How to prove that ||$\frac1{（4πit)^{n/2}}\int_{R^n}e^{i|x-y|^{2}/|4t}g(y)||L^2(R^n)$ is invariant with respect to time

Question

If u(x,t)=$\frac1{（4πit)^{n/2}}\int_{R^n}$$e^{i|x-y|^{2}/|4t}g(y)$ where x is the spatial variable, and g(y) is a real value function if the input y is real.

How to prove that $||u(x,t)||L^2(R^n)$ = $||g(x)||L^2(R^n) $ for t>0 ?

I can only understand that $||u(x,t)||L^1(R^n)$ will not change with respect to time,since u is just a convolution.

Answer 1

Denote by $T$ the operator on $L^2(\mathbb R^n)$ by $Tf(x) = e^{i|y|^2/(4t)} f(y)$. $T$ preserves $L^2$ norm. We have $$u(x,t) = \frac{1}{(4\pi i t)^{n/2}}e^{\frac{i|x|^2}{4t}} F(Tg)(x/(2t)),$$ where $F$ denotes the Fourier transforms. Using Plancherel theorem, we have $$\|u(.,t)\|_{L^2}^2 = \frac{1}{(4\pi t)^n} \|F(Tg)(./(2t))\|_{L^2}^2 = \frac{1}{(2\pi)^n} \|F(Tg)\|_{L^2}^2 = \|Tg\|_{L^2}^2 = \|g\|_{L^2}^2.$$