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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.


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?

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3 Answers 3

up vote 4 down vote accepted

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).

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Thanks for that! I was just about to add something to the same effect. –  t.b. Apr 4 '11 at 9:13
    
@xenom: Your edit is used in the statement $(c_i) \in l^1$ and is standard in any proof of $l^1$ being the dual of $c_0$. –  Vobo Apr 4 '11 at 9:14
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@Vobo: Thanks! Maybe this is silly question, but are you sure that $(c_i) \in \ell^1$? Since $T: c_0 \to \ell^1$ we have $c_i \in \ell^1$, don't we? –  xen Apr 4 '11 at 9:31
    
@xenom: You are right, I have to think about this. –  Vobo Apr 4 '11 at 10:31
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@xenom: I removed the $(c_i)$-part, as it was not necessary. Please recheck the argument above, which looks so obvious, that I wonder why I introduced the $(c_i)$-part ... –  Vobo Apr 4 '11 at 14:13

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.


Added:

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


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.


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...


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.

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@xenom: I am aware that this answer is borderline in terms of terseness. I just felt that it didn't make much sense in reproducing Delpech's approach here, (and that's what further elaboration would have amounted to). If you have further questions or want some more clarifications, please ask. –  t.b. Apr 4 '11 at 7:21
    
@Theo Buehler: Thank you very much for the reference! I will read it carefully. –  xen Apr 4 '11 at 8:40
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@xenom: Yes, it is obvious. Try to express the continuity of such an operator in terms of (semi-)norms. Basically you have $||Tx|| \leq C |x^*(x)|$ for some $x^* \in X^*$. –  Vobo Apr 5 '11 at 14:03
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@Theo Bühler: It is even simpler. Denote by $B^{\ast\ast}$ unit ball in $X^{\ast\ast}$, compact and metrizable in the weak**-topology. Hence every sequence in the unit ball in $X$ - regarded as an element of $X^{\ast\ast}$ - has a weak**-convergent subsequence, so in particular a weak**-Cauchy subsequence. And this subsequence - now regarded back as elements of X - must be a weak Cauchy sequence. Oh, and sorry, I didn't read your "pedestrian approach" in contrast to my heavy artillery. –  Vobo Apr 5 '11 at 14:53
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@Vobo: sure, but you're using Arzelà-Ascoli as well, don't you? If you unpack your argument you get exactly what I do. As I said, I wanted to do this in a completely pedestrian way. –  t.b. Apr 5 '11 at 14:54

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.

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