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Let $X$ and $Y$ be a Banach spaces and let $S\in B(Y^\ast,X^\ast)$. Do we always have $S=T^\ast$ for some $T\in B(X,Y)$?

I know that every bounded linear operator has an adjoint, also bounded linear operator. But in this case I don’t even know what can be used. Maybe you can give me some hints?

thing
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2 Answers2

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Consider the case where $X = \mathbb{R}$. Then you can identify $B(Y^*, X^*) = Y^{**}$ (say, via the map $\mathbb{R}^* \to \mathbb{R}$, $f \mapsto f(1)$), and with a bit of work you can prove that the elements of $B(Y^*, X^*)$ which are adjoints of elements of $B(X, Y)$ are exactly those in the image of the canonical map $Y \to Y^{**}$. Since this map isn't always surjective (e.g. when $Y = c_0$, the space of sequences converging to $0$), it's not true in general that every element of $B(Y^*, X^*)$ is an adjoint.

user125932
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  • Thanks. I need to think it over and understand, and then I will accept the answer. – thing Jan 03 '20 at 07:33
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    @thing Here are some details to help you. Define $T: \ell^{1}=(c_0)^{} \to \mathbb R$ by $T(a_n)=\sum_n a_n$. Suppose there exists $S: \mathbb R \to c_0$ such that $T=S^{}$. Let $S(1)=(b_n)$ so $b_n \to 0$. Using the definition adjoint show that $\sum a_n =\sum a_nb_n$ for all $(a_n) \in \ell^{1}$. Conclude that $b_n=1$ for all $n$ which is a contradiction. – Kavi Rama Murthy Jan 03 '20 at 07:38
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    I also have a more fleshed-out proof in one of the earlier edits of my answer; I edited because I wasn't sure if you would want an outline or a full solution – user125932 Jan 03 '20 at 07:48
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    @user125932 You are right. I didn't notice that OP only wants hints. I gave away too much I guess. $+1$ for your answer. – Kavi Rama Murthy Jan 03 '20 at 07:51
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In fact

Suppose that $S:Y^*\to X^*$ is linear. There exists a bounded $T:X\to Y$ such that $S=T^*$ if and only if $S$ is continuous with respect to the weak* topologies on $Y^*$ and $X^*$.

David C. Ullrich
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