On the Inverse Fourier Transform (Proof)

Proof. $$\begin{array}{rlrl}({\mathcal{F}}^{-1}(\mathcal{F}(f)))(x)& =& & \frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (ipx)(\mathcal{F}(f))(p)d^{d}p\\ & & & \text{ (executing the proposed }{\mathcal{F}}^{-1}\text{)}\\ & =& & \frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\underset{=1}{\underbrace{(\underset{ϵ\to 0}{lim}\exp (-\frac{ϵ}{2}{p}^2))}}\exp (ipx)(\mathcal{F}(f))(p)d^{d}p\\ & & & \text{ (regulator)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{p}^2)\exp (ipx)(\mathcal{F}(f))(p)d^{d}p\\ & & & \text{ (Lebesgue dominatedconvergence)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{p}^2)\exp (ipx)(\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-ipy)f(y)d^{d}y)d^{d}p\\ & & & \text{ (executing }(\mathcal{F}(f))(p)\text{)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}(\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{p}^2)\underset{\exp (-ip(y-x))}{\underbrace{\exp (ipx)\exp (-ipy)}}{d}^{d}p)f(y)d^{d}y\\ & & & \text{ (Fubini)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\underset{=:(\mathcal{F}(p\mapsto -\exp (-\frac{ϵ}{2}{p}^2))(z)}{\underbrace{(\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{p}^2)\exp (-ipz)d^{d}p)}}f(z+x)d^{d}z\\ & & & \text{ (change of variable }z\leftarrow y-x;d^{d}z\leftarrow d^{d}y\text{)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}((\mathcal{F}(p\mapsto -\exp (-\frac{ϵ}{2}{p}^2))(z))f(z+x)d^{d}z\\ & & & \text{ (Fourier operatordefinition)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\frac{1}{{ϵ}^{d/2}}\exp (-\frac{1}{2}(\frac{z}{{ϵ}^{1/2}}{)}^2)f(z+x)d^{d}z\\ & & & \text{ (previous lemma, with }z\leftarrow ϵ,p\leftarrow z\text{)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\frac{{ϵ}^{d/2}}{{ϵ}^{d/2}}\exp (-\frac{t^2}{2})f(t{ϵ}^{1/2}+x)d^{d}t\\ & & & \text{ (change of variable: }t\leftarrow z/{ϵ}^{1/2},d^{d}t\leftarrow d^{d}z/{ϵ}^{d/2}\text{)}\\ & =& & \frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{t^2}{2})\underset{=f(x)}{\underbrace{\underset{ϵ\to 0}{lim}f(t{ϵ}^{1/2}+x)}}{d}^{d}t\\ & & & \text{ (dominated convergence)}\\ & =& & f(x)\frac{1}{(2\pi {)}^{d/2}}\underset{=(2\pi {)}^{d/2}}{\underbrace{{\int }_{{\mathbb{R}}^d}\exp (-\frac{1}{2}{t}^2)d^{d}t}}\\ & & & \text{ (previous well-knownintegration result, generalized)}\\ & =& & f(x)\\ & \Rightarrow & & \boxed{{\displaystyle {\mathcal{F}}^{-1}\circ \mathcal{F}={\text{id}}_{S({\mathbb{R}}^d)}}}\end{array}$$

We’ve proven that the proposed ${\mathcal{F}}^{-1}$ is the left inverse of $\mathcal{F}$; it remains to prove it is its right inverse. But the process is very similar: we use the same regulator, dominated convergence to swap the limit and the integral, and Fubini to swap the integration order.

Hopefully, without any typos: $$\begin{array}{rlrl}(\mathcal{F}({\mathcal{F}}^{-1}(g)))(x)& =& & \frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-ixy)({\mathcal{F}}^{-1}(g))(y)d^{d}y\\ & & & \text{ (executing }\mathcal{F}\text{)}\\ & =& & \frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\underset{=1}{\underbrace{(\underset{ϵ\to 0}{lim}\exp (-\frac{ϵ}{2}{y}^2))}}\exp (-ixy)({\mathcal{F}}^{-1}(g))(y)d^{d}y\\ & & & \text{ (regulator)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{y}^2)\exp (-ixy)({\mathcal{F}}^{-1}(g))(y)d^{d}y\\ & & & \text{ (Lebesgue dominatedconvergence)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{y}^2)\exp (-ixy)(\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (ipy)g(p)d^{d}p)d^{d}y\\ & & & \text{ (executing the proposed }({\mathcal{F}}^{-1}(g))(y)\text{)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}(\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{y}^2)\underset{\exp (-iy(x-p))}{\underbrace{\exp (ipy)\exp (-ixy)}}{d}^{d}y)g(p)d^{d}p\\ & & & \text{ (Fubini)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\underset{=:(\mathcal{F}(y\mapsto -\exp (-\frac{ϵ}{2}{y}^2))(t)}{\underbrace{(\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{ϵ}{2}{y}^2)\exp (-iyt)d^{d}y)}}g(x-t)d^{d}t\\ & & & \text{ (change of variable }t\leftarrow x-p;d^{d}t\leftarrow d^{d}p\text{)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\mathcal{F}(y\mapsto -\exp (-\frac{ϵ}{2}{y}^2))(t)g(x-t)d^{d}t\\ & & & \text{ (Fourier operatordefinition)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\frac{1}{{ϵ}^{d/2}}\exp (-\frac{1}{2}(\frac{t}{{ϵ}^{1/2}}{)}^2)g(x-t)d^{d}t\\ & & & \text{ (previous lemma, with }z\leftarrow ϵ,p\leftarrow t\text{)}\\ & =& & \underset{ϵ\to 0}{lim}\frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\frac{{ϵ}^{d/2}}{{ϵ}^{d/2}}\exp (-\frac{z^2}{2})g(x-z{ϵ}^{1/2})d^{d}z\\ & & & \text{ (change of variable: }z\leftarrow t/{ϵ}^{1/2},d^{d}z\leftarrow d^{d}t/{ϵ}^{d/2}\text{)}\\ & =& & \frac{1}{(2\pi {)}^{d/2}}{\int }_{{\mathbb{R}}^d}\exp (-\frac{z^2}{2})\underset{=g(x)}{\underbrace{\underset{ϵ\to 0}{lim}g(x-z{ϵ}^{1/2})}}{d}^{d}z\\ & & & \text{ (dominated convergence)}\\ & =& & g(x)\frac{1}{(2\pi {)}^{d/2}}\underset{=(2\pi {)}^{d/2}}{\underbrace{{\int }_{{\mathbb{R}}^d}\exp (-\frac{1}{2}{z}^2)d^{d}z}}\\ & & & \text{ (previous well-knownintegration result, generalized)}\\ & =& & g(x)\\ & \Rightarrow & & \boxed{{\displaystyle \mathcal{F}\circ {\mathcal{F}}^{-1}={\text{id}}_{S({\mathbb{R}}^d)}}}\end{array}$$ ◻