Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Analytic At Infinity?

  1. Jan 8, 2012 #1
    Hi, I'm reading "Spectral Theory of Linear Operators" by John Dowson. I've seen the phrase "analytic at infinity" popping up very early in the book, but no definition is given. I wonder if anyone could tell me what the definition is or where I might find the definition and perhaps a few basic results on analyticity at infinity? The operators I'm looking at are continuous linear maps from a complex Banach space to itself, so my question is really about complex analysis.

    I've tried Google but have found no definition.

    I think "bounded at infinity" is a related concept, but again I do not know what that means. Maybe someone can help me out there too?

    Thanks.
     
  2. jcsd
  3. Jan 11, 2012 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    f(z) is analytic at infinity if f(1/z) is analytic at zero. Probably the same kind of definition can be given for "bounded at infinity" - but to be sure, you'll need to give us some context. I guess the Banach spaces you're working with are some kind of function spaces.

    Anyway, this kind of stuff should be viewed in the context of doing complex analysis on the Riemann sphere (= the complex plane plus a "point at infinity"). This might help you in your search for appropriate reading material.
     
  4. Jan 14, 2012 #3
    Thanks, I'm looking at two Banach spaces. An arbitrary infinite-dimensional Banach space X and the Banach space of all bounded linear operators from X to X. Is that enough context? I'm not sure what else to say.
     
  5. Jan 15, 2012 #4

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    No, that's not enough context! Where are the analytic functions coming in? Are you trying to define a "functional calculus"? I.e. are you trying to assign meaning to f(T) where f is an analytic function and T is a bounded operator on X?
     
  6. Jan 15, 2012 #5
    Yes that's exactly what I'm trying to do. Sorry should've said that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Analytic At Infinity?
  1. An analytic solution? (Replies: 6)

  2. Analytic functions (Replies: 3)

  3. Analytical integral (Replies: 1)

Loading...