I am trying the following problem.

Let $X$ be a Banach space that does not contain a copy of $l^1$. Show that every operator $T : X \rightarrow L^1$ is weakly compact.

I am not sure how to start. Should we use the equi-integrability criterion? Any help is appreciated!

  • 2
    $\begingroup$ I can't think of a solultion not using Rosenthal's $\ell_1$-Theorem (A Banach space does not contain a copy of $\ell_1$ iff every bounded sequence has a weakly cauchy subsequence). Here, take a bounded sequence in $X$. Pass to a weakly Cauchy subsequence. Its image is weakly Cauchy and in fact weakly convergent since $L[0,1]$ is weakly sequentially complete. (The latter fact is 5.2.10 in Albiac/Kalton. They do mention the $\ell_1$-theorem prior to this exercise, but only prove it later in the text.) $\endgroup$ – David Mitra Apr 13 at 3:47

Suppose $T:X\rightarrow L_1$ is not weakly compact. Then, using the Eberlein–Šmulian theorem, there is a bounded sequence $(x_n)$ in $X$ such that $F=\{ Tx_n : n=1,2,\ldots\}$ is not relatively weakly compact in $L_1$. This implies that $F$ contains a basic sequence $(T x_{n_k})_k$ that is equivalent to the standard unit vector basis of $\ell_1$ (by a well-known characterization of weakly compact sets in $L_1$. See e.g., Theorem 5.2.9 of Albiac and Kalton's Topics in Banach Space Theory).

Let $Z$ be the closed linear span of $\{ Tx_{n_k}: k=1,2,\ldots\}$. $Z$, of course, is $\ell_1$ is disguise with "standard basis" $(Tx_{n_k})$.

Let $Y$ be the closed linear span of the $x_{n_k}$ in $X$. Let's look at $T|_Y$.

Note $T(Y)\subseteq Z$. We now show $T|_Y$ is in fact onto $Z$. Let $z\in Z$. Then $z=\sum\limits_{k=1}^\infty \alpha_k Tx_{n_k}$ for some sequence $(\alpha_k)\in\ell_1$. Since $(\alpha_k)\in\ell_1$ and $(x_{n_k})$ is bounded,
$$y=\sum_{k=1}^\infty \alpha_k x_{n_k}=\lim_{m\rightarrow\infty}\sum_{k=1}^m\alpha_k x_{n_k}$$ is well-defined. Further $y\in Y$ and $$Ty=\lim_{m\rightarrow\infty}\sum_{k=1}^m\alpha_k Tx_{n_k}=z.$$

So $T|_Y$ is onto $Z$.

Thus $T|_Y$ is a bounded operator from the Banach space $Y$ onto a space isomorphic to $\ell_1$. This implies (see exercise 2.8 in Albiac-Kalton, e.g.) that $Y$, and thus $X$, contains a copy of $\ell_1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.