Fourier Transform Convolution and Approximate Identities

The exponential factor is continuous, hence measurable, and has modulus one. Thus the product is measurable and $\int \mathrm{\mid }f(x)e^{-2\pi ix\cdot \xi }\mathrm{\mid }dx=\int \mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$. Its componentwise integral exists, and $\mathrm{\mid }\widehat{f}(\xi )\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$.

If $g=f$ outside a measurable null set $N$, the same modulus equality proves the product with $g$ integrable, and the two products agree outside $N$. Applying integral invariance on ${\mathbb{R}}^n$ gives identical values at this $\xi$. Since $\xi$ was arbitrary, this holds at every frequency; no union of frequency-dependent exceptional sets is taken. In particular a zero class has identically zero transform.

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.

The maps $y\mapsto y+a$ and $x\mapsto Ax$ are $C^1$ diffeomorphisms of the open set ${\mathbb{R}}^n$, with determinants $1$ and $\det A\mathrm{\ne }0$. Applying F2 to $\mathrm{\mid }f\mathrm{\mid }$ proves $\mathrm{\parallel }{\tau }_{a}f{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }f\circ A{\mathrm{\parallel }}_1=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$; unit modulus gives $\mathrm{\parallel }{M}_{b}f{\mathrm{\parallel }}_1=\mathrm{\parallel }\bar{f}{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$. These operations preserve null equivalence, by the same substitution applied to indicators of null sets (or directly by its null-set proof). Countable choice is precisely the assumption inherited from this Lebesgue substitution interface.

Substituting $x=y+a$ in the absolutely convergent translation integral gives $\int f(y)e^{-2\pi i(y+a)\cdot \xi }dy=e^{-2\pi ia\cdot \xi }\widehat{f}(\xi )$. Combining exponential factors in the modulation integral gives $e^{-2\pi ix\cdot (\xi -b)}$, hence the modulation formula.

Substituting $y=Ax$ gives $x\cdot \xi =y\cdot A^{-T}\xi$ and $dx=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}dy$, proving the dilation formula. For $A=-I$ its determinant has absolute value one, proving reflection even when orientation is reversed. Finally conjugate the componentwise integral for $\widehat{f}(-\xi )$: conjugation commutes with its real and imaginary integrals and sends $e^{2\pi ix\cdot \xi }$ to $e^{-2\pi ix\cdot \xi }$. This proves the last identity. All equalities are pointwise because F1 gives absolute convergence at each frequency.

Apply [F7] to the real and imaginary components of $f$ and $g$, changing infinite values on their null sets to zero, to obtain finite Borel representatives. Euclidean Lebesgue spaces are sigma-finite (the boxes $[-k,k{]}^n$ have finite volume and cover them). F2 gives product measurability. Translation invariance and Tonelli give $\iint \mathrm{\mid }f(x-y)g(y)\mathrm{\mid }dxdy=\int \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\mid }g(y)\mathrm{\mid }dy=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1<\mathrm{\infty }$. Thus the exponential-weighted integrand is also absolutely integrable at every fixed frequency, with this same bound.

Fubini now permits exchanging the integrals in the transform of the convolution representative. The inner $x$ integral is the translation transform from F5, giving $\widehat{f\ast g}(\xi )=\int g(y)e^{-2\pi iy\cdot \xi }\widehat{f}(\xi )dy=\widehat{f}(\xi )\widehat{g}(\xi )$. The convolution is defined arbitrarily on its null exceptional set; F6 makes that choice irrelevant at every frequency. This proves the asserted equality, including when either input is zero.

