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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $T\in B(X,Y)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.