Operator from $\ell_{4}$ to $\ell_{1}$ is compact, if it's continuous.

Question

Define $T: \ell_{4} \rightarrow \ell_{1}$ as $Tx=(a_1x_1, a_2x_2, \ldots)$. I showed that $T$ is continuous if and only if $\sum \left| a_i \right|^{\frac{4}{3}} < \infty$. How can I prove that if $T$ is continuous then it's compact?

Answer 1

Every continuous linear operator $T$ from an $\ell_p$ space with $1 < p < \infty$ to $\ell_1$ is compact.

The crucial facts are that $\ell_p$ is reflexive then, and $\ell_1$ has the Schur property.

Since $\ell_p$ is reflexive, every bounded sequence $(x_n)$ has a weakly convergent subsequence $(x_{n_k})$. The image of that subsequence is weakly convergent, since $T$ is also continuous when both spaces are endowed with their weak topology. By the Schur property, the weakly convergent sequence $T(x_{n_k})$ is norm-convergent.

Thus the image of the unit ball in $\ell_p$ is relatively sequentially compact, and since $\ell_1$ is a Banach space, it is relatively compact, i.e. $T$ is compact.