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.
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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 $(x_n^{**})\in \ell_2(X)$.
answered Jan 22, 2020 at 8:20
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$\begingroup$ We have $\frac{x_n^{**}}{n} \in l_2(X) $, then $\sum|| \frac{x_n^{**}}{n}||^{2} < \infty $. Taking account that $\sum|| \frac{x_n^{**}}{n}||^{2} \leq \sum|| x_n^{**}||^{2}$, so perhaps $\sum|| x_n^{**}||^{2}=\infty$. $\endgroup$ Commented Jan 22, 2020 at 8:46
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$\begingroup$ $(x_n^{**}/n)\in \ell_2(X)\Rightarrow x_n^{**}\in X$ for each $n$. Moreover $(x_n^{**})\in \ell_2(X^{**})$ and $x_n^{**}\in X$ for each $n$ implies $(x_n^{**})\in \ell_2(X)$. $\endgroup$ Commented Jan 22, 2020 at 9:20
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$\begingroup$ Please, a nother quetion Mr. Gonzalez. If $(x_n ^{**}) \in l_2 (X^{**})$ and $(x_n^{**}) \in X$, then $(x_n ^{**}) \in l_2 (X^{**}) \cap X $. How to conclude that $(x_n ^{**}) \in l_2 (X)$. $\endgroup$ Commented Feb 7, 2020 at 18:20