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I think it goes like this. \begin{align*} P \big(\overline{P(\overline{f} \circ \varphi_\lambda}) \big) (w) & = \int_\Omega \overline{P(\overline{f} \circ \varphi_\lambda}) (z) \overline{k_w(z)} dV(z) …

I asked the polydisc experts for a reference and in fact it is known. It was proved by by J.P. Rosay in this paper. It is in french but it shouldn't be difficult to understand.

No, I don't think so. For an explicit example, for $v,u\in L^2(\mathbb{R})$, let me use the notation $v\otimes w$ for the rank $1$ operator $(v\otimes w) (f) = (w,f) v$. Then consider any fixed $g\in …

Suppose that $A\in M_d(\mathbb{Z})$ is a $d \times d$ matrix with non zero determinant and suppose that $\mathbb{T}^d$ is the $d$-dimensional torus. Then one can define an operator on $L^2(\mathbb{T}^ …

If a sequence is weakly separated, i.e. there exists a multiplier $\varphi_{ij}$ of multiplier norm at most one such that $\varphi_{ij}(\lambda_i)=\varepsilon, \varphi_{ij}(\lambda_j)=0$, then necessa …

Let $A:D_A \subseteq H \to K$ a linear closed surjctive operator between two Hilbert spaces $H$ and $K$. One would expect that in such a situation there must exist a bounded right inverse of $A$, name …

It seems that I have found a counter example myself. For the Hilbert matrix $$ H_\lambda:= \big( \frac{1}{1-\lambda+k+n} \big)_{k,n\geq 0}, \lambda < 1 $$ Rosenblum in "On the Hilbert Matrix I, Pro …

There might be an obvious answer to the question, but it doesn't come to mind. Suppose we have an infinite matrix $A=(a_{ij})$, which defines a bounded linear operator on $\ell^p$, i.e. for all seque …