The fourier transform preserves norms for a square-integrable function proof

Question

Can you provide proof for "The Fourier transform preserves norms for a square-integrable function"?

Answer 1

• The Schwartz functions are dense in $L^2$: because for $f\in L^2$ for $T$ large $\|f-f 1_{[T,-T]}\|_2$ is small and for $n$ large $\|f 1_{[T,-T]}-n e^{-\pi n^2 x^2}\ast f 1_{[T,-T]}\|_2$ is small ($\ast$ is the convolution).

• The finite linear combination of Gaussian functions $e^{-a (x+b)^2+icx}$ are dense in the Schwartz functions thus in $L^2$ (do a convolution with $n e^{-\pi n^2 x^2}$ again and approximate the Riemann integral by a Riemann sum)

• For such a finite linear combination of Gaussians functions let $(Wg)(y)=\int_{-\infty}^\infty e^{-2i\pi xy}g(x)dx,(W^* g)(y)=\int_{-\infty}^\infty e^{2i\pi xy}g(x)dx$, the known Fourier transform of the Gaussian gives that $W^* Wg =g$. Also it is easily seen that $\langle Wg,h\rangle=\langle g,W^*h\rangle$ thus $$\|Wg\|_2^2=\langle Wg,Wg\rangle=\langle g,W^* Wg\rangle=\langle g,g\rangle = \|g\|_2^2$$

• Take a sequence $g_n$ of such finite linear combination of Gaussians such that $\|f-g_n\|_2\to 0$ then its Fourier transform $Wf$ is defined as the limit in $L^2$ of the sequence $Wg_n$ (which is Cauchy by the previous result) so it is by definition that $\|f\|_2=\|Wf\|_2$.

It works the same way when replacing the Gaussians by any kind of functions for which you proved the Fourier inversion theorem (for example the functions such that $h,h'\in C^0\cap L^1$, from the Dirichlet kernel).