functions. I have never seen that terminology, but if my guess is correct then they would be the same thing. The particular kernel for $H^2$ is called the Szegő kernel. – Jonas Meyer Feb 3 '17 at 17:50
• @JonasMeyer Thank you!! Actually I've consulted the notes of a classmate and it turns out I have "(Szegö)" missing after "receives the name of Riesz kernel" in my notes, so that solves my problem. =) – user378947 Feb 3 '17 at 18:12
• @JonasMeyer As an aside, you say that 'by knowing the kernel function you can "reproduce" the space' only as a synonym of 'by taking inner products you "reproduce" the function values', or it does have a precise meaning? I mean, how could you construct a RKHS if I give you only a kernel $k$? – user378947 Feb 3 '17 at 18:18
• It has another precise meaning as indicated here: en.wikipedia.org/wiki/… (I realize these comments may be close to answers but I have to do something else). – Jonas Meyer Feb 3 '17 at 20:34
• @JonasMeyer Oh, I'm sorry; I had started to read that wiki article but didn't imagine the reason they are called reproducing not to be at the beginning... Thank you so much for the reference!! And yes, your comments are rather an answer but no worry. I will be glad to accept an answer if you write it in the future though. :) – user378947 Feb 3 '17 at 20:55

1. Given $f\in H^2$, if you know the kernel you can "reproduce" all function values of $f$ using $k$, because $f(z)=\langle f,k_z\rangle$.
2. Given the domain, in this case $\mathbb D$, and the kernel function $k:\mathbb D\times \mathbb D\to \mathbb C$ you can "reproduce" the function space as the unique Hilbert space of functions containing the functions $k_a:z\to k(a,z)$ whose evaluations are given by inner products with the functions $k_a$. This is called the Moore–Aronszajn theorem and given much more precision and elaboration on Wikipedia.
Some special spaces' kernels have special names; yours is called the Szegő kernel, and there are other Szegő kernels for generalizations of $H^2$ on other domains. There are also the Bergman kernels for spaces of holomorphic functions with inner product given by integration with respect to area measure. Perhaps there is some class of kernels called "Riesz kernels" of which the Szegő kernel is an example.