Compactness of a bounded operator $T\colon c_0 \to \ell^1$

Question

Pitt Theorem says that any bounded linear operator $T\colon \ell^r \to \ell^p$, $1 \leq p < r < \infty$, or $T\colon c_0 \to \ell^p$ is compact.

I know how to prove this in case $\ell^r \to \ell^p$, and $c_0 \to \ell^p$, where $p > 1$. Main idea in the first case is that $\ell^r$ is reflexive and hence closed ball $B_{\ell^r}$ is weakly compact. In the second case we could just use Schauder Theorem ($T$ is compact if and only if $T^*$ is compact).

The only case left is $T\colon c_0 \to \ell^1$. I have tried something like this:

By Schauder Theorem we need to prove that $T^*\colon \ell^\infty \to \ell^1$ is compact. By Banach-Alaoglu Theorem we know that $B_{\ell^\infty}$ is compact in the $weak^{*}$ topology on $\ell^\infty$. Moreover, we know, since $\ell^1$ is separable, that $B_{\ell^\infty}$ is metrizable. Hence, it is enough to prove that if $(x_n)$ is a $weak{}^{*}$ convergent (say, to $x$) sequence in $B_{\ell^\infty}$ then $(Tx_n)$ converges (to $Tx$, I think). Since Schur Theorem (weak and norm convergence is the same in $\ell^1$) we only need to show that $(Tx_n)$ converges weakly in $\ell^1$.

And here I stuck. Could you give me any ideas or references? In every book I have looked so far this particular case was omitted.

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Edit (4.4.2011): I found in Diestel's Sequences and series in Banach spaces (chap. VII, Exercise 2(ii)) something like this:

A bounded operator $T: c_0 \to X$ is compact if and only if every subseries of $\sum_{n=1}^\infty Te_n$ is convergent, where $(e_n)$ is canonical basis for $c_0$.

I know how to prove this, but how we can show that operators $T: c_0 \to \ell^1$ possess the subseries property?

Answer 1

A very nice and short proof of Pitt's theorem (a bit more than a page and covering the case you're interested in as well) was recently given by Sylvain Delpech MR review here, online article here.

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Added:

Let me adress your questions in the comments in a pedestrian way since I find this more illuminating than appealing to heavy artillery.

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Lemma. Let $X$ be a Banach space with separable dual space $X'$. Then every bounded sequence $(x_{i})_i$ has a weak Cauchy subsequence.

Proof. Let $(\phi_{j})_{j}$ be a dense sequence in $X'$ and suppose $\|x_{i}\| \leq C$ for all $i$.

1. (Diagonal trick)
   Since $|\langle x_{i}, \phi_{1} \rangle| \leq C \|\phi_{1}\|$, we may extract a subsequence $(x_{i}^{(1)})_i$ of $(x_{i})_i$ such that $\langle x_{i}^{(1)} ,\phi_{1}\rangle$ converges. Assume by induction we have constructed a subsequence $(x_{i}^{(n)})_{i}$ such that $\langle x_{i}^{(n)}, \phi_{j} \rangle$ converges for $j = 1, \ldots, n$. Then, as $|\langle x_{i}^{(n)}, \phi_{n+1} \rangle| \leq C \|\phi_{n+1}\|$ we find a subsequence $(x_{i}^{(n+1)})_{i}$ of $(x_{i}^{(n)})_{i}$ such that $\langle x_{i}^{(n+1)}, \phi_{n+1} \rangle$ converges. Now put $y_i = x_{i}^{(i)}$ and observe that $\langle y_i, \phi_j \rangle$ converges for all $j$.

2. (Triangle inequality)
   The sequence $(y_j)_j$ is a weak Cauchy sequence: Let $\phi \in X'$ be arbitrary and let $\varepsilon \gt 0$. Choose $j$ such that $\|\phi_j - \phi\| \lt \varepsilon$. For $m,n$ large enough we then have $|\langle y_n - y_m, \phi_j \rangle| \lt \varepsilon$, hence \begin{align*} |\langle y_{n} - y_{m}, \phi \rangle| & \leq |\langle y_{n}, \phi - \phi_j \rangle| + |\langle y_n - y_m, \phi_j \rangle| + |\langle y_m, \phi_j - \phi\rangle| \\ & \lt (2C + 1) \varepsilon \end{align*} and thus $(y_j)_j$ is indeed a weak Cauchy sequence.

