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For the stated angular-frequency Fourier transform convention, the convolution theorem givesThe Fourier inversion theorem and the assumed absolute integrability therefore imply

Since $|-i|=1$ and the principal argument is $-\pi /2$,All logarithms are $-\pi i/2+2\pi in$, $n\in \mathbb{Z}$.

By the Fourier inversion theorem and Fubini's theorem,Thus multiplication of Fourier transforms corresponds to the convolution $h=f\ast g$, which is the convolution theorem in the convention used here.

Use the conventionThe Fourier inversion theorem says that, when $f,\hat{f}\in L^1({\mathbb{R}}^d)$,for almost every $x$.

Sincethe Fubini's theorem and Fourier inversion giveThus the Plancherel theorem identity in this convention is

The inverse integral is a continuous function of $x$, since $\hat{f}\in L^1$ permits dominated convergence. If $f$ is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the continuous version of Fourier inversion.

PutThis function is bounded near zero and is $O(u^{-2})$ at infinity, so $H\in L^1(\mathbb{R})$ and the defining integral for $g$ converges absolutely.

For the triangular functiondirect integration givesThe Fourier inversion theorem, or equivalently the Fourier transform of a triangular function, therefore yieldsIn particular $g(x)=0$ whenever $|x|>1$. Finally,This also agrees with the Plancherel theorem applied to $h$ and $H$.

We prove the estimate first for $u$ in the Schwartz space. By the Fourier inversion theorem and the Cauchy-Schwarz inequality,The integral is finite because $n>n/2$.

For $0<\alpha \le 1$, the elementary boundgivesThe last integral is finite near zero for every $\alpha >0$ and at infinity exactly when $2\alpha <n$. Choose, for example, $\alpha =1/4$, which works for every $n\ge 1$. We obtainDensity of the Schwartz space in $H^n$ extends the estimate and supplies a unique Hölder-continuous representative. Thereforecontinuously. This is the Fourier proof of Hölder regularity from a Sobolev norm.

Choose $ϵ=1/6$ and a sufficiently large absolute constant $k$. Part (iii)'s size bound givesLetand choose a maximal dissociated set $\Lambda \subseteq \Gamma$. The entropy form of the Chang theorem givesand maximality gives $\Gamma \subseteq \mathrm{Span}(\Lambda )$.

Put $B=B(\Lambda ,1/(6|\Lambda |))$. If $y\in B$ and $\gamma \in \Gamma$, expressing $\gamma$ as a product of characters in $\Lambda$ and their inverses givesFor $f=1_{-A}\ast {\mu }_A$, Parseval identity givesAlso $\hat{\mu }(\gamma )=|\widehat{{\mu }_Y}(\gamma ){|}^{2k}$. Fourier inversion theorem therefore yieldsonce $k$ is large enough.

If $\Lambda$ is empty, interpret $B$ as $G$; the same Fourier estimate, using only the second term, is even stronger.

Part (iii) gives $\Vert f\ast \mu -f{\Vert }_{\infty }\le 1/6$. Applying this at $x$ and $x+y$ and using the last estimate givesFinally $f(0)=1$. Hence $f(y)\ge 1/3>0$ for every $y\in B$, while the support of $f$ is $A-A$. Consequently

For a primitive nonprincipal Dirichlet character modulo $q$, finite Fourier inversion theorem and the primitive Gauss magnitude give this bound uniformly in the interval. Bound the exponential sums by $C\Vert a/q{\Vert }^{-1}$ and sum the resulting harmonic series. The general Gauss identity follows from the prime-power identity by the Chinese remainder theorem.

When the continuous Fourier transform $F$ of an integrable function vanishes outside $[-\pi ,\pi ]$, its periodic Fourier coefficients are the integer samples of $f$, with the sign of the index reversed. Projecting $F$ in $L^2[-\pi ,\pi ]$ and then applying the Fourier inversion theorem gives the displayed sinc function reconstruction. The Cauchy-Schwarz inequality controls the reconstruction error uniformly in $t$. Bessel's inequality bounds the shifted sinc coefficient vector, giving absolute uniform tails when the sample sequence belongs to ${\ell }^2$.