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Let $X, Y$ be Banach spaces, $S \in B(Y^{*}, X^{*})$. Does such operator $T \in B(X, Y)$ exist so that $T^{*}=S$?

I suppose that the answer should be - no. Are there any hints that might help in constructing a counterexample?

Any help would be much appreciated.

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  • $\begingroup$ Look at $S^{\ast}$. If $S = T^{\ast}$, what conditions on $S^{\ast}$ does that impose? $\endgroup$ – Daniel Fischer Feb 24 '16 at 21:05
  • $\begingroup$ @DanielFischer Since, $||S^{*}||=||S||$, then $||S^{*}||=||T||$. $\endgroup$ – hyperkahler Feb 24 '16 at 21:16
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    $\begingroup$ True, but that doesn't give you much to work with. What do you know about operators of the form $T^{\ast\ast}$? $\endgroup$ – Daniel Fischer Feb 24 '16 at 21:24
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Such an operator exists if and only if $S$ is weakly*-to-weakly* continuous. This is always possible if $X$ and $Y$ are reflexive.

As for concrete examples of operators that do not have a "pre-joint", let $X$ be a separable, reflexive Banach space. Recall that $c_0^*\cong \ell_1$ and take any bounded linear surjection $S\colon (c_0)^*\to X^*$. It does not have a pre-joint because if it had, it would be an isomorphic embedding, but a reflexive Banach space cannot embed into $c_0$.

Here is another, perhaps more exciting example. Note that $C[0,1]^*$ contains a complemented copy of $L_1$. Let $P\colon C[0,1]^*\to C[0,1]^*$ be a projection onto any such subspace. If there were an operator $T$ such that $T^*=P$ then $T$ would be a projection too (that is $T^2=T$), but $L_1$ is not isomorphic to a dual space because separable dual spaces have the Radon–Nikodym property which $L_1$ clearly does not have.

If you do not like this example, note that $C[0,1]^*$ contains a copy of the non-separable space $\ell_1([0,1])$ (the closed linear span of Dirac delta measures) which is complemented. Again, the projection onto $\ell_1([0,1])$ does not have a pre-joint because if it had, it would be a projection with non-separable range.

There are indecomposable spaces with dual isomorphic to $\ell_1$. A space is indecomposable if the only complemented subspaces are either finite- or cofinite-dimensional. Now, take any projection on $\ell_1$ whose range is infinite-dimensional and has infinite codimension. In this case there is no chance for a pre-joint.

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  • $\begingroup$ I understood nearly nothing. $T \in B(X,Y)$ means two things : that $T$ is a bounded operator $\bar{X}\to \bar{Y}$ (the completions of $X,Y$ if they are not complete), and that $T$ is an operator $X \to Y$ not only $\bar{X}\to \bar{Y}$ . which one is your answer about ? $\endgroup$ – reuns Feb 24 '16 at 21:58
  • $\begingroup$ there are two questions : if the banach spaces are completes, is the adjoint operator bounded ? and if the Banach spaces are not completes, is the adjoint operator also an operator on those un-complete spaces, or does it send some points of $X$ to the completion of $Y$ (so outside of $Y$) ? $\endgroup$ – reuns Feb 24 '16 at 22:02
  • $\begingroup$ yes sorry, but this doesn't explain why your answer is impossible to understand. you rely on even more abstract concepts without saying what is the difficult point $\endgroup$ – reuns Feb 24 '16 at 22:10
  • $\begingroup$ @user1952009 check my first example (after edit). $\endgroup$ – Tomek Kania Feb 24 '16 at 22:11
  • $\begingroup$ for you it seems to be a complicated topology subject, whereas it is also a very simple question on linear operators. is all the point is about $X \cong X^{**}, Y \cong Y^{**}$ ? $\endgroup$ – reuns Feb 24 '16 at 22:21
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For a normed space $E$, let $\Phi_E \colon E \to E^{\ast\ast}$ denote the canonical embedding. The key to the following is the identity

$$\Phi_Y \circ T = T^{\ast\ast} \circ \Phi_X\tag{1}$$

for all $T \in B(X,Y)$. So we have $T^{\ast\ast}(\Phi_X(X)) \subseteq \Phi_Y(Y)$ and find the necessary condition

$$S^{\ast}(\Phi_X(X)) \subseteq \Phi_Y(Y)\tag{2}$$

for $S\in B(Y^{\ast}, X^{\ast})$ to be the transpose of some $T\in B(X,Y)$.

We note that $(2)$ is also sufficient for the existence of a $T$ with $S = T^{\ast}$, using $(1)$ we can then define $T = \Phi_Y^{-1} \circ S^{\ast} \circ \Phi_X$. It is a routine verification that $T$ is then a well-defined continuous linear operator with $T^{\ast} = S$.

Hence there always is such a $T$ if $Y$ is a reflexive Banach space or $X = \{0\}$.

If $Y$ is not reflexive and $X \neq \{0\}$, we can construct an $S\in B(Y^{\ast}, X^{\ast})$ that isn't a transpose by choosing an $\eta \in Y^{\ast\ast}$ and a $\xi \in X^{\ast}\setminus \{0\}$, and setting

$$S(\lambda) = \eta(\lambda)\cdot \xi.\tag{3}$$

For $S^{\ast}$, we find that

$$S^{\ast}(\psi) = \psi(\xi)\cdot \eta.\tag{4}$$

Since $\xi \neq 0$, there is an $x\in X$ with $\xi(x) \neq 0$, and then we have

$$S^{\ast}(\Phi_X(x)) = \Phi_X(x)(\xi)\cdot \eta = \xi(x)\cdot \eta \notin \Phi_Y(Y),$$

so $S$ cannot be the transpose of any $T\in B(X,Y)$.

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