Compact operators, space of sequences

Question

Let $\phi\in\ell^\infty$. For $p\in[1,\infty]$, define $M_\phi:\ell^p\to\ell^p$ by

$$M_\phi(f)=\phi f.$$

Show that $\Vert M_\phi\Vert=\Vert\phi\Vert_\infty$, and $M_\phi$ is compact if and only if $\phi\in c_0$, i.e. $\phi$ is a sequence that converges to $0$.

I only have problem with the part "$\phi\in c_0$ $\Rightarrow$ $M_\phi$ compact.

I tried to prove by contradiction, assume $M_\phi$ is not compact, then there is a bounded sequence $(f_n)_{n\in\mathbb{N}}$ in $\ell^p$ s.t. $(\phi f_n)_{n\in\mathbb{N}}$ has no convergent subsequence, then it also has no Cauchy subsequence. Then we can define

$$t:=\inf_{m\neq n}\Vert \phi(f_m-f_n)\Vert_p>0$$

At this point I am stuck, I tried to find some $n\neq m$ s.t. $\Vert \phi(f_m-f_n)\Vert_p<t$, then we get a contradiction. Can someone give some hints? Thanks!

P.S. It is required to use the definition of compactness to prove this question.

Answer 1

First I would point out that the $t$ you have defined may in fact be zero, even if the sequence $M_\phi (f_n)$ has no Cauchy subsequence - all you would need is for some pair $f_n,f_m$ to map to the same sequence under $M_\phi$.

Now, to prove the claim, one relatively easy way is to express $M_\phi$ as the norm limit of a sequence of finite rank operators.

If you don't want to appeal to this fact, you can prove the claim directly. You should consider a bounded set $B$ (say, the unit ball) in $\ell^p$, and show that the set $M_\phi(B)$ is totally bounded by covering it with a finite number of $\varepsilon$-balls for arbitrary $\varepsilon>0$. Hint: think about what multiplication by $\phi \in c_0$ will do to the tail of a unit-norm element of $\ell^p$.

Edit: details on what I mean above. Let $B$ be the unit ball in $\ell^p$ and let $\varepsilon>0$; we seek a finite collection of $\varepsilon$-balls covering $M_\phi(B)$.

As $\phi\in c_0$ we can find $N$ so that $\vert \phi_n \vert\leq \varepsilon/2$ for all $n>N$. Now consider each element of $M_\phi(B)$ as having two pieces: the first $N$ entries, and the rest (the tail). By the argument that gives $\Vert M_\phi \Vert = \Vert \phi\Vert_\infty$, the $p$-norm of the tail will be less than $\varepsilon/2$. In other words, any element of $M_\phi(B)$ is less than $\varepsilon/2$ away from a finite-dimensional (i.e. $N$-dimensional) bounded subset of $\ell^p$, call it $F$.

Finite dimensional and bounded means that we can cover $F$ by a finite number of $\varepsilon/2$-balls. Now, any element of $M_\phi(B)$ differs by at most $\varepsilon/2$ from some element of $F$, which in turn differs by at most $\varepsilon/2$ one of the finitely many centers. The triangle inequality gives us that an arbitrary element of $M_\phi (B)$ differs by at most $\varepsilon$ from one of the finitely many centers - i.e. we've covered $M_\phi(B)$ by a finite collection of $\varepsilon$-balls.

Answer 2

If $\boldsymbol\varphi\in\ell^\infty(\mathbb N)\smallsetminus c_0(\mathbb N)$, then there is an $\varepsilon>0$, and a subsequence $\{\varphi(k_n)\}_{n\in\mathbb N}$, such that $$ \lvert \varphi(k_n)\rvert\ge \varepsilon. $$ Let now $$ \boldsymbol{u}_n=\boldsymbol{e}_{k_n}, $$ where $\boldsymbol{e}_{n}$ is the sequence is which is zero for all $k\in\mathbb N$, except $k=n$, where it takes the value 1. Clearly, $\boldsymbol{e}_{n}\in\ell^p(\mathbb N)$, for all $p$, and $\|\boldsymbol{e}_{n}\|_p=1$. Now $\{\boldsymbol\varphi\boldsymbol{u}_n\}_{n\in\mathbb N}$ does not have a convergent subsequence as $$ \|\boldsymbol\varphi\boldsymbol{u}_m-\boldsymbol\varphi\boldsymbol{u}_n\|_p\ge 2^{1/p}\varepsilon, $$ for all $m\ne n$.