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I want to show that if $X$ and $Y$ are two Banach spaces, and $T : X \to Y$ is an isomorphism, then $$ X \textrm{ reflexive} \iff Y \textrm{ reflexive}. $$ I saw several proof, but i cannot comprehend, some are working with the dual of $T$, but i am not sure what the dual of $T$ should be?

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Does "(norm) isomorphic" mean isometrically isomorphic, i.e. there's a bijective linear isometry between them? –  Chris Eagle Jan 6 '13 at 23:58
    
You probably mean the adjoint of $T$. Any bounded linear operator $T:X\to Y$ has an adjoint $T^*: Y^*\to X^*$ which is naturally defined by $T^*(\phi)=\phi\circ T$ (recall that the elements of $Y^*$ are functions on $Y$, so it's natural to compose them with a map into $Y$). –  user53153 Jan 7 '13 at 1:34
    
No, just an isomorphism, not necessarily an isometry. –  Stefan Jan 7 '13 at 10:31
    
Yes, with dual of $T$ i mean the adjoint. –  Stefan Jan 7 '13 at 10:42
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1 Answer 1

Let $J_X:X\to X^{**}$ be the canonical isometry defined by $J(x)(f)=f(x)$. By defition X is reflexive if J is surjective.

$$\begin{array}{ccc}X&\xrightarrow{T}& Y\\\downarrow{J_X} &&\downarrow{J_Y}\\X^{**}&\xrightarrow{T^{**}}& Y^{**}\end{array}$$,

We define $T^*:Y^*\to X^*$ as above and $T^{**}$ analogously: $T^{**}G=GT^*$,

The diagram above commute, therefore if $J_X$ is surjetive then $J_Y$ also is surjective

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