Let $X$ be a Banach space non reflexive and $T$ from $l\_2(X)$ to $l\_2(X)$ a bounded operator defined by:
$$T(x\_n )=\frac{x\_n }{n}.$$
We know that :
$$T^{\*\*-1}(l\_2(X))=\{x\_n^{\*\*} \in l\_2(X^{\*\*}) : \frac{x\_n^{\*\*}}{n} \in l\_2 (X)\}.$$
To prove that $T$ is Tauberian, it suffices to prove that $T^{\*\*−1}(l\_2(X))\subset l\_2(X)$. I.e., we will check that: $\sum \|x\_n^{\*\*}\|^{2} < \infty.$
Please help me to solve this problem.



Tauberian operators
You can proceed as follows. Let $(x\_n^{\*\*})\in \ell\_2(X^{\*\*})$. Then
$$(x\_n^{\*\*})\in T^{\*\*-1}(\ell\_2(X))\Rightarrow T^{\*\*}(x\_n^{\*\*})= 
(\frac{x\_n^{\*\*}}{n})\in \ell\_2(X),$$ 
hence $(x\_n^{\*\*})\in \ell\_2(X)$.


