The L1 transform is bounded and uniformly continuous

Reconstruct first the integral interface used here. Augment any finite disjoint display of a nonnegative simple function by the complement with coefficient $0$. Intersections of two augmented displays partition the whole space and have equal coefficients on nonempty cells, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants and the sets $\{u_j\ge cs\}$, $0<c<1$, give monotone convergence; increasing simple approximations give nonnegative additivity, and positive/negative plus real/imaginary decompositions give finite complex $L^1$ linearity. Thus the integrable functions $f(x)e^{-2\pi ix\cdot \xi }$ and $g(x)e^{-2\pi ix\cdot \xi }$ have linear combination $(af+bg)(x)e^{-2\pi ix\cdot \xi }$, and integrating gives $\widehat{af+bg}=a\widehat{f}+b\widehat{g}$. The pointwise estimate in F1, with a right side independent of frequency, gives boundedness and the asserted supremum bound.

The preceding MCT also gives Fatou by applying it to ${\inf }_{j\ge m}{v}_j\uparrow {\mathrm{lim\hspace{0.17em}inf}}_j{v}_j$. If $u_j\to u$ almost everywhere and $\mathrm{\mid }{u}_j\mathrm{\mid }\le g\in L^1$, Fatou applied to $2g-\mathrm{\mid }{u}_j-u\mathrm{\mid }\ge 0$ gives ${\mathrm{lim\hspace{0.17em}sup}}_j\int \mathrm{\mid }{u}_j-u\mathrm{\mid }\le 0$; hence $L^1$ convergence, and the local finite linearity gives convergence of integrals. This proves the exact dominated-convergence clause used below without [F2]'s affected foundation. Factoring the exponentials yields $\widehat{f}(\xi +h)-\widehat{f}(\xi )=\mathcal{F}(f(e^{-2\pi ix\cdot h}-1))(\xi )$. F1 therefore gives the bound $\mathrm{\mid }\widehat{f}(\xi +h)-\widehat{f}(\xi )\mathrm{\mid }\le I(h)$, where $I(h)=\int \mathrm{\mid }f(x)\mathrm{\mid }\mathrm{\mid }{e}^{-2\pi ix\cdot h}-1\mathrm{\mid }dx$. For real $u$, F5--F6 and F8 give $\mathrm{\mid }\sin u\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$, $\mathrm{\mid }\cos u-1\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$, and hence $\mathrm{\mid }{e}^{iu}-1\mathrm{\mid }\le 2\mathrm{\mid }u\mathrm{\mid }$; F5 also bounds this modulus by two. Applying the locally proved dominated convergence to the explicit integer-ball tails, dominated by $\mathrm{\mid }f\mathrm{\mid }$, choose an integer $R\ge 1$ with ${\int }_{\mathrm{\mid }x\mathrm{\mid }>R}\mathrm{\mid }f\mathrm{\mid }<ϵ\mathrm{/}4$. On $\mathrm{\mid }x\mathrm{\mid }\le R$, F7 gives $\mathrm{\mid }2\pi x\cdot h\mathrm{\mid }\le 2\pi R\mathrm{\mid }h\mathrm{\mid }$, so the single choice $\delta =ϵ\mathrm{/}(8\pi R(1+\mathrm{\parallel }f{\mathrm{\parallel }}_1))$ makes $\mathrm{\mid }{e}^{-2\pi ix\cdot h}-1\mathrm{\mid }<ϵ\mathrm{/}(2(1+\mathrm{\parallel }f{\mathrm{\parallel }}_1))$ whenever $\mathrm{\mid }h\mathrm{\mid }<\delta$. The inside integral is below $ϵ\mathrm{/}2$ and the outside integral below $ϵ\mathrm{/}2$, so $I(h)<ϵ$. Consequently, for every $ϵ>0$ there is $\delta >0$ such that $\mathrm{\mid }h\mathrm{\mid }<\delta$ implies the difference is below $ϵ$ for every $\xi$. This is uniform continuity.