Every operator $T : X \rightarrow L^1$ is weakly compact.

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!

• 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.) – 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 $$\ell_1$$.