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.

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    $\begingroup$ 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. $\endgroup$
    – GEdgar
    Nov 27, 2021 at 1:20
  • $\begingroup$ Related post: math.stackexchange.com/questions/1832836/… $\endgroup$ Nov 27, 2021 at 14:58

2 Answers 2


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.

  • $\begingroup$ Nice answer! Lovely that you got a characterisation. $\endgroup$ Nov 27, 2021 at 10:20
  • $\begingroup$ @QuantumSpace Thank you very much:) $\endgroup$ Nov 27, 2021 at 10:31
  • $\begingroup$ @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. $\endgroup$
    – Mrcrg
    Dec 9, 2021 at 17:13
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    $\begingroup$ @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 $\endgroup$ Dec 9, 2021 at 19:24
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    $\begingroup$ @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 $p$) $\endgroup$ 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.

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    $\begingroup$ Thanks for the confirmation, I appreciate it! $\endgroup$ Nov 27, 2021 at 10:45

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