Dual of the product is isometric to the product of the dual

Question

Let $X,Y$ be Banach spaces. Let $Z = X\times Y$ equipped with the $p$-norm, where $||(x,y)||_Z = (||x||_X^p + ||y||_Y^p)^{1/p}$.

Suppose $X^* \times Y^*$is equipped with the $q$-norm where $1/p+1/q = 1.$

Then there exists $\phi:X^* \times Y^* \longrightarrow Z^*$, an isometric isomorphism.

For $f \in X^*, g \in Y^*$, the obvious map I can think of is $\phi(f,g)(x,y) = f(x) + g(y)$.

With holder's inequality I managed to show that $||\phi(f,g)||\leq ||(f,g)||$. But I'm not sure how to show the other direction.

Answer 1

First suppose $||f|| = ||g|| = 1$. Take $x \in X, y \in Y$ with $||x||=||y|| = (\frac{1}{2})^{1/p}$ and $f(x) = ||f||\cdot ||x||, f(y) = ||f|| \cdot ||y||$ (this is by definition, possibly after multiplying $x$ and $y$ by some number on the unit circle). Then $$|\phi(f,g)(x,y)| = |f(x)+g(y)| = ||f||\cdot||x||+||g||\cdot||y|| = (\frac{1}{2})^p2 = 2^{1/q} = (||f||^q+||g||^q)^{1/q}(||x||^p+||y||^p)^{1/p}.$$ Therefore, $||\phi(f,g)(x,y)|| \ge ||(f,g)||$, as desired.

Now since $\phi$ is linear, for arbitrary $f,g$, we can apply the preceding argument to $\frac{f}{||f||}$ and $\frac{g}{||g||}$ (I'll leave it to you to see what happens if $f = 0$ or $g = 0$).