Let $H$ be any Hilbert space. How can we prove that any bounded linear operator $T\colon H \to \ell^1$ is compact?
If we use the fact that the space $\ell^1$ has Schur property (norm and weak convergence is the same), then we need to show that for a sequence $(x_n) \subset H$ such that $||x_n|| \leq 1$ the sequence $(Tx_n) \subset \ell^1$ contains weakly convergent subsequence. But I do not know how I should do this.
What property of a Hilbert spaces I need to use?
First, we can assume $H$ is separable, since any sequence in $H$ is contained in a closed separable subspace.
Because $\ell^1$ has the Schur property, we just need to show that for every $x_n$ in the closed unit ball of $H$, $Tx_n$ has a weakly convergent subsequence. Since for every $f \in {\ell^1}^*$, $fT$ is a continuous linear functional on $H$, it's enough to show that $x_n$ has a weakly convergent subsequence. Thus it's enough to show that the closed unit ball of $H$ is weakly sequentially compact. Since $H$ is self-dual and separable, the sequential Banach-Alaoglu theorem tells us exactly what we want.