Write $u=r+is$, $v=p+iq$. Apply F1 to the four real pairs $(r,p),(s,q),(r,q),(s,p)$. Subtract the second identity from the first and add $i$ times the sum of the last two. Since $u^{\mathrm{\prime }}v=(r^{\mathrm{\prime }}p-s^{\mathrm{\prime }}q)+i(r^{\mathrm{\prime }}q+s^{\mathrm{\prime }}p)$ and $u{v}^{\mathrm{\prime }}=(r{p}^{\mathrm{\prime }}-s{q}^{\mathrm{\prime }})+i(r{q}^{\mathrm{\prime }}+s{p}^{\mathrm{\prime }})$, the result is the complex integration-by-parts identity. All integrands are continuous on the compact interval and hence bounded and Riemann integrable; F2 changes each real integral to a Lebesgue integral. Applying F3 to $r,s$ and the same F2 gives the complex FTC. This is the sole countable-choice use here.

For the whole-line assertion apply step 1.1 on $[-N,N]$. The truncated products converge pointwise to $u^{\mathrm{\prime }}v$ and $u{v}^{\mathrm{\prime }}$ and have integrable majorants $\mathrm{\mid }{u}^{\mathrm{\prime }}v\mathrm{\mid }$ and $\mathrm{\mid }u{v}^{\mathrm{\prime }}\mathrm{\mid }$. F4 gives convergence of both integrals; the boundary term $u(N)v(N)-u(-N)v(-N)$ tends to zero by the two assumed limits. Passing to the limit proves the assertion. No separate integrability of $u$ or $v$ is required.

For $c>0$ and integer $m\ge 0$, F3 bounds $\mathrm{\mid }x{\mathrm{\mid }}^m{e}^{-c{x}^2\mathrm{/}2}$ on the tails, and it is bounded on a compact middle interval by continuity. Thus $\mathrm{\mid }x{\mathrm{\mid }}^m{e}^{-c{x}^2}\le C{e}^{-c{x}^2\mathrm{/}2}$. F1, F2 on both half-lines (reflect the negative half), and F7 give integrability of these majorants and $\int e^{-\pi x^2}dx=1$. In several dimensions bound a polynomial by a finite sum of monomials and factor the Gaussian; successive nonnegative integration gives the product of the finite one-dimensional bounds.

Put $G(\xi )=\int e^{-\pi x^2}{e}^{-2\pi ix\xi }dx$ in dimension one. F4 applies on every frequency interval with derivative majorant $2\pi \mathrm{\mid }x\mathrm{\mid }{e}^{-\pi x^2}$ from step 1.1. Hence $G^{\mathrm{\prime }}(\xi )=-2\pi i\int x{e}^{-\pi x^2}{e}^{-2\pi ix\xi }dx$. Apply F5 to $u=e^{-\pi x^2}$ and $v=e^{-2\pi ix\xi }$: both derivative products are integrable by step 1.1 and $uv\to 0$ at both ends. Since $u^{\mathrm{\prime }}=-2\pi xu$ and $v^{\mathrm{\prime }}=-2\pi i\xi v$, it follows that $\int xuv=-i\xi G(\xi )$, and therefore $G^{\mathrm{\prime }}=-2\pi \xi G$.

The product rule gives $(e^{\pi {\xi }^2}G(\xi ){)}^{\mathrm{\prime }}=0$. Applying the finite-interval complex FTC to its real and imaginary parts shows this product is constant, equal to $G(0)=1$. Thus $G(\xi )=e^{-\pi {\xi }^2}$. F6 tensors this formula in $n$ coordinates; absolute integrability is supplied by step 1.1. Finally substitute $y=\sqrt{t}x$ using F7; the Jacobian is $t^{n\mathrm{/}2}$ and the frequency becomes $\xi \mathrm{/}\sqrt{t}$. This gives exactly the claimed formula. Countable choice is inherited from F2 and F7, and the argument uses neither later Schwartz theory nor a Fourier inversion theorem.

For $\xi \mathrm{\ne }0$ set $h=\xi \mathrm{/}(2\mathrm{\mid }\xi {\mathrm{\mid }}^2)$. Then $h\cdot \xi =1\mathrm{/}2$, so $\widehat{{\tau }_{h}f-f}(\xi )=-2\widehat{f}(\xi )$ by F2 and linearity. The bound in F1 gives $2\mathrm{\mid }\widehat{f}(\xi )\mathrm{\mid }\le \mathrm{\parallel }{\tau }_{h}f-f{\mathrm{\parallel }}_1$.

