Get 20M+ Full-Text Papers For Less Than $1.50/day. Start a 14-Day Trial for You or Your Team.

Learn More →

A Gleason–Kahane–Żelazko theorem for reproducing kernel Hilbert spaces

A Gleason–Kahane–Żelazko theorem for reproducing kernel Hilbert spaces We establish the following Hilbert‐space analog of the Gleason–Kahane–Żelazko theorem. If H${\mathcal {H}}$ is a reproducing kernel Hilbert space with a normalized complete Pick kernel, and if Λ$\Lambda$ is a linear functional on H${\mathcal {H}}$ such that Λ(1)=1$\Lambda (1)=1$ and Λ(f)≠0$\Lambda (f)\ne 0$ for all cyclic functions f∈H$f\in {\mathcal {H}}$, then Λ$\Lambda$ is multiplicative, in the sense that Λ(fg)=Λ(f)Λ(g)$\Lambda (fg)=\Lambda (f)\Lambda (g)$ for all f,g∈H$f,g\in {\mathcal {H}}$ such that fg∈H$fg\in {\mathcal {H}}$. Moreover Λ$\Lambda$ is automatically continuous. We give examples to show that the theorem fails if the hypothesis of a complete Pick kernel is omitted. We also discuss conditions under which Λ$\Lambda$ has to be a point evaluation. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Bulletin of the London Mathematical Society Wiley

A Gleason–Kahane–Żelazko theorem for reproducing kernel Hilbert spaces

Loading next page...
 
/lp/wiley/a-gleason-kahane-elazko-theorem-for-reproducing-kernel-hilbert-spaces-r19sLzN5Fq

References (27)

Publisher
Wiley
Copyright
© 2022 London Mathematical Society.
ISSN
0024-6093
eISSN
1469-2120
DOI
10.1112/blms.12618
Publisher site
See Article on Publisher Site

Abstract

We establish the following Hilbert‐space analog of the Gleason–Kahane–Żelazko theorem. If H${\mathcal {H}}$ is a reproducing kernel Hilbert space with a normalized complete Pick kernel, and if Λ$\Lambda$ is a linear functional on H${\mathcal {H}}$ such that Λ(1)=1$\Lambda (1)=1$ and Λ(f)≠0$\Lambda (f)\ne 0$ for all cyclic functions f∈H$f\in {\mathcal {H}}$, then Λ$\Lambda$ is multiplicative, in the sense that Λ(fg)=Λ(f)Λ(g)$\Lambda (fg)=\Lambda (f)\Lambda (g)$ for all f,g∈H$f,g\in {\mathcal {H}}$ such that fg∈H$fg\in {\mathcal {H}}$. Moreover Λ$\Lambda$ is automatically continuous. We give examples to show that the theorem fails if the hypothesis of a complete Pick kernel is omitted. We also discuss conditions under which Λ$\Lambda$ has to be a point evaluation.

Journal

Bulletin of the London Mathematical SocietyWiley

Published: Jun 1, 2022

There are no references for this article.