prove that this double transpose $T''$ it's not injective.

Question

Prove that if $T''$ is injective then $T$ it's injective, but the converse it's not true.

I proved the first part " $T''$ injective implies $T$ injective".

It's easy to prove that $T''$ is injective if and only if the image of $T'$ is dense on $X'$. So If I prove that the $cl({Ran(T')})$ it's a proper subset of $X'$ then I'm done. But even this it's difficult.

Answer 1

Let $R_X$ and $L_X$ be the operators of right and left shift on the space $X$, where $X\in\{c_0, \ell_1,\ell_\infty\}$. Show that $L_{c_0}'=R_{\ell_1}$ and $R_{\ell_1}'=L_{\ell_\infty}$. Note that $T=L_{c_0}-1_{c_0}$, so $T''=L_{\ell_\infty}-1_{\ell_\infty}$. Now $T''$ is not injective becase $\ker (T'')=\operatorname{span}\{(1,1,\ldots)\}\neq\{0\}$.