Given $ϵ>0$, F3 supplies $\delta >0$ with $\mathrm{\parallel }{\tau }_{h}f-f{\mathrm{\parallel }}_1<2ϵ$ for $\mathrm{\mid }h\mathrm{\mid }<\delta$. If $\mathrm{\mid }\xi \mathrm{\mid }>1\mathrm{/}(2\delta )$, the explicit $h$ from step 1.1 satisfies that condition, hence $\mathrm{\mid }\widehat{f}(\xi )\mathrm{\mid }<ϵ$. Continuity is already F1. This is the asserted $C_0$ property. Countable choice is inherited from F2 and F3, not from the explicit selection of $h$.

Apply F1 with $s=1\mathrm{/}t$ and multiply by $t^{-n\mathrm{/}2}$ to obtain $\widehat{k_t}(\xi )=e^{-\pi t\mathrm{\mid }\xi {\mathrm{\mid }}^2}$. At $\xi =0$ this gives mass one; positivity gives the same $L^1$ norm. Applying F1 with $s=t$ at frequency $-z$ proves the displayed inverse integral without an inversion theorem.

Write $y=x\mathrm{/}\sqrt{t}$. F2 gives ${\int }_{\mathrm{\mid }x\mathrm{\mid }>\delta }{k}_t(x)dx={\int }_{\mathrm{\mid }y\mathrm{\mid }>\delta \mathrm{/}\sqrt{t}}{e}^{-\pi \mathrm{\mid }y{\mathrm{\mid }}^2}dy$. This tends to zero: F3 applied to the explicit integer-radius tails with majorant the integrable Gaussian gives their convergence to zero, and monotonicity bounds every sufficiently small t-tail by any fixed integer tail. All three conditions in F4 now hold with norm bound one. Countable choice is inherited from the Gaussian and substitution results.

Balls are open by the triangle inequality, so F1 establishes measurability before the cube bounds F2 are used. Put $v_n=\mathrm{\mid }B(0,1)\mathrm{\mid }\in (0,\mathrm{\infty })$. F3 gives $\mathrm{\mid }B(0,r)\mathrm{\mid }=v_n{r}^n$ and $\mathrm{\mid }B(0,r)\setminus B(0,r\mathrm{/}2)\mathrm{\mid }=v_n(1-2^{-n})r^n$. On that annulus $\mathrm{\Phi }(\mathrm{\mid }y\mathrm{\mid })\ge \mathrm{\Phi }(r)$; hence $r^n\mathrm{\Phi }(r)\le J_r\mathrm{/}[v_n(1-2^{-n})]\to 0$ as $r\to \mathrm{\infty }$, where $J_r$ is the integral over $\mathrm{\mid }y\mathrm{\mid }\ge r\mathrm{/}2$. Integrability implies these tails tend to zero, by countable additivity on explicit integer annuli. Also disjoint annuli $2^{j-1}\le \mathrm{\mid }y\mathrm{\mid }<2^j$, $j\ge 0$, give ${\sum }_{j\ge 0}{2}^{jn}\mathrm{\Phi }(2^j)\le J\mathrm{/}[v_n(1-2^{-n})]$.

Write $H(y)=\mathrm{\mid }f(x-y)-a\mathrm{\mid }$. Boundedness of $\mathrm{\Phi }$ makes the integral with $f$ absolutely convergent: it is at most ${\epsilon }^{-n}\mathrm{\Phi }(0)\mathrm{\parallel }f{\mathrm{\parallel }}_1$ by reflection and translation of Lebesgue measure. Scaling the defining integral of K gives $\int {\epsilon }^{-n}K(y\mathrm{/}\epsilon )dy=1$. Fact F5 applies to these affine diffeomorphisms and supplies both substitutions. Thus the absolute error is at most $\int {\epsilon }^{-n}\mathrm{\Phi }(\mathrm{\mid }y\mathrm{\mid }\mathrm{/}\epsilon )H(y)dy$.

