= Solution
Use the convention
$$
\widehat f(u)=\int_{\mathbb R^d}e^{iu\cdot x}f(x)\,dx.
$$
The <Fourier inversion theorem> says that, when $f,\widehat f\in L^1(\mathbb R^d)$,
$$
f(x)=\frac1{(2\pi)^d}
\int_{\mathbb R^d}e^{-iu\cdot x}\widehat f(u)\,du
$$
for almost every $x$.
Since
$$
\overline{\widehat f(u)}
=\int_{\mathbb R^d}e^{-iu\cdot x}\overline{f(x)}\,dx,
$$
the <Fubini's theorem> and Fourier inversion give
$$
\begin{aligned}
\int_{\mathbb R^d}|\widehat f(u)|^2\,du
&=\int_{\mathbb R^d}\widehat f(u)
\int_{\mathbb R^d}e^{-iu\cdot x}\overline{f(x)}\,dx\,du\\
&=\int_{\mathbb R^d}\overline{f(x)}
\int_{\mathbb R^d}e^{-iu\cdot x}\widehat f(u)\,du\,dx\\
&=(2\pi)^d\int_{\mathbb R^d}|f(x)|^2\,dx.
\end{aligned}
$$
Thus the <Plancherel theorem> identity in this convention is
$$
\boxed{\|\widehat f\|_2^2=(2\pi)^d\|f\|_2^2.}
$$
The inverse integral is a continuous function of $x$, since $\widehat f\in L^1$ permits dominated convergence. If $f$ is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the <continuous version of Fourier inversion>.
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