Of course, this is nothing but the usual Arzelà-Ascoli argument.

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Further edit

Vobo points out in a comment below that one can prove the lemma in a blow by saying: Since $X'$ is separable, the closed unit ball $B_{X''}$ in $X''$ with the weak$^{\ast}$-topology is metrizable and it is compact by Alaoğlu. Hence every sequence in $B_{X''}$ has a weak$^{\ast}$-convergent subsequence, in particular it is weak$^{\ast}$-Cauchy. But it is a tautology that a sequence from $B_X$ is weak$^{\ast}$-Cauchy in $B_{X''}$ if and only if it is weakly Cauchy in $B_X$, so the lemma follows.

But what have we actually done in this argument? First, we use a diagonal argument to show that $B_{X''}$ is compact (to prove Alaoğlu we need the ultrafilter lemma, Tychonoff, Arzelà-Ascoli or whatever). Then we use separability of $X'$ to see that the weak$^{\ast}$-topology on $B_{X''}$ is second countable, hence metrizable by Urysohn. To do this a bit more explicitly, we can construct a metric on $B_{X''}$ by choosing a dense sequence $\{\phi_{n}\}$ in $B_{X'}$ and putting $d(x'',y'') = \sum 2^{-n} \frac{|\langle x'' - y'', \phi_n\rangle|}{1 + |\langle x'' - y'', \phi_n\rangle|}$. That this metric induces the weak$^{\ast}$-topology essentially is the argument in 2. above. Then we remember that compact metrizable spaces are sequentially compact and use a bit of fancy language. Unpacking all this and simplifying, we get exactly the argument I gave above. Conversely, understanding the argument above gives us a way of really understanding what is going on in all these theorems as well, so I don't think we've lost anything. On the contrary.

Of course, I agree that calling these theorems heavy artillery is exaggerating a bit...

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As for the second question, a linear operator between locally convex spaces $(E,\{|\cdot|_p\}_p)$ and $(F,\{|\cdot|_q\}_q)$ is continuous if and only if for each $q$ there exist $p_1,\ldots,p_n$ and $C \gt 0$ such that for all $x \in E$ we have $|Tx|_q \leq C\sum |x|_{p_j}$. Given that we know that $T$ is weak-norm continuous we have $\|Tx\| \leq C \sum |\langle x, \phi_j \rangle|$ for some $\phi_1, \ldots, \phi_n \in X'$ so that for a weak Cauchy sequence $(x_i)_i$ we have for $n,m$ large enough that $\|T(x_n - x_m)\| \leq C \sum |\langle x_n - x_m, \phi_j\rangle| \leq C \varepsilon$ and hence $(T(x_i))_i$ is Cauchy in norm.

Answer 2

In my opinion, the proof strategy is straightforward.

Let $(x_n)$ be a sequence in $B_{c_0}$, the unit ball of $c_0$. As $((Tx_n)(1))_n$ is a bounded sequence of scalars, there is a bounded subsequence converging to some scalar y_1. Now, inductively, you have a sub-sub-... sequence, call it for simplicity $(x_m)$ of $(x_n)$, and an element $y = (y_i)_i$ such that for each $i \in \mathbb{N}$ $(T(x_m)(i)) \to y_i$ as $m \to \infty$.

Now $\sum_i |y_i| \leq \sum_i |y_i - T(x_m)(i)| + \sum_i |T(x_m)(i)|$. The latter series is bounded by $||T||$ independently of $m$. The first series can be made arbitrary small for large $m$ by construction. Hence $y \in l^1$ and $T(x_m) \to y$ as $m\to \infty$ in norm (the first series).

Answer 3

Any operator $T:c_0\to X$ fails to be compact only if it fails to be strictly singular (see Albiac and Kalton, Theorem $2.4.10$). This would mean there is an infinite-dimensional subspace $E\leq c_0$ so that $T|_E$ is an embedding. But $c_0$ is self-saturated, so there is some $F\leq E$ isomorphic to $c_0$, and $T|_F$ is an embedding. So if $T:c_0\to X$ fails to be compact, $X$ contains a copy of $c_0$. A corollary is that all operators $T:c_0\to \ell_p$ are compact, $1\leq p<\infty$.

In fact, this is a characterization of spaces which contain $c_0$. $X$ does not contain a copy of $c_0$ if and only if all operators $T:c_0\to X$ are compact.