Let $E$, $F$ be Banach spaces and $T:E\longrightarrow F$ be a bounded linear operator. Suppose $T$ is injective, is $T^{\prime\prime}$ necessarily injective?
Remark. My attempts at proving this were to prove the results
Claim 1. $S$ injective $\Rightarrow$ $S'$ has norm-dense image, for all operators, $S$, between Banach spaces.
Claim 2. $S$ has norm-dense image $\Rightarrow$ $S'$ injective.
from which the above easily follows. However Claim 1 seems to be invalid. Instead, I only know that
Claim 1´. $S$ injective $\Rightarrow$ $S'$ has weak*-dense image
and this result is useless here, unless
Claim 2´. $S'$ has weak*-dense image $\Rightarrow$ $S''$ injective.
holds true.
Consider $T:\ell^1\to \ell^1$ which sends a sequence $(a_n)$ to the sequence $(a_n/n)$. The dual map is then $T':\ell^\infty\to\ell^\infty$ which again sends a sequence $(a_n)$ to $(a_n/n)$. However, the image of $T'$ is not norm-dense, since every sequence in the image of $S'$ converges to $0$. Thus the double dual $T''$ is not injective.