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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 proofs, but I cannot comprehend them, 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

2 Answers 2

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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I feel like the other answer is a bit sketchy, and possibly is begging the question (by assuming that the diagram commutes). So here's my own proof:

If we have a bounded operator $f:X \to Y$ then $f^*:Y^* \to X^*$ and $f^{**} = (f^*)^*: X^{**} \to Y^{**}$ then $$f^{**}(i_X x) = i_Y f(x)$$ because $$f^{**}(i_X x)(\psi) = i_X x (f^*(\psi)) = f^* (\psi)(x) = \psi(fx) = i_Y(fx)(\psi)$$

where $\psi \in Y^*$ and $i_X, i_Y$ are the canonical injective maps from $X,Y$ to $X^{**},Y^{**}$ respectively. We say that $X$ is reflexive (by definition) if $X \cong X^{**}$ isometrically via $i_X$ (so it suffices to show that $i_X$ is onto, as we know it's an isometry).

Now assume $X$ is reflexive. If $f$ is an isomorphism then $f^*$ is an isomorphism (because $(f^{-1})^*=(f^*)^{-1}$ and is linear) and so likewise $f^{**}$ is an isomorphism. But isomorphisms are necessarily onto, so $f^{**} \circ i_X$ is onto and hence given any $\xi \in Y^{**}$ we can find an $x\in X$ such that the first $=$ below is true, then the second $=$ below was proven above:

$$(\xi) (\psi)=f^{**}(i_Xx)(\psi) = i_Y (fx)(\psi)$$

So this shows that $i_Y$ is onto, hence $Y$ is reflexive.

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By symmetry, of course, we get the converse. –  Squirtle Oct 13 at 2:05

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