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Suppose $f$, $g$ $\in$ $\mathcal{S}$ the space of Schwartz functions and consider the Fourier transform $\mathcal{F}_m:\mathcal{S}\to\mathcal{S}$ defined by

\begin{equation} \mathcal{F}_m(f)(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dy\exp(-ixy)f(y). \end{equation}

In physics, the Plancherel formula

\begin{equation} (\mathcal{F}_m(f),\mathcal{F}_m(g))=(f,g) \end{equation}

where

\begin{equation} (f,g)=\int_{-\infty}^{\infty}dxf(x)^*g(x),\tag{1} \end{equation}

is typically proven as follows:

\begin{equation} (\mathcal{F}_m(f),\mathcal{F}_m(g))=\frac{1}{2\pi}\int{dx}\left(\int{dy}_1\exp(-ixy_1)f(y_1)\right)^*\left(\int{dy}_2\exp(-ixy_2)g(y_2)\right)\tag{2} \end{equation}

\begin{equation} =\frac{1}{2\pi}\int\int{dy}_1dy_2f(y_1)^*g(y_2)\int{dx}\exp(ix(y_1-y_2))\tag{3} \end{equation}

\begin{equation} =\int\int{dy}_1dy_2f(y_1)^*g(y_2)\delta(y_1-y_2)\tag{4} \end{equation}

\begin{equation} =\int{dy_2}f(y_2)^*g(y_2)\tag{5} \end{equation}

\begin{equation} =(f,g).\tag{6} \end{equation}

My question is not about the mathematically specious introduction of the Dirac delta function, but rather what goes wrong between (2) and (3). Since $f$, $g$ $\in$ $\mathcal{S}$, all the integrals in (2) seem well defined. However, in (3) the last integral over $dx$ seems ill-defined, which suggests that passing from (2) to (3) is not justified. Since the exchange of the $x$ and $y_1$, $y_2$ integrals was the main operation performed between (2) and (3), I am guessing there is some kind of misapplication of Fubini's theorem going on. Is it the case that the conditions necessary to exchange the integrals are not met when going from (2) to (3) in the manner shown here or is something else going on?

Araq
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    right, the conditions for Fubini are not met. If they were, a Dirac delta would never appear “as a function”. Also, despite the notation (and despite what people often say), Dirac deltas do not belong inside integrals either (unless you treat it as a measure, which is not what the notation here refers to). This whole proof is circular at best, and wrong at worst. The fact that Fourier’s inversion theorem holds is what allows us to give meaning to the statement that the Fourier transform (in the sense of tempered distributions) of $1$ is the Dirac delta. I would suggest reading Folland. – peek-a-boo May 06 '23 at 17:47
  • Thank you. Is this a way to see that the Fubini conditions are not met? Roughly check to see if the integral over the modulus of the integrand is finite: Let $I=\lvert(\exp(-ixy_1)f(y_1))^*\exp(-ixy_2)g(y_2)\rvert=\lvert{f(y_1)}\rvert\lvert{g(y_2)}\rvert$, then $\int{dx}\int{dy_1}\int{dy_2}{I}=\int{dx}\int{dy_1}\int{dy_2}\lvert{f(y_1)}\rvert\lvert{g(y_2)}\rvert$. Since $f$, $g$ $\in$ $\mathcal{S}$ the y integrals are finite, but the x integral diverges, since it is an infinite integral over a constant. – Araq May 06 '23 at 21:12
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    correct (unless of course one of $f$ or $g$ is identically zero, but this is the exceptional case). – peek-a-boo May 06 '23 at 22:05

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