Reflexivity of $X \times Y$

Question

I want to prove the following

Theorem. Let $X,Y$ be reflexive. Then $X \times Y$ is reflexive.

Here my try.

Proof. Let $J_X, J_Y$ be the canonical injections of $X$ onto $X''$ and of $Y$ onto $Y''$ respectively. Define $J := ( J_X, J_Y )$ (in the sense that $J(x,y) = ( J_X (x), J_Y (y) )$ for all $(x,y) \in X\times Y$). Thus $J : X\times Y \to X'' \times Y''$ and $J$ is surjective, so that $X'' \times Y''$ is reflexive.

Now, if we show that $X'' \times Y''$ and $(X \times Y)''$ are isometrically isomorphic, we have done. Assume that $K : X \times Y \to (X \times Y)''$ is the canonical injection of $X\times Y$ into $(X \times Y)''$. Since $J$ is an isometric isomorphism, we have that $K \circ J^{-1} : X'' \times Y'' \to (X \times Y)''$ is an isometric isomorphism.

Does it work? I'm not really sure about my last statements.

Edit - Last part deleted (I answered myself!)

Answer 1

I post two proofs, the first completes the one in my question and the other is deduced from references in comments.

Proposition. $K \circ J^{-1}$ in proof of the theorem in question is an isometric isomorphism.

Proof. Since $K$ is an isometry, follows that $K \circ J^{-1}$ is an isometry. Now we have to show that it is surjective. Given an element $H \in ( X \times Y )''$, we can construct an element of $X'' \times Y''$ by imposing \begin{equation} H(f(x,\cdot)) =: u(x), H(f(\cdot, y)) =: v(y) \end{equation}

and definining $L := (u, v)$, in the sense that $L(x,y) = (u(x), v(y))$ for all $(x,y) \in X\times Y$. So to each element in $(X \times Y)''$ is associated a pair in $X'' \times Y''$. Hence $K\circ J^{-1}$ is surjective. $\square$

Here is another proof (taken and translated from reference provided by @GiuseppeNegro), based on the following

Lemma. Let $X$, $Y$ be Banach spaces and $X', Y'$ their duals. Then $(X \times Y)'$ and $X' \times Y'$ are isometrically isomorphic.

Proof. Let $T: X' \times Y' \to (X \times Y)'$ be defined by

\begin{equation} \langle T(f,g) , (x,y) \rangle := \langle f,x \rangle + \langle g,y \rangle \quad f \in X', \, g \in Y' \end{equation}

Then define $\phi : X \times Y \to \mathbb K$ by

\begin{equation} \phi( x,y ) := \langle f,x \rangle + \langle g,y \rangle, \quad x \in X, \, y \in Y. \end{equation}

It is clear that $T$ is linear. Moreover, it is surjective. In fact, given $\phi \in ( X \times Y)'$, we can define functionals

\begin{equation} f(x) := \langle \phi, (x,0) \rangle, \, g(y) := \langle \phi, (0,y) \rangle, \quad x \in X, \, y \in Y , \end{equation}

so that $f \in X'$, $g \in Y'$ and $\phi = T(x,y)$. Now it remains to show that $T$ is an isometry. For $(x,y) \in X \times Y$ we have

\begin{equation} \lvert \phi(x,y) \rvert \leq \lvert \langle f , x \rangle \rvert + \lvert \langle g , y \rangle \rvert \leq \lVert f \rVert \lVert x \rVert + \lVert g \rVert \lVert y \rVert \leq (\lVert f \rVert ^2 + \lVert g \rVert^2 )^{1/2}(\lVert x \rVert ^2 + \lVert y \rVert^2 )^{1/2} , \end{equation}

and by Cauchy-Schwarz inequality for the euclidean norm in $\mathbb{R}^2$,

\begin{equation} \lVert \phi \rVert \leq \lVert (f,g) \rVert \lVert (x,y) \rVert. \end{equation}

Conversely, we prove that, for all $\epsilon > 0$, $\lVert \phi \rVert \geq \lVert (f,g)\rVert - 2\epsilon$. Let $\epsilon > 0$ arbitrary. By definition of $\lVert f \rVert$, there exists $x_0 \in X$ such that $\lVert x_0 \rVert = 1$ and $\lvert \langle f, x_0 \rangle \rvert \geq \lVert f \rVert - \epsilon$. Let $\theta \in \mathbb R$ such that $\langle f, x_0 \rangle = \lvert \langle f, x_0 \rangle \rvert e^{i \theta}$ and set $x = e^{-i\theta} \lVert f \rVert x_0$. Then $\lVert v \rVert = \lVert f \rVert$ and

\begin{equation} \langle f, x \rangle = e^{-i\theta}\lVert f \rVert x_0 = \lVert f \rVert \langle f, x_0 \rangle \geq \lVert f \rVert ( \lVert f \rVert - \epsilon ) = \lVert f \rVert^2 - \epsilon \lVert x \rVert. \end{equation}

A similar construction can be carried out also for $g$, so that

\begin{equation} \langle \phi, (x,y) \rangle = \langle f, x \rangle + \langle g,y\rangle \geq \lVert f \rVert^2 + \lVert g \rVert^2 - ( \lVert x \rVert + \lVert y \rVert ) = ( \lVert f \rVert^2 + \lVert g \rVert^2 )^{1/2} ( \lVert x \rVert^2 + \lVert y \rVert^2 )^{1/2} - ( \lVert x \rVert + \lVert y \rVert ) \geq \lVert ( f,g) \rVert \lVert ( x,y) \rVert - \epsilon \lVert ( x,y) \rVert \end{equation}

and hence the result. $\square$

The identification required now follows by iteration.

In retrospect, I believe that the idea is the same for both the proofs. If my proof was correct (I think it should be now), it is much simpler than the other, because it is easier to verify that the operators involved are isometries. Please, inform me about any mistake you'll find. (my examination in functional analysis is very near!)