= Solution
Use the <Fourier transform> convention
$$
\widehat f(s)=\int_{\mathbb R}f(x)e^{isx}dx,\qquad
f(x)=\frac1{2\pi}\int_{\mathbb R}\widehat f(s)e^{-isx}ds.
$$
This convention also gives the sign of the inverse transform in 15A. The <Parseval identity>, in its bilinear form, is
$$
\boxed{\int_{\mathbb R}f(x)\overline{g(x)}dx
=\frac1{2\pi}\int_{\mathbb R}\widehat f(s)\overline{\widehat g(s)}ds.}
$$
In particular $\int|f|^2=(2\pi)^{-1}\int|\widehat f|^2$. 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 $\varepsilon>0$, form
$$
I_\varepsilon=\frac1{2\pi}\int\widehat f(s)\overline{\widehat g(s)}e^{-\varepsilon s^2}ds.
$$
Absolute integrability allows <Fubini's theorem> to replace the transformed factors by their defining integrals. The elementary Gaussian integral gives
$$
K_\varepsilon(t)=\frac1{2\pi}\int e^{ist-\varepsilon s^2}ds
=\frac1{\sqrt{4\pi\varepsilon}}e^{-t^2/(4\varepsilon)},
$$
so $I_\varepsilon=\int f(x)\overline{(K_\varepsilon*g)(x)}dx$, with <convolution> $(h*g)(x)=\int h(x-y)g(y)dy$. 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 $K_\varepsilon*g\to g$ 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 $I_\varepsilon\to(2\pi)^{-1}\int\widehat f\overline{\widehat g}$. 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.
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