Decay of high momentum fluctuations

Question

Let $\chi(k)=e^{-k^2}$ and define $$ \Gamma(x)=\int_{\mathbb{R}^d} e^{ikx} \frac{1}{k^2}(\chi(k)^2-\chi(Lk)^2) $$ where $L>1$. Then how would one prove that $\Gamma$ decays exponentially, i.e., $$ |\Gamma(x)| \le A_d e^{-|x|/L} $$ for some constant $A_d$

Answer 1

By the Fundamental Theorem of Calculus $$ e^{-2k^2}-e^{-2L^2k^2}=\int_{1}^{L^2}dt\ 2k^2 e^{-2tk^2}\ . $$ Insert that and use Fubini to get $$ \Gamma(x)=2\int_{1}^{L^2}dt\ \int_{\mathbb{R}^d}d^dk\ e^{ikx-2tk^2}\ . $$ Then use the formula for the Fourier transform of a Gaussian to obtain $$ \Gamma(x)=2\left(\frac{\pi}{2}\right)^{\frac{d}{2}}\int_{1}^{L^2} t^{-\frac{d}{2}}\ e^{-\frac{x^2}{8t}}\ . $$ By expanding $\frac{1}{8t}(|x|-4\sqrt{t})^2\ge 0$ we get the lower bound $$ \frac{x^2}{8t}\ge\frac{|x|}{\sqrt{t}}-2\ge \frac{|x|}{L}-2 $$ which can be used to get the bounds on the integral $$ 0\le \Gamma(x)\le A_{d,L}e^{-\frac{|x|}{L}} $$ with $$ A_{d,L}=2e^2\left(\frac{\pi}{2}\right)^{\frac{d}{2}}\int_{1}^{L^2} t^{-\frac{d}{2}}\ . $$ Finally, if $d>2$ one can get rid of the $L$ dependence in the constant $A$ because the last integral converges at $L=\infty$.