On the Inverse Fourier Transform (Proof)

Proof. \[\begin{aligned} (\mathcal{F}^{-1}(\mathcal{F}(f)))(x) &=&& \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d}\exp{(ipx)}(\mathcal{F}(f))(p)d^dp \\ ~ &~&& \textcolor{blue}{\text{ (executing the proposed $\mathcal{F}^{-1}$)}} \\ ~ &=&& \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \underbrace{ \left(\lim_{\epsilon\rightarrow0}\exp{(-\frac\epsilon2p^2)}\right) }_{=1} \exp{(ipx)}(\mathcal{F}(f))(p)d^dp \\ ~ &~&& \textcolor{blue}{\text{ (regulator)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2p^2)} \exp{(ipx)}(\mathcal{F}(f))(p)d^dp \\ ~ &~&& \textcolor{blue}{\text{ (Lebesgue dominated convergence)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2p^2)}\exp{(ipx)} \left( \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-ipy)}f(y) d^dy \right)d^dp \\ ~ &~&& \textcolor{blue}{\text{ (executing $(\mathcal{F}(f))(p)$)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \left( \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2p^2)} \underbrace{\exp{(ipx)}\exp{(-ipy)}}_{\exp{(-ip(y-x))}} d^dp \right)f(y)d^dy \\ ~ &~&& \textcolor{blue}{\text{ (Fubini)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \underbrace{\left( \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2p^2)}\exp{(-ipz)} d^dp \right)}_{=:(\mathcal{F}(p\mapsto -\exp{(-\frac\epsilon2p^2}))(z)} f(z+x)d^dz \\ ~ &~&& \textcolor{blue}{\text{ (change of variable $z \leftarrow y-x; d^dz \leftarrow d^dy$)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \left( (\mathcal{F}(p\mapsto -\exp{(-\frac\epsilon2p^2}))(z) \right) f(z+x)d^dz \\ ~ &~&& \textcolor{blue}{\text{ (Fourier operator definition)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \frac{1}{\epsilon^{d/2}}\exp{ (-\frac12(\frac{z}{\epsilon^{1/2}})^2) }f(z+x)d^dz \\ ~ &~&& \textcolor{blue}{\text{ (previous lemma, with $z \leftarrow \epsilon, p \leftarrow z$)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \frac{\epsilon^{d/2}}{\epsilon^{d/2}}\exp{(-\frac{t^2}2)} f(t\epsilon^{1/2}+x)d^dt \\ ~ &~&& \textcolor{blue}{\text{ (change of variable: $t \leftarrow z/\epsilon^{1/2}, d^dt \leftarrow d^dz/\epsilon^{d/2}$)}} \\ ~ &=&& \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac{t^2}2)}\underbrace{ \lim_{\epsilon\rightarrow0}f(t\epsilon^{1/2}+x) }_{=f(x)}d^dt \\ ~ &~&& \textcolor{blue}{\text{ (dominated convergence)}} \\ ~ &=&& f(x)\frac1{(2\pi)^{d/2}}\underbrace{ \int_{\mathbb{R}^d}\exp{(-\frac12t^2)}d^dt }_{=(2\pi)^{d/2}} \\ ~ &~&& \textcolor{blue}{\text{ (previous well-known integration result, generalized)}} \\ ~ &=&& f(x) \\ ~ &\Rightarrow&& \boxed{\mathcal{F}^{-1}\circ\mathcal{F} = \text{id}_{S(\mathbb{R}^d)}} \\ \end{aligned}\]

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{aligned} (\mathcal{F}(\mathcal{F}^{-1}(g)))(x) &=&& \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d}\exp{(-ixy)}(\mathcal{F}^{-1}(g))(y)d^dy \\ ~ &~&& \textcolor{blue}{\text{ (executing $\mathcal{F}$)}} \\ ~ &=&& \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \underbrace{ \left(\lim_{\epsilon\rightarrow0}\exp{(-\frac\epsilon2y^2)}\right) }_{=1} \exp{(-ixy)}(\mathcal{F}^{-1}(g))(y)d^dy \\ ~ &~&& \textcolor{blue}{\text{ (regulator)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2y^2)} \exp{(-ixy)}(\mathcal{F}^{-1}(g))(y)d^dy \\ ~ &~&& \textcolor{blue}{\text{ (Lebesgue dominated convergence)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2y^2)}\exp{(-ixy)} \left( \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(ipy)}g(p) d^dp \right)d^dy \\ ~ &~&& \textcolor{blue}{\text{ (executing the proposed $(\mathcal{F}^{-1}(g))(y)$)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \left( \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2y^2)} \underbrace{\exp{(ipy)}\exp{(-ixy)}}_{\exp{(-iy(x-p))}} d^dy \right)g(p)d^dp \\ ~ &~&& \textcolor{blue}{\text{ (Fubini)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \underbrace{\left( \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac\epsilon2y^2)}\exp{(-iyt)} d^dy \right)}_{=:(\mathcal{F}(y\mapsto -\exp{(-\frac\epsilon2y^2}))(t)} g(x-t)d^dt \\ ~ &~&& \textcolor{blue}{\text{ (change of variable $t \leftarrow x-p; d^dt \leftarrow d^dp$)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \mathcal{F}(y\mapsto -\exp{(-\frac\epsilon2y^2}))(t) g(x-t)d^dt \\ ~ &~&& \textcolor{blue}{\text{ (Fourier operator definition)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \frac{1}{\epsilon^{d/2}}\exp{ (-\frac12(\frac{t}{\epsilon^{1/2}})^2) }g(x-t)d^dt \\ ~ &~&& \textcolor{blue}{\text{ (previous lemma, with $z \leftarrow \epsilon, p \leftarrow t$)}} \\ ~ &=&& \lim_{\epsilon\rightarrow0}\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \frac{\epsilon^{d/2}}{\epsilon^{d/2}}\exp{(-\frac{z^2}2)} g(x-z\epsilon^{1/2})d^dz \\ ~ &~&& \textcolor{blue}{\text{ (change of variable: $z \leftarrow t/\epsilon^{1/2}, d^dz \leftarrow d^dt/\epsilon^{d/2}$)}} \\ ~ &=&& \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-\frac{z^2}2)}\underbrace{ \lim_{\epsilon\rightarrow0}g(x-z\epsilon^{1/2}) }_{=g(x)}d^dz \\ ~ &~&& \textcolor{blue}{\text{ (dominated convergence)}} \\ ~ &=&& g(x)\frac1{(2\pi)^{d/2}}\underbrace{ \int_{\mathbb{R}^d}\exp{(-\frac12z^2)}d^dz }_{=(2\pi)^{d/2}} \\ ~ &~&& \textcolor{blue}{\text{ (previous well-known integration result, generalized)}} \\ ~ &=&& g(x) \\ ~ &\Rightarrow&& \boxed{\mathcal{F}\circ\mathcal{F}^{-1} = \text{id}_{S(\mathbb{R}^d)}} \\ \end{aligned}\] ◻