= Solution
Set $\widetilde\phi(x)=\overline{\phi(-x)}$. This is a <Schwartz function> and its <Fourier transform> is $\widehat{\widetilde\phi}(\xi)=\overline{\widehat\phi(\xi)}$. The <convolution theorem> gives
$$
\widehat{\phi*\widetilde\phi}(\xi)=|\widehat\phi(\xi)|^2.
$$
The <convolution> of two <Schwartz functions> is again a <Schwartz function>: differentiating under the <integral> and expanding $x^m=(y+(x-y))^m$ bounds every weighted <derivative> by products of integrable and bounded weighted <derivatives>. Evaluate <Fourier inversion> for this <convolution> at zero:
$$
\boxed{\int_{\mathbb R}|\phi(x)|^2\,dx
=(\phi*\widetilde\phi)(0)
=\frac1{2\pi}\int_{\mathbb R}|\widehat\phi(\xi)|^2\,d\xi.}
$$
The <Schwartz space> is dense in the <L2 space>. For $f\in L^2(\mathbb R)$ choose <Schwartz functions> $\phi_j\to f$ in the <L2 norm>. The identity makes $\widehat\phi_j$ a <Cauchy sequence> in the <L2 space>, and its limit defines $\mathcal Ff$, independently of the approximation. It agrees with the <distributional Fourier transform>, because <Cauchy-Schwarz inequality> makes convergence in the <L2 norm> imply convergence when paired with every <Schwartz function>. For $f\in L^1\cap L^2$ it therefore agrees <almost everywhere> with the ordinary <Fourier transform>.
The extension is onto: its image is closed by the norm identity and contains the <Schwartz space>, since <Fourier inversion> maps that space to itself. This proves <Plancherel theorem>:
$$
\boxed{\|\mathcal Ff\|_2^2=2\pi\|f\|_2^2,\qquad
(2\pi)^{-1/2}\mathcal F:L^2(\mathbb R)\longrightarrow L^2(\mathbb R)\text{ is unitary}.}
$$
Its inverse is the extension of the <inverse Fourier transform>, with limits understood in the <L2 norm>.
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