Given $\eta >0$, fix $\delta >0$ such that $A(r)\le \eta r^n$ for $0<r<\delta$. For $\epsilon <\delta \mathrm{/}2$, the central ball contributes at most $\mathrm{\Phi }(0){\epsilon }^{-n}A(\epsilon )\le \eta \mathrm{\Phi }(0)$. Each dyadic shell $2^j\epsilon \le \mathrm{\mid }y\mathrm{\mid }<2^{j+1}\epsilon$ whose lower radius is below $\delta \mathrm{/}2$ contributes at most ${\epsilon }^{-n}\mathrm{\Phi }(2^j)A(2^{j+1}\epsilon )\le \eta 2^n{2}^{jn}\mathrm{\Phi }(2^j)$. Their sum is bounded by $\eta C$, where $C=\mathrm{\Phi }(0)+2^{n}J\mathrm{/}[v_n(1-2^{-n})]$, independently of $\epsilon$. These regions cover $\mathrm{\mid }y\mathrm{\mid }<\delta \mathrm{/}2$.

On $\mathrm{\mid }y\mathrm{\mid }\ge \delta \mathrm{/}2$, the contribution from $\mathrm{\mid }f(x-y)\mathrm{\mid }$ is at most ${\epsilon }^{-n}\mathrm{\Phi }(\delta \mathrm{/}(2\epsilon ))\mathrm{\parallel }f{\mathrm{\parallel }}_1\to 0$ by step 1.1. The contribution from $\mathrm{\mid }a\mathrm{\mid }$ is at most $\mathrm{\mid }a\mathrm{\mid }{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge \delta \mathrm{/}(2\epsilon )}\mathrm{\Phi }(\mathrm{\mid }z\mathrm{\mid })dz\to 0$ by integrability and scaling. The total error therefore has limit superior at most $\eta C$. Letting $\eta \downarrow 0$ proves convergence. The proof uses the stated countable-choice Euclidean measure interfaces and explicit shells, with no full AC or choice of witnesses at different points.

By F1 and Gaussian integrability, the defining integral for $S_{t}f(x)$ is absolutely convergent. Inserting the definition of $\widehat{f}$, the product integrand has modulus $\mathrm{\mid }f(y)\mathrm{\mid }{e}^{-\pi t\mathrm{\mid }\xi {\mathrm{\mid }}^2}$, whose double integral is $\mathrm{\parallel }f{\mathrm{\parallel }}_1{t}^{-n\mathrm{/}2}<\mathrm{\infty }$. Use a Borel representative of f as supplied in F5's convolution construction; the resulting integrand is product measurable. F4 therefore gives $S_{t}f(x)=\int f(y)[\int e^{-\pi t\mathrm{\mid }\xi {\mathrm{\mid }}^2}{e}^{2\pi i(x-y)\cdot \xi }d\xi ]dy=\int f(y)k_t(x-y)dy$ by F2. This last integral exists at every x since k_t is bounded; substitution $y\mapsto x-y$ gives the stated convolution convention.

F2 and F5 now give $\mathrm{\parallel }{S}_{t}f-f{\mathrm{\parallel }}_1\to 0$. For pointwise recovery, take $K(y)=e^{-\pi \mathrm{\mid }y{\mathrm{\mid }}^2}$, $\mathrm{\Phi }(r)=e^{-\pi r^2}$ and $\epsilon =\sqrt{t}$ in F3. This profile is bounded, nonnegative and decreasing, its radial integral is one, and the dilated kernel is exactly k_t. Thus at every specified Lebesgue value a, $S_{t}f(x)\to a$. The two assertions use different estimates; norm convergence alone has not been used to infer pointwise convergence.

