1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Dual Norm Spaces (isomorphism/isometric)

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data
    If X and Y are normed spaces, define [itex]\alpha : X^* x X^*\rightarrow (X x X)^*[/itex] by [itex]\alpha(f,g)(x,y) = f(x)+g(y)[/itex].
    Then [itex]\alpha[/itex] is an isometric isomorphism if we use the norm [itex]||(x,y)|| = max(||x||,||y||)[/itex] on [itex]X x Y[/itex], the corresponding operator norm on [itex](X x Y)^*[/itex], and the norm [itex]||(f,g)||=||f||+||g||[/itex] on [itex]X^* x Y^*[/itex].

    2. Relevant equations
    [itex]||x||=sup[|f(x)|:f\in X^*, ||f||\leq 1][/itex]
    [itex]||f||=sup[|f(x)|:x\in X, ||x||\leq 1][/itex]

    3. The attempt at a solution

    to show it is isomorphism,
    suppose [itex]\alpha (f,g) = f(x) + g(y) = m(x) + n(y) = \alpha (m,n)[/itex]
    but this can only happen if f(x)=m(x) and g(y)=n(y) since they depend on x and y respectfully.

    but i am having trouble with proving it is isometric
    here is my attempt:
    I assume i want to show that ||(f,g)||=||(x,y)||

    [itex]||(f,g)||=sup[|f(x)|:x\in X, ||x||\leq 1] + sup[|g(y)|:y\in Y, ||y||\leq 1]
    =sup[|f(x)|+|g(y)|:x\in X, y\in Y, ||y||\leq 1, ||x||\leq 1][/itex]

    [itex]||(x,y)||=max(sup[|f(x)|:f\in X^*, ||f||\leq 1],sup[|g(y)|:g\in X^*, ||f||\leq 1])[/itex]

    but I don't know what else to do.
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted