Fourier pairing for a finite measure and Schwartz data

Joint measurability and absolute integrability. The characters $e^{2\pi ix\cdot \omega }$ are Borel on ${\mathbb{R}}^{2n}$ by the Cartesian form, the $1$-Lipschitz sine and cosine, and the closure rules of [F7]; the projections $(x,\omega )\mapsto x$, $(x,\omega )\mapsto \omega$ are Borel, so $G(\omega )$, $\overline{F(x)}$ and $\widehat{F}(\omega )$ pull back to Borel functions, and $p(x,\omega ):=e^{2\pi ix\cdot \omega }G(\omega )\overline{F(x)}$, $q(z,\omega ):=e^{2\pi iz\cdot \omega }F(x-z)$ for fixed $x$ are Borel by [F7]. For $p$, using $\mathrm{\mid }{e}^{2\pi ix\cdot \omega }\mathrm{\mid }=1$ and $\mathrm{\mid }G\mathrm{\mid }\le \mathrm{\parallel }G{\mathrm{\parallel }}_{\mathrm{\infty }}$, $\mathrm{\mid }F\mathrm{\mid }\le \mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}$, with $\mathrm{\parallel }F{\mathrm{\parallel }}_1<\mathrm{\infty }$ from [F4], ${\int }_{{\mathbb{R}}^n}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_{{\mathbb{R}}^n}\mathrm{\mid }p\mathrm{\mid }d\mathrm{\mid }\mu \mathrm{\mid }dx=\mathrm{\parallel }F{\mathrm{\parallel }}_1\text{\hspace{0.17em}⁣}\int \mathrm{\mid }G\mathrm{\mid }d\mathrm{\mid }\mu \mathrm{\mid }\le \mathrm{\parallel }F{\mathrm{\parallel }}_1\mathrm{\parallel }G{\mathrm{\parallel }}_{\mathrm{\infty }}M<\mathrm{\infty },$ and for $q$, translation invariance gives $\int \int \mathrm{\mid }q\mathrm{\mid }d\mathrm{\mid }\mu \mathrm{\mid }dz=\mathrm{\parallel }F{\mathrm{\parallel }}_{1}M<\mathrm{\infty }$. Thus both kernels are integrable against ${\lambda }_n\times \mathrm{\mid }\mu \mathrm{\mid }$. Fubini against the complex measure follows first for simple functions by linearity and then by approximation in this absolute-integral norm, using [F8]; this licenses the interchange below.

The clause (iii). By definition $\check{\mu }(x)=\widehat{\mu }(-x)$, so $\mathrm{\mid }\check{\mu }(x)\mathrm{\mid }=\mathrm{\mid }\widehat{\mu }(-x)\mathrm{\mid }\le M$ for every $x$ by [F2], and $\mathrm{\mid }\check{\mu }(x)-\check{\mu }(y)\mathrm{\mid }=\mathrm{\mid }\widehat{\mu }(-x)-\widehat{\mu }(-y)\mathrm{\mid }$ tends to $0$ uniformly as $\mathrm{\mid }x-y\mathrm{\mid }\to 0$ because $\widehat{\mu }$ is uniformly continuous [F2].

The identity (i). By definition of $(G\mu {)}^{\vee }$ and step 1.1, $\int (G\mu {)}^{\vee }(x)\overline{F(x)}dx=\int \int p(x,\omega )d\mu (\omega )dx$, and Fubini [F3] rewrites this iterated integral as $\int [\int p(x,\omega )dx]d\mu (\omega )=\int G(\omega )[\int e^{2\pi ix\cdot \omega }\overline{F(x)}dx]d\mu (\omega )$, because $G(\omega )$ does not depend on $x$. By the conjugation law of [F5] with $\xi =-\omega$, $\int e^{2\pi ix\cdot \omega }\overline{F(x)}dx=\widehat{\bar{F}}(-\omega )=\overline{\widehat{F}(\omega )}$, so the last expression is $\int G(\omega )\overline{\widehat{F}(\omega )}d\mu (\omega )$. This proves (i).

The convolution identity (ii). Fix $x\in {\mathbb{R}}^n$ and put $F_x:=F(-\cdot )$, so that ${\tau }_x{F}_x(z)=F_x(z-x)=F(x-z)$. By the translation and reflection laws of [F5], for every $\omega$, $\int F(x-z)e^{2\pi iz\cdot \omega }dz=\widehat{{\tau }_x{F}_x}(-\omega )=e^{-2\pi ix\cdot (-\omega )}\widehat{F_x}(-\omega )=e^{2\pi ix\cdot \omega }\widehat{F}(\omega )\mathrm{.}$ Substituting this into the definition of $(\widehat{F}\mu {)}^{\vee }$ and applying Fubini [F3], which is licensed by step 1.1, gives $(\widehat{F}\mu {)}^{\vee }(x)=\int e^{2\pi ix\cdot \omega }\widehat{F}(\omega )d\mu (\omega )=\int \int q(z,\omega )dzd\mu (\omega )=\int F(x-z)[\int e^{2\pi iz\cdot \omega }d\mu (\omega )]dz\mathrm{.}$ The inner integral is $\check{\mu }(z)$ by its defining formula, so the last expression is $\int F(x-z)\check{\mu }(z)dz=(F\ast \check{\mu })(x)$ by [F6]; this holds for every $x$.

Bounded continuity. By [F2], $\mathrm{\mid }\check{\mu }(z)\mathrm{\mid }\le M$ for every $z$, so $\mathrm{\mid }F\ast \check{\mu }(x)\mathrm{\mid }\le \mathrm{\parallel }F{\mathrm{\parallel }}_{1}M<\mathrm{\infty }$; for the other side, $\mathrm{\mid }(\widehat{F}\mu {)}^{\vee }(x)\mathrm{\mid }=\mid \int e^{2\pi ix\cdot \omega }\widehat{F}(\omega )d\mu (\omega )\mid \le \int \mathrm{\mid }\widehat{F}\mathrm{\mid }d\mathrm{\mid }\mu \mathrm{\mid }\le \mathrm{\parallel }\widehat{F}{\mathrm{\parallel }}_{\mathrm{\infty }}M\le \mathrm{\parallel }F{\mathrm{\parallel }}_{1}M<\mathrm{\infty }$ by the modulus bound of [F8] and [F4], and the equality of step 2.2 therefore holds between two bounded functions. For continuity of $F\ast \check{\mu }$, let ${\omega }_{\check{\mu }}$ be the modulus of uniform continuity of [F2]: for all $x,x^{\mathrm{\prime }}$, $\mathrm{\mid }F\ast \check{\mu }(x)-F\ast \check{\mu }(x^{\mathrm{\prime }})\mathrm{\mid }\le \int \mathrm{\mid }F(z)\mathrm{\mid }\mathrm{\mid }\check{\mu }(x-z)-\check{\mu }(x^{\mathrm{\prime }}-z)\mathrm{\mid }dz\le \mathrm{\parallel }F{\mathrm{\parallel }}_1{\omega }_{\check{\mu }}(\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid })⟶0$ as $x^{\mathrm{\prime }}\to x$; hence $F\ast \check{\mu }$ is continuous, and by step 2.2 so is $(\widehat{F}\mu {)}^{\vee }$. Thus the identity of (ii) holds as everywhere-defined bounded continuous functions.

Conclusion. Step 2.1 proves (i); steps 2.2 and 3.1 prove (ii) as an identity of everywhere-defined bounded continuous functions; step 1.2 proves (iii). Countable Choice is spent only through the sigma-finite Fubini/Tonelli interfaces and the transform law of the cited suppliers, whose own hypotheses carry it.