$X$ is reflexive $\iff$ the transpose map is an isometric isomorphism

Question

I don't understand the following exercise; We are supposed to show that a normed linear space $X$ is reflexive if and only if the transpose map $\;t: B(X,X) \to B(X^*, X^*)$ is an isometric isomorphism. I could already show the first implication, that the transpose map is an isometric isomorphism if the space X is reflexive. But i really don't have any idea to prove the other direction. Can anybody help me?

Answer 1

The answer below can be summarized somewhat into a more concise answer, but I'm keeping it spread out because I think the individual pieces and their proofs are interesting in their own right.

In the real case, we'll use transpose and adjoint interchangeably. In this case, the transpose map is always a linear isometric embedding. Whether or not it's an isomorphism only depends on whether it's surjective.

Claim $1$: A member $f$ of $X^{**}$ is a member of (the canonical image of) $X$ (in $X^{**})$ iff it is weak$^*$ continuous.

Proof of Claim $1$: Let $J:X\to X^{**}$ be the canonical inclusion. If $f=J(x)$, then for any weak$^*$-convergent net $(x^*_\lambda)$ in $X^*$ with limit $x^*$, $$f(x^*_\lambda)=x^*_\lambda(x)\to x^*(x)=f(x^*),$$ so we have weak$^*$-continuity. Conversely, assume $f\in X^{**}\setminus J(X)$. By the Hahn-Banach separation theorem, there exists $g\in X^{***}$ such that $g(f)=1$ and $g|_{J(X)}\equiv 0$. Let $D$ be the set of all finite, non-empty subsets of $X$, ordered by inclusion. For each $F\in D$, by Helly's Lemma (or local reflexivity, or Goldstine's theorem), since $g(J(x))=0$ for all $x\in F$, there exists $x^*_F\in X^*$ such that $x^*_F(x)=0$ for all $x\in F$ and such that $f(x^*_F)=g(f)=1$. Then $(x_F^*)_{F\in D}$ is weak$^*$-convergent to $0$ in $X^*$, but $\lim_F f(x^*_F)=\lim_F 1=1$. So $f$ is not weak$^*$-continuous. $\square$

Claim $2$: Let $E,F$ be Banach spaces. Then an operator $S:F^*\to E^*$ is the adjoint (transpose)? of an operator $T:E\to F$ iff $S$ is weak$^*$-weak$^*$ continuous iff $S^*(E)\subset F$, where $E$ and $F$ are identified with their canonical images in $E^{**}$, $F^{**}$, respectively.

Proof of Claim $2$: Let $J_E:E\to E^{**}$ and $J_F:F\to F^{**}$ be the canonical inclusions. Assume $S=T^*$ for some $T:E\to F$. Then $$S^*(J_E(E))=T^{**}(J_E(E))=J_F(T(E))\subset J_F(F).$$ Or, identifying $J_E(E)$ with $E$ and $J_F(F)$ with $F$, $S^*(E)\subset F$, as stated above. Conversely, if $S^*(J_E(E))\subset J_F(F)$, then define $T=J_F^{-1}\circ S^*\circ J_E$. This is well-defined, because the range of $S^*$ is contained in the domain $J_F(F)$ of $J_F^{-1}$. We note that $T^*=S$. Indeed, for $x\in E$ and $y^*\in F^*$, \begin{align*} \langle y^*,J_F^{-1}S^* J_E(x)\rangle & = \langle S^*J_E(x),y^*\rangle=\langle J_E(x),Sy^*\rangle =\langle Sy^*,x\rangle.\end{align*} Therefore $S:F^*\to E^*$ is the adjoint of an operator $T:E\to F$ iff $S^*(E)\subset F$.

Next, suppose that $S=T^*$. Suppose that $S$ is weak$^*$-weak$^*$ continuous. Fix $e\in E\subset E^{**}$ and let $(f^*_\lambda)\subset F^*$ be weak$^*$-convergent to some $f^*\in F^*$. Then by weak$^*$-weak$^*$-continuity, $(Sf^*_\lambda)$ is weak$^*$-convergent to $Sf^*$, and \begin{align*} \langle S^*e,f^*_\lambda\rangle & = \langle Sf^*_\lambda,e\rangle\to \langle Sf^*,e\rangle=\langle S^*e,f\rangle.\end{align*} This shows that $S^*e$ is weak$^*$-continuous (as a function from $F^*$ into the scalar field), and therefore $S^*e\in F$ by Claim $1$. Thus $S^*(E)\subset F$. Conversely, suppose that $S$ is not weak$^*$-weak$^*$ continuous. We know all adjoints are weak$^*$-weak$^*$ continuous, so $S$ cannot be an adjoint, but we prove it another way just for completeness of the picture. Since $S$ is not weak$^*$-weak$^*$-continuous, there exists a net $(f^*_\lambda)\subset F^*$ which is weak$^*$-convergent to some $f^*$, but $(Sf^*_\lambda)$ is not weak$^*$-convergent to $Sf^*$. Since $(Sf^*_\lambda)$ is not weak$^*$-convergent to $Sf^*$, there exists $e\in E$ such that $(\langle Sf^*_\lambda,e\rangle)_\lambda=(\langle S^*e,f^*_\lambda\rangle)$ does not converge to $\langle Sf^*,e\rangle=\langle S^*e,f^*\rangle$. This shows that $S^*e$ is not weak$^*$-continuous, and is therefore not in $F$ by Claim $1$. Thereofore $S(E)\subset F$ iff $S$ is weak$^*$-weak$^*$-continuous. $\square$

