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

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.

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 $$g = 0$$).