$T$ has not a closed range

Question

Let $$T = {\rm diag}(0, 1, 0, \frac{1}{2!}, 0, \frac{1}{3!}, \dots)$$

Clearly $T$ is a positive operator on the Hilbert space $\ell_{\mathbb{N}^*}^2(\mathbb{C})$.

I want to prove that $T$ has not a closed range.

Answer 1

Write $\{e_n\}$ for the canonical basis. You have that $Te_{2n-1}=0$ for all $n$, so $\operatorname{ran}T\subset X$, where $$X=\overline{\operatorname{span}}\{e_{2n}:\ n\in\mathbb N\}.$$

For any $n$, $Te_{2n}=\tfrac1{n!}\,e_{2n}.$. So $$e_{2n}=T(n!\,e_{2n})\in \operatorname{ran}T.$$ So $\operatorname{ran}T$ is dense in $X$. But there are elements in $X$ that are no in $\operatorname{ran}T$. Consider for instance $$ x=\sum_{n=1}^\infty \tfrac1n\,e_{2n}. $$ To have $x\in\operatorname{ran}T$, say $x=Ty$, where $y=\sum_{n=1}^\infty c_ne_{2n}$ and $\sum_n|c_n|^2<\infty$. From $x=Ty$ we get $$ \tfrac1n\,e_{2n}=\tfrac1{n!}\,c_n\,e_{2n}, $$ which would require $c_n=(n-1)!$.