Since $\widehat{f}\in L^1$, F2 applied to it shows that $g(x)=\mathcal{F}(\widehat{f})(-x)$ is bounded and uniformly continuous. For fixed $x$ and the explicit sequence $t_m=1\mathrm{/}(m+1)$, $m\in \mathbb{N}$, the damped integrands converge pointwise to $\widehat{f}(\xi )e^{2\pi ix\cdot \xi }$ and have majorant $\mathrm{\mid }\widehat{f}\mathrm{\mid }$. F3 therefore gives $S_{t_m}f(x)\to g(x)$.

At any Lebesgue point with value a, F1 gives the same sequence limit a, so $g(x)=a$. Since $f\in L^1$ is locally integrable, F4 gives such points with a=f(x) outside a null set (for complex inputs apply the real conclusion to both components and bound the complex oscillation by their sum). Hence $g=f$ almost everywhere. In particular g represents the original $L^1$ class. Countable choice is precisely inherited from F1 and F4; no statement about all values of an arbitrary representative follows.

F1 and the bound F2 applied to $\widehat{f}$ give $\mathrm{\mid }{f}_c(x)\mathrm{\mid }\le \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1$, hence $\mathrm{\parallel }{f}_{c}g{\mathrm{\parallel }}_1\le \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1<\mathrm{\infty }$. Changing either input on a null set changes its product only on the union of those two sets, so the integrable product class is well-defined. Also the convolution integral in the conclusion is absolutely convergent at every frequency, bounded by $\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$ using F2.

Insert the inverse integral for $f_c$ into $\int f_c(x)g(x)e^{-2\pi ix\cdot \xi }dx$. The product integrand has modulus $\mathrm{\mid }\widehat{f}(\eta )\mathrm{\mid }\mathrm{\mid }g(x)\mathrm{\mid }$ with double integral $\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$; measurability follows from coordinate pullbacks and the continuous exponential. F3 exchanges the integrals, giving $\int \widehat{f}(\eta )[\int g(x)e^{-2\pi ix\cdot (\xi -\eta )}dx]d\eta$, the required expression. Exchanging the roles of f and g proves the symmetric variant under its stated hypothesis. Countable choice is inherited from F1. No assertion that arbitrary products of two integrable functions are integrable is used.

Set $h=f-g$. By linearity, $\widehat{h}=0$. If equality was given only almost everywhere, continuity still implies this everywhere: a nonzero value would remain bounded away from zero on an open ball, which contains a positive-volume box and cannot be null. Thus the transform of h is the zero integrable function.

F2 applies to h, since both h and its zero transform are integrable. It gives $h(x)=\int 0d\xi =0$ almost everywhere, so $f=g$ as classes. Countable choice is inherited from inversion (and the Euclidean measure interface in the optional almost-everywhere hypothesis).

Put $M={\sup }_{\epsilon }\mathrm{\parallel }{K}_{\epsilon }{\mathrm{\parallel }}_1$. Unit mass gives ${\widehat{K}}_{\epsilon }(\xi )-1=\int K_{\epsilon }(y)(e^{-2\pi iy\cdot \xi }-1)dy$. For fixed $\xi$, its modulus is bounded by $M{\sup }_{\mathrm{\mid }y\mathrm{\mid }<\delta }\mathrm{\mid }{e}^{-2\pi iy\cdot \xi }-1\mathrm{\mid }+2{\int }_{\mathrm{\mid }y\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_{\epsilon }(y)\mathrm{\mid }dy$. The latter tail tends to zero (bound it by the defining tail outside $\delta \mathrm{/}2$), and the first term tends to zero with $\delta$. This proves the pointwise multiplier limit.

F1 applies to every stated finite p and gives norm convergence, including p=1. By F2 and F3, $\mathrm{\parallel }\widehat{f}{\widehat{K}}_{\epsilon }-\widehat{f}{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }\mathcal{F}(f\ast K_{\epsilon }-f){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f\ast K_{\epsilon }-f{\mathrm{\parallel }}_1\to 0$. Thus uniform transform convergence follows from norm approximation, not merely the pointwise multiplier limit. Countable choice is inherited from F1 and F2.