# Dual Norm Spaces (isomorphism/isometric)

1. Oct 18, 2011

### CornMuffin

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

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

3. The attempt at a solution

to show it is isomorphism,
suppose $\alpha (f,g) = f(x) + g(y) = m(x) + n(y) = \alpha (m,n)$
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)||

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

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

but I don't know what else to do.