(adsbygoogle = window.adsbygoogle || []).push({}); 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]

also

[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.

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# Homework Help: Dual Norm Spaces (isomorphism/isometric)

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