= Solution
First take a <Schwartz function>, so that <Fourier inversion> and the exchange of integrals are justified. Inserting the inverse transform of the <complex conjugate> gives
$$
\begin{aligned}
\int_{\mathbb R}|f(x)|^2dx
&=\frac1{2\pi}\int_{\mathbb R}\!\int_{\mathbb R} f(x)\overline{\widetilde f(k)}e^{-ikx}\,dk\,dx\\
&=\frac1{2\pi}\int_{\mathbb R}\overline{\widetilde f(k)}\left[\int_{\mathbb R}f(x)e^{-ikx}dx\right]dk\\
&=\boxed{\frac1{2\pi}\int_{\mathbb R}|\widetilde f(k)|^2dk.}
\end{aligned}
$$
For $f\in L^1\cap L^2$, approximate it by <Schwartz functions> in both <norms>. The identity just proved makes their transforms a <Cauchy sequence> in <L2 space>, while convergence in <L1 space> makes the transforms converge uniformly to the defining transform of $f$. Thus the two transform limits agree, and passing to the $L^2$ <norms> proves the finite <Parseval identity>. This is the normalization of the <Plancherel theorem> appropriate to the paper.
There is a qualification to the printed hypothesis: absolute integrability alone does not guarantee a finite square integral. For instance $f(x)=x^{-3/4}$ on $(0,1)$ and zero elsewhere belongs to $L^1$ but not $L^2$. For an $L^1$ function of this kind the transform is not in $L^2$ either, by <L2 density from a square-integrable Fourier transform>, so both sides are infinite if the identity is interpreted in that extended sense. The finite equality requires square integrability as well.
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