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Martin Sleziak
Martin Sleziak
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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 knownknow 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.$$\sum \|x_n^{**}\|^{2} < \infty.$ Please help me to solve this problem.

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 known 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.

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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mahamed-beghdadi
mahamed-beghdadi
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Tauberian operators

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 known 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.