Carleson operator and measurable linearisation

Assume The Axiom of Choice for the countable-choice hypotheses of the Schwartz Fourier and integrability suppliers. Use the transform $\widehat{f}(\xi )=\int f(x)e^{-2\pi ix\xi }dx$ of Fourier transform on complex L1 classes, with absolute convergence and representative independence from The integral transform is representative independent. For $f\in \mathcal{S}(\mathbb{R})$ as in Schwartz space and its seminorms, Fourier transform acts continuously on Schwartz space and Schwartz derivatives are integrable give $f,\widehat{f}\in L^1$. Define $T_{a}f(x)={\int }_{-\mathrm{\infty }}^a\widehat{f}(\xi )e^{2\pi ix\xi }d\xi ,C_{\mathbb{R}}f(x)={\sup }_{a\in \mathbb{R}}\mathrm{\mid }{T}_{a}f(x)\mathrm{\mid }\mathrm{.}$ These integrals converge absolutely at every x and satisfy $\mathrm{\mid }{T}_{a}f(x)\mathrm{\mid }\le \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1$. The cutoff function is continuous in a: since $\widehat{f}$ is bounded, $\mathrm{\mid }{T}_{a}f(x)-T_{b}f(x)\mathrm{\mid }\le \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }a-b\mathrm{\mid }$. Hence the supremum equals the supremum over rational a.

For fixed a, $T_{a}f$ is continuous in x. To check this without an unproved interchange, first truncate its absolutely integrable frequency tail beyond $\mathrm{\mid }\xi \mathrm{\mid }>R$ to make that tail contribution to any difference less than a prescribed epsilon. On the remaining interval, $\mathrm{\mid }{e}^{2\pi ix\xi }-e^{2\pi iy\xi }\mathrm{\mid }\le 2\pi R\mathrm{\mid }x-y\mathrm{\mid }$, so the integral difference is bounded by $2\pi R\mathrm{\mid }x-y\mathrm{\mid }\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1$. Thus the countable rational supremum is measurable.

For a nonempty finite list of rational cutoffs $q_1,\dots ,q_M$, put $C_{M}f={\max }_{1\le j\le M}\mathrm{\mid }{T}_{q_j}f\mathrm{\mid }$. Choose the least maximizing index j(x); its level sets are finite intersections of measurable comparison sets, so $a_f(x)=q_{j(x)}$ is measurable. Then $C_{M}f(x)=\mathrm{\mid }{T}_{a_f(x)}f(x)\mathrm{\mid }$. For a fixed measurable selector a, the map $f\mapsto T_{a(x)}f(x)$ is linear; when the selector was chosen from f, it is frozen before invoking a uniform operator estimate. No single selector is claimed to attain the unrestricted real supremum.

For $h(x)=T_{a_f(x)}f(x)$ put $u(x)=h(x)\mathrm{/}\mathrm{\mid }h(x)\mathrm{\mid }$ when $h(x)\mathrm{\ne }0$ and $u(x)=1$ otherwise. This is a measurable unimodular phase and $h\bar{u}=\mathrm{\mid }h\mathrm{\mid }$. Thus, on a measurable finite-measure testing set E, pairing h with $u{\mathbf{1}}_E$ gives ${\int }_E\mathrm{\mid }h\mathrm{\mid }$. The phase at zero is fixed explicitly. All finite selectors use least-index rules; AC is inherited from the stated analytic suppliers, not from selecting maximizers.