Paley projection is not absolutely summing

Question

It is well known that the operator $P: H^{1}(\mathbb{T}) \to \ell_2$, given by $Pf= (\hat{f}(2^{n}))_{n \in \mathbb{N}}$ is bounded and its restriction $P_{|A(\mathbb{D})}: A(\mathbb{D}) \to \ell_2$ is absolutely summing. I feel that $P$ is not absolutely summing but I am not able to prove it; probably missing something obvious. So here is my question: is Paley projection $P:H^{1}(\mathbb{T}) \to \ell_2$ absolutely summing?

Answer 1

If I'm reading Wojtaszczyk's book "Banach spaces for analysts" correctly, any absolutely summing operator maps weakly convergent sequences to norm convergent sequences. The Paley projection does not: if you denote by $(f_k)$ the trigonometric system and by $(e_n)$ the canonical basis of $\ell_2$, then $Pf_{2^n}=e_n$ for all $n\geq 0$, but $f_{2^n}\to 0$ weakly in $H^1$ and $e_n$ does not tend to $0$ in norm. So $P$ is not absolutely summing.