 Solution

Set $\tilde{\varphi }(x)=\overline{\varphi (-x)}$. This is a Schwartz function and its Fourier transform is $\hat{\tilde{\varphi }}(\xi )=\overline{\hat{\varphi }(\xi )}$. The convolution theorem givesThe 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:The Schwartz space is dense in the L2 space. For $f\in L^2(\mathbb{R})$ choose Schwartz functions ${\varphi }_j\to f$ in the L2 norm. The identity makes ${\hat{\varphi }}_j$ a Cauchy sequence in the L2 space, and its limit defines $\mathcal{F}f$, 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:Its inverse is the extension of the inverse Fourier transform, with limits understood in the L2 norm.