Is the map sends $T$ to $T^*$ adjoint of $T$ surjective?

Let $$B(X)$$ denotes the set of all bounded linear operators from $$X$$ to $$X$$, where $$X$$ is a Banach space. Same is defined for the set $$B(X^*)$$, where $$X^*$$ denotes the set of all bounded linear functionals of $$X$$, that is all bounded linear functionals from $$X$$ to the field $$\mathbb F$$. In other words, $$X^*$$ is the dual space of $$X$$.

Let us now define a map $$\alpha :B(X) \to B(X^*)$$ by. $$\alpha(T)=T^*, \quad T \in B(X)$$ where $$T^*$$ is the Banach space adjoint of the operator $$T$$. That is, for Banach spaces $$X,Y$$ and $$T\in B(X,Y)$$ the adjoint operator $$T^*: Y^* \to X^*$$ is defined as $$T^*(y^*)(x)=y^*(T(x)),~~~y^* \in Y^*,~~x \in X.$$ Now note that, since $$\|T\|=\|T^*\|$$, the map $$\alpha$$ is an isometry and also injective. Can I show that $$\alpha$$ is a linear isomorphism? Linearity of $$\alpha$$ comes from the linearity of adjoint operators but I am not able show that $$\alpha$$ is surjective. One of my friends told me $$\alpha$$ may not be surjective but can't give any argument.

• So: the question is whether every member of $B(X^*)$ is the adjoint of some member of $B(X)$. Hint For a counterexample, you must take $X$ not reflexive. Nov 27, 2021 at 1:20
• Related post: math.stackexchange.com/questions/1832836/… Nov 27, 2021 at 14:58

In the case that $$X$$ is reflexive, I believe this is true, but I haven't checked it out; but I'm pretty sure that the map $$B(X^*)\to B(X^{**})$$ composed with the canonical isometric isomorphism $$B(X^{**})\cong B(X)$$ will give you the desired surjectivity.

However, the well-defined linear isometry $$B(X)\to B(X^*)$$, $$T\mapsto T^*$$ is not surjective in general. Take a non-reflexive Banach space $$X$$ (like $$c_0$$, or $$\ell^1$$) and let $$j:X\to X^{**}$$ be the canonical inclusion. Since $$j$$ is not surjective, let $$\chi\in X^{**}$$ be an element outside of $$j(X)$$. Fix a non-zero functional $$\psi_0\in X^*$$ and define an operator $$P:X^*\to X^*$$ by $$P(\phi):=\chi(\phi)\cdot\psi_0$$. This is a well-defined, bounded operator (and actually its range is one dimensional, but we dont care about it).

Now assume that $$P$$ is in the range of $$\alpha:B(X)\to B(X^*)$$, so there exists a bounded operator $$T\in B(X)$$ such that $$P=T^*$$, i.e. $$P(\phi)=T^*\phi=\phi\circ T$$ for all $$\phi\in X^*$$, so $$\phi(Tx)=\chi(\phi)\cdot\psi_0(x)$$ for all $$x\in X$$ and all $$\phi\in X^*$$. Since $$\psi_0$$ is non-zero, find $$x_0\in X$$ such that $$\psi_0(x_0)\ne0$$. We then have $$\phi(Tx_0)=\chi(\phi)\cdot\psi_0(x_0)$$ and this is true for all $$\phi\in X^*$$; set $$\psi_0(x_0):=\lambda\in\mathbb{C}\setminus\{0\}$$. We have just shown that $$\chi=\frac{1}{\lambda}j_{Tx_0}=j_{\frac{1}{\lambda}Tx_0}\in j(X)$$, which is a contradiction.

Note that this answer works for any non-reflexive space; in other words, if the claim in my first paragraph is true (which i think it is) we get the following:

Corollary: Let $$X$$ be a Banach space. The canonical linear isometry $$B(X)\to B(X^*)$$, $$T\mapsto T^*$$ is surjective if and only if $$X$$ is reflexive.

Comment: I've spent a few hours thinking about this, so I have to give some credit to GEdgar. I knew the counter-example would come from the non-reflexive world from the first moment I started on this, but I was trying to give an answer specifically for $$X=c_0$$; after seeing GEdgar's comment I realized there is no need to go in a specific space, the abstraction actually helps here.

• Nice answer! Lovely that you got a characterisation. Nov 27, 2021 at 10:20
• @QuantumSpace Thank you very much:) Nov 27, 2021 at 10:31
• @JustDroppedIn For reference, your corollary is true. I found this result in the book "Introduction to Functional Analysis" by Meise and Vogt, proposition 9.2. Dec 9, 2021 at 17:13
• @Mrcrg you mean $X$ is reflexive implies $B(X)\to B(X^*)$ is surjectve? this is obvious if you play around with the fact that the canonical embedding $X\to X^{**}$ is surjective Dec 9, 2021 at 19:24
• @Mrcrg but this is exactly what I prove: I prove $X$ not reflexive implies $T\mapsto T^*$ not surjective, which is equivalent to what you just wrote ($p\implies q$ is equivalent to negation of $q$ implies negation of $p$) Dec 9, 2021 at 19:47

This is an addition to the excellent answer of @JustDroppedIn, where I provide full details of the claim he sketches.

Theorem: If $$X$$ is reflexive, the map $$\varphi: B(X) \to B(X^*): T \mapsto T^*$$ is surjective.

Proof: Consider the canonical embedding $$j: X \to X^{**}: x \mapsto \operatorname{ev}_x$$. Since $$X$$ is reflexive, $$j$$ is an isometric isomorphism. As mentioned by @JustDroppedIn, this implies that we have a natural map $$\psi: B(X^*) \to B(X^{**}) \to B(X)$$ given by $$\psi(T) = j^{-1} T^* j, \quad T \in B(X^*).$$ We claim that $$\varphi \psi = \operatorname{id}_{B(X^*)}$$ and surjectivity of $$\varphi$$ will follow.

For this, fix $$T \in B(X^*)$$. We have to show that $$(j^{-1}T^* j)^* = T.$$ Fix $$f\in X^*$$. It suffices to show $$(j^{-1}T^*j)^*(f) = T(f) \in X^*$$ so fix $$x \in X$$, and note that it suffices to show $$((j^{-1} T j)^*(f))(x) = (T(f))(x).$$

From here, the proof is a simple calculation. Start by noting that $$\operatorname{ev}_x \circ T = \operatorname{ev}_y$$ for some $$y \in X$$. Hence, $$(T(f))(x) = \operatorname{ev}_x(T(f)) = f(y)$$ and thus \begin{align*}((j^{-1}Tj)^*(f))(x)&=f(j^{-1}T^*j(x))= f(j^{-1}(\operatorname{ev_x}\circ T))= f(y) = (T(f))(x)\end{align*} and we are done.

• Thanks for the confirmation, I appreciate it! Nov 27, 2021 at 10:45