No bounded linear surjectivity from $\ell^2(\mathbb{N})$ to $\ell^1(\mathbb{N})$

Question

I would like to show that a bounded linear operator $T: \ell^2(\mathbb{N}) \rightarrow \ell^1(\mathbb{N})$ cannot be surjective.

If we assume surjectivity, then by the Open Mapping Theorem, $T$ is an open map, i.e. for every open $O \subset \ell^2(\mathbb{N})$, its image $T(O)$ is open in $\ell^1(\mathbb{N})$.

Is it then possible to contradict Baire's Theorem by constructing a sequence of open dense subsets of $\ell^1(\mathbb{N})$, whose intersection is not dense?

Any help pointing me into the right direction would be appreciated.

Answer 1

If $T: \ell^2(\mathbb N) \to \ell^1(\mathbb N)$ is a surjective bounded linear transformation, then the adjoint $T^*: (\ell^1(\mathbb N))^* \to (\ell^2(\mathbb N))^*$ is an isomorphism onto a closed linear subspace. But $(\ell^1(\mathbb N)^* \sim \ell^\infty(\mathbb N)$ is not separable, while $\ell^2(\mathbb N)^* \sim \ell^2(\mathbb N)$ is separable, and so is any of its subspaces.

Answer 2

As $\ell_2$ is reflexive, every operator $T\colon \ell_2\to \ell_1$ is weakly compact. Thus the unit ball $B_{\ell_1}$ of $\ell_1$ would be weakly compact but the canonical basis of $\ell_1$ is in $B_{\ell_1}$ but it does not have a weakly convergent subsequence.

Actually, by Pitt's theorem, every bounded linear operator $T\colon \ell_2\to\ell_1$ is compact.