On the Beurling-Ahlfors transform's weak-type constant
The Beurling-Ahlfors transform, denoted by \(S\), is defined on \(L^{p}(\mathbb{C})\), \(1\leq p <\infty\), by \[ Sf(z)=-\frac{1}{\pi}\,\int_{\mathbb{C}}\frac{f(w)}{(z-w)^2}\,dw, \] where \(dw\) is the \(2\)-dimensional Lebesgue measure and the integral is understood as a Cauchy principal value. For a function \(f\) on a measure space \((X,\,\mu)\) the weak ``norm'' of \(f\) is defined as \[ \|f\|_{w}:=\sup_{\lambda>0}\mu(|f|\geq \lambda)\lambda. \] For an operator \(T\) defined on \(L^{1}(X,\,\mu)\), but not necessarily bounded, the weak ``norm'' of \(T\) is defined as \[ \parallel T\parallel_{w}:=\sup_{f\in L^{1},\, f\neq 0}\,\frac{\|Tf\|_{w}}{\|f\|_{1} }. \] In the paper under review, the author considers the operator \[ \Lambda^{*}f(x):=\int_{x}^{\infty}\frac{f(y)}{y}\,dy-f(x), \quad 0<x<\infty, \] defined on \(L^{p}(0,\,\infty)\) for all \(1\leq p<\infty\) and first shows that \(\Lambda^{*}\) is the adjoint of the operator \[ \Lambda f(x):=\frac{1}{x}\,\int_{0}^{x} f(y)\,dy-f(x), \quad 0<x<\infty. \] Then, the author proves the following theorems: 1) If \(f\in L^{p}(\mathbb{R}^{+},\,\mathbb{R}),\;1\leq p<\infty\) and \(F(z)=(z/\bar{z})f(|z|^2)\), then \[ SF(z)=-\Lambda^{*}f(|z|^2). \] 2) If \(f\in L^{1}(\mathbb{R}^{+},\,\mathbb{R})\) with \(\int_{0}^{\infty} f(x)dx=0\) , then \[ \frac{|\{x\in \mathbb{R}^{+}: |f(x)|\geq \lambda\}|\lambda}{\|\Lambda f\|_{1}}\leq \frac{1}{\log 2} \,. \] There exist \(f\) and \(\lambda\) for which equality can be attained. The author also shows that 2) is equivalent to: 3) \(\displaystyle \|\Lambda ^{*}\|_{w}=\frac{1}{\log 2}\) and so \(\displaystyle \|S\|_{w}\geq\frac{1}{\log 2}\).
scientific article; zbMATH DE number 5792257