Conditions for a given operator being compact.

Question

So I was given this question in class, and I thought it looked easy enough at first glance, but actually trying to do it, I have gotten quite stuck on the "only if" part.

Let $T\colon \ell_\infty \rightarrow \ell_\infty$ be defined in the following way; fix a sequence $c = (c(n))_{n\in\mathbb N}\in \ell_\infty$. For $x = (x(n))_{n\in\mathbb N}, (Tx)(n) = c(n)x(n)$ for ${n\in\mathbb N}$.

Show that $T$ is compact if and only if $\lim_{n\to\infty}c(n) = 0$.

I thought about using the fact that any sequence in $\overline{TE_1}$ must have a convergent subsequence, but can't see how to use it?

Any helpful hints are very welcome!

Answer 1

Take $$ x_n = \sum_{j=1}^n e_j \in \ell_{\infty} $$ where $$e_j (k) =\begin{cases} 1 \mbox{ if } j=k \\ 0 \mbox{ if } j\neq k \end{cases}.$$

The sequence $(Tx_n )$ must have a convergent subsequence hence: for every $\varepsilon >0$ there exists $k$ such that for every $j,m\geq k$ we have $$||Tx_{n_m} -Tx_{n_j}||\leq \varepsilon$$ but the last inequality shows that $$\lim_{n\to\infty} c(n) =0.$$

Answer 2

Suppose that $c(n)\not \to 0$ as $n\to \infty$. Then passing to a subsequence if necessary, we conculde that for some $\delta>0$ we have $|c(n)|>\delta$ for all $n$. Then $(e_n)_{n=1}^\infty$ is a bounded sequence such that $(Te_n)_{n=1}^\infty$ has no convergent subsequence.

Consider now the case where $c(n)\to 0$ as $n\to \infty$. Let $((x_n^k)_{k=1}^\infty)_{n=1}^\infty$ be a bounded sequence in $\ell_\infty$. The sequence $((c(k)x_n^k)_{k=1}^\infty)_{n=1}^\infty$ has a convergent subsequence by the standard diagonalisation argument.