Is the multiplication operator on $\ell^p$ compact?

Question

Let $p\in[1,\infty]$, $a\in\ell^\infty$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ Note that $\left\|T\right\|=\left\|a\right\|_{\ell^\infty}$. It's easy to see that if $a\in\ell^1$, then $T$ is nuclear and hence compact.

Can we more generally conclude that $T$ is compact?

Assume $p<\infty$. The intuitive approach would be to define $$T_n:\ell^p\to\ell^p\;,\;\;\;x\mapsto(a_1x_1,\ldots,a_nx_n,0,\ldots)$$ for $n\in\mathbb N$ and to show that $\left\|T_n-T\right\|\to0$. Now, $$\left\|(T_n-T)x\right\|^p=\sum_{i=n+1}^\infty|a_ix_i|^p\tag1$$ for all $x\in\ell^p$ and $n\in\mathbb N$. The right-hand side of $(1)$ clearly converges to $0$ as $n\to\infty$, but this only yields convergence in the strong operator topology.

On the other hand, if $(a_n)_{n\in\mathbb N}$ would be nonincreasing and $a_n\to0$ (e.g. $a_n=\frac1n$), then $$\left\|(T_n-T)x\right\|^p\le|a_{n+1}|^p\left\|x\right\|_{\ell^p}^p\tag2$$ for all $x\in\ell^p$ and $n\in\mathbb N$ and we could immediately conclude $\left\|T_n-T\right\|\to0$.

EDIT: Maybe the general result can be obtained from the convergence in strong operator topology (which holds by $(1)$) by noting that $\ell^p$ has the approximation property; but I'm not sure whether this is true.

Answer 1

First of all, you cannot conclude that an operator is compact if it is the strong limit of finite rank operators, for example in any Hilbert space the identity operator is the strong limit of a net of finite-rank projections, but of course it is compact if and only if the dimension of our space is finite.

As Timur Bakiev says, the multiplication operator $T:\ell^p\to\ell^p$ is compact if and only if the sequence of $\ell^\infty$ that acts on $\ell^p$ belongs to $c_0$.

You already showed the converse: In your notation, $$\|(T_n-T)x\|^p=\sum_{k=n+1}^\infty|a_k|^p\cdot|x_k|^p\leq\sup_{k\geq n+1}|a_k|^p\cdot\|x\|^p$$ so $\|T_n-T\|\leq\sup_{k\geq n+1}|a_k|\to0$.

Conversely, suppose that $T$ is compact. We will use the following lemma from real analysis: if $(t_n)$ is a sequence of real numbers and every subsequence of $(t_n)$ has a further subsequence converging to $t$, then $(t_n)$ converges to $t$. See this post for example: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.

By definition of compactness, if $(x_n)\subset\ell^p$ is a bounded sequence, $(Tx_n)$ has a convergent subsequence in $\ell^p$. Fix indexes $n_1<n_2<n_3<\dots$ and consider the elements $e_k\in\ell^p$ having $0$ everywhere and $1$ in the $n_k$ position. Then $\|e_k\|=1$, so this sequence is bounded and therefore $(Te_k)$ has a convergent subsequence in $\ell^p$. To simplify my notation, I will assume that $n_1=1, n_2=2,\dots$ but the method below can be applied to arbitrary subsequences too, you can verify this yourself easily.

So as I said, let $e_n$ be the sequence having $0$ everywhere and $1$ in the $n$-th position, so $\|e_n\|=1$ for all $n$. Note that $Te_n$ is the sequence that has $0$ everywhere and $a_n$ in the $n$-th position, so $\|Te_n\|_p=|a_n|$. Now $(Te_n)$ has a convergent subsequence, say $(Te_{n_k})$ and say that $Te_{n_k}\to y\in\ell^p$, thus $\|Te_{n_k}-y\|_p^p\to0$. If $y=(y_i)$, then $$\|Te_{n_k}-y\|_p^p=\sum_{i=1}^{n_k-1}|y_i|^p+|a_{n_k}-y_{n_k}|^p+\sum_{i=n_k+1}^\infty|y_i|^p$$ Fix an integer $m$. For $\varepsilon>0$ choose $k>m$ so large so that $\|Te_{n_k}-y\|_p^p<\varepsilon$.

Then $|y_m|^p\leq\sum_{i\neq n_k}|y_i|^p+|a_{n_k}-y_{n_k}|^p=\|Te_{n_k}-y\|_p^p<\varepsilon$ and as $\varepsilon>0$ was arbitrary, we conclude that $y_m=0$, so $y=0$ since $m$ was arbitrary before fixing. We conclude that $|a_{n_k}|\to0$.

Now as I said, this can be done for any subsequence (this is why I fixed $n_1<n_2<\dots$ in the beginning, nevermind that I eventually went wLOG to the case $n_1=1,n_2=2\dots$), so the lemma will yield that $|a_n|\to0$, i.e. $a_n\in c_0$.

Answer 2

The set of all compact operators between Banach spaces is closed in norm topology.

A multiplication operator on $l^p$ is compact if and only if the corresponding sequence is an element of $c_0$.

Answer 3

Here's another proof of necessity. Let $T_\alpha$ be a multiplication operator and $\alpha = \{\alpha_n\} \in l^\infty$ be the corresponding sequence.

Let $\alpha \notin c_0$. Then there exists $\varepsilon > 0$ s.t. for any $N \in \mathbb{N}$ we can find $n>N$ for which $|a_n| > \varepsilon$. Hence we can assume $|a_n| > \varepsilon$ for all $n \in \mathbb{N}$. This assumption is made for simplicity, but if we take the subspace of $l^p$ corresponding to the chosen subsequence, it will be isometrically isomorphic to $l^p$. A restriction of $T_\alpha$ to this subspace is again a compact operator. Then $T_\alpha(\mathbb{B}_1) \supset \mathbb{B}_\varepsilon$ ($\mathbb{B}_1$ is a unit ball).

Indeed, let $x \in l^p$, $||x|| \geq 1$. Then $$ ||T_\alpha x||_p = \sqrt[p]{\sum_{n=1}^\infty |\alpha_n x_n|^p} > \sqrt[p]{\sum_{n=1}^\infty \varepsilon^p |x_n|^p} = \varepsilon ||x||_p \geq \varepsilon, $$ and the statement follows since our operator is surjective.

But $\mathbb{B}_\varepsilon$ is not relatively compact in infinite dimensional space. Hence $T_\alpha$ is not compact.