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If we know that there is a (topological) isomorphism between two Banach spaces $X,Y$ called $\phi \in L(X,Y)$. Then the appropriate isomorphism between the dual spaces $X',Y'$ is given by $\phi' \in L(Y',X')$.

I was wondering: What is the fastest way to see that this $ \phi'$ is actually an isomorphism? Does anybody here have a good idea to show this fast? I mean sure, you could show that it is onto and injective, but this does not seem to be a fast idea.

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  • $\begingroup$ Plus one. Any reason for not adding the banach-spaces tag? $\endgroup$
    – Rudy the Reindeer
    Apr 15, 2014 at 18:52
  • $\begingroup$ Maybe the word "fast" needs a definition? $\endgroup$
    – Tomás
    Apr 15, 2014 at 18:54
  • $\begingroup$ @MattN. my list of tags was exhausted ;-) $\endgroup$
    – user66906
    Apr 15, 2014 at 19:02
  • $\begingroup$ @Tomás most fast is defined as the infimum over all possible number of symbols to prove this. does this serve to you as a definition? ;-) $\endgroup$
    – user66906
    Apr 15, 2014 at 19:02
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    $\begingroup$ The proof given in this answer here seems fairly short. $\endgroup$
    – Rudy the Reindeer
    Apr 15, 2014 at 19:45

1 Answer 1

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We can easily construct an inverse to $\phi^*$, it is given by $(\phi^*)^{-1}=(\phi^{-1})^*$, indeed we have: $$\left<(\phi^{-1})^*\phi^*v,x\right>=\left<\phi^*v,\phi^{-1}(x)\right>=\left<v,\phi\phi^{-1}(x)\right>=\left<v,x\right>$$ for all $x\in X$ and $v\in X^*$. Thus $(\phi^{-1})^*\phi^*=1$, and similarly $\phi^*(\phi^{-1})^*=1$.

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    $\begingroup$ what is $\langle,\rangle$? $\endgroup$
    – user66906
    Apr 16, 2014 at 0:23
  • $\begingroup$ Daniel, there is no inner product, the spaces in the question are Banach spaces. For a proof see e.g. the link I give in the comments to the question. $\endgroup$
    – Rudy the Reindeer
    Apr 16, 2014 at 5:58
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    $\begingroup$ @MattN. It is not an inner product, it is just the natural pairing between $X$ and $X^*$, i.e. $\left<v,x\right>=v(x)$ for $v\in X^*$ and $x\in X$. $\endgroup$
    – Daniel Robert-Nicoud
    Apr 16, 2014 at 16:56
  • $\begingroup$ Ok, good. I take back my comment! : ) $\endgroup$
    – Rudy the Reindeer
    Apr 16, 2014 at 17:10

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