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Let $X,Y$ be two Banach spaces and $S:Y^* \to X^*$ be a bounded linear operator. Is there always bounded linear $T: X\to Y$ such that $S=T^*?$

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  • $\begingroup$ No. An operator is an adjoint operator if and only if it is weak*-weak* continuous. $\endgroup$ – David Mitra Jan 25 '15 at 16:44
  • $\begingroup$ @DavidMitra Is there any easy counterexample? I don't know what is weak*-weak* continuous. $\endgroup$ – Hautr Jan 25 '15 at 16:47
  • $\begingroup$ See my comment in this post. The post also has a reference to the fact mentioned in my previous comment. $\endgroup$ – David Mitra Jan 25 '15 at 16:51
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Let $e_n$ denote the element of $\ell_1$ whose $m$'th coordinate is $0$ if $m\ne n$ and $1$ if $m=n$.

To see more directly why the bounded linear map, referenced in my comment to the main post, $T:\ell_1\rightarrow\ell_1$ defined by $$Te_1=e_1;\ Te_n=e_1+e_n, n>1$$ is not an adjoint operator, consider the sequence $(e_n)$ in $\ell_1$. Note for any $x\in c_0$, we have $\lim\limits_{n\rightarrow\infty} e_n(x)= 0$.

Suppose $T$ were the adjoint of $S:c_0\rightarrow c_0$. By the definition of the adjoint, for any $x\in c_0$ and all $i$, we have $$ (S^*e_i)(x) =e_i(Sx).\tag{1}$$ Now note for $x\in c_0$, we have $$\lim\limits_{n\rightarrow\infty} e_n(Sx) =0,\tag{2}$$ and $$\lim\limits_{n\rightarrow\infty} ( S^*e_n)(x) =\lim\limits_{n\rightarrow\infty} (Te_n)(x)=x_1\tag{3}$$ By $(1)$, the limits $(2)$ and $(3)$ should be the same. But this isn't the case if the first coordinate of $x$ is non-zero.

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