 Solution

Use the Fourier transform conventionThis convention also gives the sign of the inverse transform in 15A. The Parseval identity, in its bilinear form, isIn particular $\int |f{|}^2=(2\pi {)}^{-1}\int |\hat{f}{|}^2$. Initially take $f,g$ to be Schwartz functions; the formula then extends to $L^2$ with Fourier transforms interpreted by completion.

Here is a proof which also justifies the needed limit operations. For $\epsilon >0$, formAbsolute integrability allows Fubini's theorem to replace the transformed factors by their defining integrals. The elementary Gaussian integral givesso $I_{\epsilon }=\int f(x)\overline{(K_{\epsilon }\ast g)(x)}dx$, with convolution $(h\ast g)(x)=\int h(x-y)g(y)dy$. The kernel is nonnegative, has integral one, and its mass outside any fixed neighborhood of zero tends to zero. Since a Schwartz function is bounded and uniformly continuous, splitting the convolution difference into small and large $|x-y|$ proves $K_{\epsilon }\ast g\to g$ uniformly. As $f\in L^1$, the last integral consequently tends to $\int f\overline{g}$.

On the spectral side, repeated integration by parts makes the transforms rapidly decreasing; their product is integrable. The dominated convergence theorem therefore gives $I_{\epsilon }\to (2\pi {)}^{-1}\int \hat{f}\overline{\hat{g}}$. Equating the two limits proves the displayed identity. For general $L^2$ functions, approximate by Schwartz functions. The norm identity makes their transforms Cauchy in $L^2$ with the prescribed factor; the limit defines the $L^2$ Fourier transform independently of the approximation. Taking limits in the bilinear identity proves it for all $L^2$ pairs. Thus the proof includes the hypotheses and avoids unjustified exchanges of nonabsolutely convergent integrals.