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!
 A: 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 $\ell_1$.
