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$X,Y$ 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!)

Dean Miller
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rusca91
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  • Maybe you would change the question title in somewhat more pertinent/attractive, in order to get wider resonance and be more useful to other users. – user91126 Feb 05 '14 at 15:31
  • Do you have any idea? – rusca91 Feb 05 '14 at 15:34
  • I'd choiche something like "How to prove $(X \times Y)^* \simeq X^* \times Y^* ?$". However, considering formulas in title are deprecated, "How to prove the dual of a product of two Banach spaces is isometrically isomorphic to the product of the duals?". (maybe a little long! For my part, I prefer something like the first one.) – user91126 Feb 05 '14 at 15:44
  • Ok, I'll write the first one! Thanks for the advice! – rusca91 Feb 05 '14 at 15:47

1 Answers1

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You need to show

  1. $J$ is linear,
  2. for all $(f,g)\in X^\ast\times Y^\ast$ you have $\lVert J(f,g)\rVert = \lVert (f,g)\rVert_\ast$ (that's the "isometric" bit, which implies injectivity), and
  3. $J$ is surjective (with the injectivity and linearity, that makes $J$ and isomprohism); that is in this case as easily done by explicitly giving the inverse of $J$.

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 $J$.

Daniel Fischer
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  • I know that your answer is already complete and it is really clear! But can I ask you one more thing about the exercise? I don't know what $u,v$ I have to choose in order to obtain the second inequality $||J(f,g)||\geq||(f,g)||_*$. I have some problems when it comes to the choice of particular elements in order to proof some inequatilies... – rusca91 Feb 05 '14 at 20:15
  • Suppose that $\lVert f\rVert \geqslant \lVert g\rVert$. Then $\lVert (f,g)\rVert = \lVert f\rVert$. Remember that $\lVert f\rVert = \sup\limits_{\lVert u\rVert = 1} \lvert f(u)\rvert$. – Daniel Fischer Feb 05 '14 at 20:40
  • So if I choose $u$ : $|f(u)|=||f||$ and $v=0$, then I have $|f(u)+g(v)|\geq |f(u)|=||f||=||(f,g)||_*$? (I don't know why but I'm pretty sure it is wrong...) – rusca91 Feb 05 '14 at 20:48
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    The only thing that's wrong(ish) is that there does not always exist a $u$ with $\lVert u\rVert = 1$ and $\lvert f(u)\rvert = \lVert f\rVert$. But you can come arbitrarily close by the definition, so that gives you $\lVert J(f,g)\rVert \geqslant \lVert f\rVert$. Choosing $u = 0$ (if you don't like to "suppose $\lVert f\rVert \geqslant \lVert g\rVert$") gives you $\lVert J(f,g)\rVert \geqslant \lVert g\rVert$, so altogether $\lVert J(f,g)\rVert \geqslant \max {\lVert f\rVert,\lVert g\rVert} = \lVert(f,g)\rVert_\ast$. – Daniel Fischer Feb 05 '14 at 21:03