Reflexivity is invariant under isomorphism

Question

Let $X$ and $Y$ be two normed space. We suppose that $T\colon X\to Y$ is an isomorphism. It's true or false that

$$X\;\text{Reflexive}\implies Y\;\text{Reflexive}?$$

I would like to answer yes, but I can't prove it. Could anyone suggest me something?

Answer 1

$\newcommand{\J}{\mathcal{J}}$Of course this is true: linear isomorphism makes linear properties of spaces indistinguishable.

Let $X,Y$ be normed linear spaces and let $T:X\cong Y$ be an isomorphism of normed linear spaces.

We are given that the canonical embedding $\J_X:X\hookrightarrow X^{\ast\ast}$ is an isomorphism. We want to show $\J_Y:Y\hookrightarrow Y^{\ast\ast}$ is an isomorphism. We only need to check bijectivity. But $T^{\ast\ast}:X^{\ast\ast}\to Y^{\ast\ast}$ and $T$ are both bijective so it only suffices to check that $\J_Y\circ T=T^{\ast\ast}\circ\J_X:X\to Y^{\ast\ast}$.

Fix $x\in X$, $\phi\in Y^\ast$. It suffices to show that $T^{\ast\ast}(\J_X(x))(\phi)=\J_Y(T(x))(\phi)$, or that: $$\J_X(x)(\phi\circ T)=\phi(T(x))$$

Which is true.

Answer 2

Hints:

1. Construct the map $T':X'\to Y'$ given by $T'(f)(k)=k$ for every $f\in X', k\in K$ ($K$ is the field). Prove that $T'$ is well defined i.e. $T(f)$ is bounded if $f$ is bounded, and is indeed an isomorphism.
2. Construct the isomorphism $T'':X''\to Y''$ (similarly to 1.)
3. As $X$ is reflexive, the canonical map $M_X:X\to X''$ is bijective.
4. Take the canonical map $M_Y:Y\to Y''$. Prove that $T''\circ M_X=M_Y\circ T$. Therefore, $M_Y=T''\circ M_X\circ T^{-1}$ is also bijective.