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Let's prove your result by contraposition. Assume $X$ is not reflexive and select $x^{**}\in X^{**}\setminus X$. Fix any $0\neq x^*\in X^*$ (non-reflexivity implies $X\neq \{0\}$) and define $x^{*}\otimes x^{**}:X^*\to X^*$ by $x^{*}\otimes x^{**} (y^*)=x^{**}(y^*)x^*$. I claim that $x^{*}\otimes x^{**}$ is not the transpose of any operator on $X$, which shows that the transpose operator is not surjective, and therefore not an isometric isomorphism. To show that $x^*\otimes x^{**}$ is not the adjoint of any operator on $X$, by Claim $2$, we only need to show that it's not weak$^*$-weak$^*$ continuous. Fix any $y\in X$ such that $x^*(y)=1$ (which we may do, since $x^*\neq 0$). Note that $T:X^*\to \mathbb{R}$ given by $Ty^*=y^*(y)$ is weak$^*$-continuous by Claim $1$. If $x^*\otimes x^{**}$ were weak$^*$-weak$^*$-continuous, then the composition $T\circ (x^*\otimes x^{**})$ would be weak$^*$-continuous. But $T\circ (x^*\otimes x^{**})(y^*)=T(x^{**}(y)x^*)=x^{**}(y)x^*(y)=x^{**}(y)$, so $T\circ (x^*\otimes x^{**})=x^{**}$, which we know is not weak$^*$-continuous.

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Let's write a shorter proof which includes these ideas. We work by contradiction. Assume $t:B(X,X)\to B(X^*,X^*)$ is surjective and that $X$ is not reflexive. Fix $0\neq x^*\in X^*$ and $x^{**}\in X^{**}\setminus X$. Fix $y\in X$ such that $x^*(y)=1$. By the Hahn-Banach theorem, there exists $x^{***}\in X^{***}$ such that $x^{***}(x^{**})=1$ and $x^{***}|_X\equiv 0$. Let $D$ be the set of all non-empty, finite subsets of $X$ ordered by reverse inclusion. By Helly's lemma, for each $F\in D$, since $x^{***}(x^{**})=1$ and $x^{***}(x)=0$ for each $x\in F$, there exists $x^*_F\in X^*$ such that $x^{**}(x^*_F)=1$ and $x^*_F(x)=0$ for all $x\in F$. Clearly $(x^*_F)$ is weak$^*$-convergent to $0$ in $X^*$. Define $S:X^*\to X^*$ by $Sy^*=x^{**}(y^*)x^*$. Since we are assuming $t:B(X,X)\to B(X^*,X^*)$ is surjective, there exists $T:X\to X$ such that $T^*=S$. Then \begin{align*} \lim_F Sx^*_F(y) & = \lim_F x^{**}(x^*_F)x^*(y) = \lim_F x^{**}(x^*_F)=\lim_F 1=1,\end{align*} but $$\lim_F Sx^*_F(y) = \lim_F x^*_F(Sy)=0,$$ since $(x^*_F)$ is weak$^*$-convergent to $0$ and $Sy\in X$. This contradiction shows that $t:B(X,X)\to B(X^*,X^*)$ cannot be surjective if $X$ is not reflexive.

An even shorter proof can go by showing that if $x^*\otimes x^{**}$ is the transpose of something in $B(X,X)$, it must be something of the form $y\otimes y^*:X\to X$ (where $y\otimes y^*(z)=y^*(z)y$) for some $y\in X$ and $y^*\in X^*$, and in this case there must exist a scalar $a\neq 0$ such that $y=ax^{**}$ and $y^*=a^{-1}x^*$. But since $x^{**}\in X^{**}\setminus X$ and $y\in X$, we get another contradiction. I'll leave the details of that one to you.

Answer 2

Suppose $t$ is an isomorphism. Let $\varphi\in X^{**}$. We want to show that $\varphi=J(x_0)$ for some $x_0\in X$ where $J:X\to X^{**}$ is the canonical inclusion. If $X=0$, there is nothing to show. Otherwise by the Hahn-Banach theorem, there exists $f\in X^*\setminus\{0\}$. Fix $x\in X$ with $f(x)\ne0$. Define $T\in{\mathcal B}(X^*,X^*)$ by $T(g) = \varphi(g)f$. By assumption, $T=t(S)=S^*$ for some $S\in\mathcal B(X,X)$. Let $g\in X^*$. Then $T(g)=S^*(g)$, so $\varphi(g)f(x) = T(g)(x)=(S^*(g))(x)=g(S(x))$, so $\varphi(g) = g\left(\frac{S(x)}{f(x)}\right)$. So we see that $\varphi=J(x_0)$ where $x_0=\frac{S(x)}{f(x)}$, so $J$ is surjective, i.e. $X$ is reflexive.