"isomorphic" normed spaces and reflexivity Let X, Y be normed spaces and suppose that there exists an bijective isometry between them.
And if X is reflexive, then it is intuitively clear that Y is reflexive also. But, when I tried to prove it, I am just stuck and can't figure out how to proceed. It is just extremely frustrating...
Could anyone help me with this?
 A: Given a bijective linear isometry $T:X\rightarrow Y$, the dual map $T^*:Y^*\rightarrow X^*$ is also a bijective linear isometry. (This follows from the fact that $T^*$ has inverse $(T^{-1})^*$ and $\|T^*\|=\|T\|$.) From this, we have that $T^{**}:X^{**}\rightarrow Y^{**}$ is a bijective linear isometry as well.
Let $J_X:X\rightarrow X^{**}$ is the canonical embedding of $X$ into its bidual, which we're assuming is surjective. If $\phi\in Y^{**}$, then there exists a unique $\psi \in X^{**}$ such that $T^{**}(\psi)=\phi$ (since $T^{**}$ is a bijective isometry). As $J_X$ is surjective, there is an $x\in X$ such that $\psi=J_X(x)$, so $\phi=T^{**}(J_X(x))$.
Now choose any $f\in Y^*$. From the above, $\phi(f)=T^{**}(J_X(x))(f)$. By definition of $T^{**}$ and of $J_X$, this is equal to $(J_X(x))(T^*(f))=(T^*(f))(x)=f(T(x))$. Notice that this is $J_Y(T(x))(f)$. As this holds for every $f\in Y^*$, we have that $J_Y(T(x))=\phi$.
This means that, for any $\phi\in Y^{**}$, we can find a $y\in Y$ (the element $T(x)$ in the above paragraph) such that $\phi=J_Y(y)$, so $J_Y$ is surjective, which means precisely that $Y$ is reflexive.
A: Given a continuous mapping $F:X\rightarrow Y$ it induces continuous map $F^{**}:X^{**}\rightarrow Y^{**}$ given by the formula
$$[F^{**}(x^{**})](y^*):=x^{**}(F^*(y^*)).$$
This assigment is functorial, i.e. it has two properties:
$(id_X)^{**}=id_{X^{**}}$ and $(G\circ F)^{**}=G^{**}\circ F^{**}.$
Additionaly for every normed space $X$ we have continuos mapping $J_X:X\rightarrow X^{**}$ given by formula $J_X(x)(x^*)=x^*(x).$ This assigment is such that for every mapping $F:X\rightarrow Y$ we have that $J_{Y}\circ F=F^{**}\circ J_X,$ i.e the following diagram commutes
\begin{array}{cc}\phantom{\dfrac{a}{b}}X & \xrightarrow{\Large J_X} & X^{**}\phantom{\dfrac{a}{b}}\\ F \downarrow && \downarrow F^{**}\\ Y & \xrightarrow{ \Large J_Y }& Y^{**}\end{array}
You have that $X$ and $Y$ are isomorphic and $X$ is reflexive. Hence there exists isomorphism $F:X\rightarrow Y$ and $J_X$ is an isomorphism as well. From functoriality we get that $F^{**}$ is an isomorphism (cause $(F^{-1})^{**}=(F^{**})^{-1}$)and form diagram you get that $J_Y$ is composition of isomorphism, so isomorhism as well. What means that $Y$ is reflexive.
