0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $(x_n^{**})\in \ell_2(X)$.

$\endgroup$
3
  • $\begingroup$ 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 $\sum|| x_n^{**}||^{2}=\infty$. $\endgroup$ Jan 22, 2020 at 8:46
  • $\begingroup$ $(x_n^{**}/n)\in \ell_2(X)\Rightarrow x_n^{**}\in X$ for each $n$. Moreover $(x_n^{**})\in \ell_2(X^{**})$ and $x_n^{**}\in X$ for each $n$ implies $(x_n^{**})\in \ell_2(X)$. $\endgroup$ Jan 22, 2020 at 9:20
  • $\begingroup$ 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 $(x_n ^{**}) \in l_2 (X)$. $\endgroup$ Feb 7, 2020 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.