$X,Y$ are Banach (real) spaces. Consider the following Banach space:
$$
X \times Y = \{(u,v)\,:\,u\in X,v\in Y\}, \qquad \|(u,v)\|=\|u\|_X+\|v\|_Y
$$
And $X^*\times Y^*=\{(f,g)\,:f\in X^*, y\in Y^*\}$ with this properties:
$$
(f,g)+(f',g')=(f+f',g+g'),\qquad\lambda(f,g)=(\lambda f,\lambda g)\quad\lambda\in\mathbb{R}
$$
$$
\|(f,g)\|_*=\max\{\|f\|_{X^*},\|g\|_{Y^*}\}
$$
During the first part of the exercise I proved that $(X^*\times Y^*,\|\cdot\|_*)$ is Banach. Now I have to prove that the application $J: X^*\times Y^*\rightarrow(X\times Y)^*$ is an isometric isomorphism. With:
$$
J((f,g))(u,v)=f(u)+g(v)
$$
But i'm a bit confused…
In order to show that $J$ is an isometry I have to prove that:
$$
\|J(f,g)\|=\|f+g\|=\|(f,g)\|_*
$$
But I don't know how to consider $\|J(f,g)\|$ even considering the definition:
$$
\|J(f,g)\|=\sup_{\|(u,v)\|=1} |J(f,g)(u,v)|
$$
And in order to prove that $J$ is isometric should I prove that exists $J^{-1}$?
(I hope I wrote it correctly!)
Best Answer
You need to show
The first point is hopefully unproblematic.
For the second, note that
$$\lVert J(f,g)\rVert = \sup_{\lVert (u,v)\rVert = 1} \lvert J(f,g)(u,v)\rvert = \sup_{\lVert (u,v)\rVert = 1} \lvert f(u) + g(v)\rvert,$$
and you have $\lvert f(u)+g(v)\rvert \leqslant \lvert f(u)\rvert + \lvert g(v)\rvert \leqslant \lVert f\rVert\cdot\lVert u\rVert + \lVert g\rVert\cdot \lVert v\rVert$, which together with $\lVert u\rVert + \lVert v\rVert = 1$ gives you $\lVert J(f,g)\rVert \leqslant \lVert (f,g)\rVert_\ast$. For the inequality in the other direction, $\lVert J(f,g)\rVert \geqslant \lVert (f,g)\rVert_\ast$, find suitable $u$ and $v$ to conclude $\lVert J(f,g)\rVert \geqslant \lVert(f,g)\rVert_\ast - \varepsilon$ for any arbitrarily given $\varepsilon > 0$.
For the third point, what is a natural (linear) map $(X\times Y)^\ast \to X^\ast \times Y^\ast$? The first that I can think of is the inverse of $J$.