Proving $T$ is a compact operator and thus has not a closed range

Question

Consider the following operator on $\ell_{\mathbb{N}^*}^2(\mathbb C)$:$$T=\begin{bmatrix}1\\&\dfrac1{2!}\\&&\dfrac1{3!}\\&&&\ddots\end{bmatrix}.$$

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

To do this I try to prove that $T$ is compact.

Answer 1

To see that $T$ is not onto, take $y=(1,1/2,1/3,\ldots)$. If $Tx=y$ then we have, for each $n$, $$ \frac1{n!}\,x_n=\frac1n. $$ This forces $x_n=(n-1)!$, so $x$ would be unbounded and thus far from being in $\ell^2$. So $T$ is not surjective.

On the other hand, if $\{e_n\}$ denotes the canonical basis we have $Te_n=e_n/n!$, so $e_n=n!\,Te_n\in\operatorname{ran}T$. Thus the range of $T$ contains the span of $\{e_n\}$, which is dense, and so $\operatorname{ran}T$ is dense.

Answer 2

Any finitely non-zero sequence is in the range and the range is not all of $\ell ^{2}$ so it is not closed. [ If $x_n=0$ for $n \geq m$ then $((x_1) (1!),(x_2) (2!),...)$ is in $\ell^{2}$ and its image is $(x_n)$. For any sequence $(y_n)$ in the range we have $|y_n| \leq \frac C {n!}$ for some constant $C$ , so $T$ is not surjective. $T$ is also compact but I think this argument is much simpler].

Answer 3

For compactness you may also use the following

Theorem. A set $ \mathcal{F} \subseteq \ell^p (\mathbb{N}) $, $p \in [1,+\infty)$, is relatively compact if and only if $\mathcal{F}$ is bounded, closed and for every $\epsilon > 0$ there esists $n_\epsilon \in \mathbb{N}$ such that for every $x = \{x_n\}_{n \ge 1} \in \mathcal{F}$ and every $n > n_\epsilon$ we have $$ \sum_{k \ge n} |x_k|^p < \epsilon.$$

Your operator acts on some orthonormal basis $ \{ e_n\} _{n \ge 1} \subseteq \ell^2 $in this way: $T e_n = e_n / n! $. If you take any element $x \in \ell^2 $ with $ \|x\|_{\ell^2}\le 1$, by standard theory of Hilbert spaces we have that $$ x = \sum_{n \ge 1} (x,e_n) e_n;$$ to conclude observe that $$ \|Tx\|^2 = \left\| \sum_{n \ge 1} (x,e_n) \frac{e_n}{n!} \right\|^2 = \sum_{n \ge 1} \frac{|(x,e_n)|^2}{n!^2} \le \|x\|^2 \sum_{n \ge 1} \frac{1}{n!^2}=\sum_{n \ge 1}\frac{1}{n!^2}. $$Being the latter a convergent series, its tail is arbitrarily small, and this gives the uniform control required by the aforementioned theorem.