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In Convergence of a integral - heat Kernel and dirac delta function

Why $$\lim_{t\to 0+}\int_{|x|>\delta}K_t(x)|\varphi(x)-\varphi(0)|\,dx=0?$$

eraldcoil
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As $|\varphi(x)-\varphi(0)|$ is bounded uniformly, we can replace this by some constant $B$, and it suffices to show that $\lim_{t\to 0^{+}}\int_{|x|>\delta}K_t(x)dx=0$. One way to do this is to do a change of variables to $u=\frac{x}{\sqrt{t}}$. Then, by the change of variables formula, $$ \int_{|x|>\delta}K_t(x)dx=\frac{1}{(4\pi)^{n/2}}\int_{|u|>\delta/\sqrt{t}}e^{-|u|^2/4} du $$ Note that $\delta/\sqrt{t}$ goes to $\infty$, so the RHS goes to zero as $t\to 0^+$ because the integral of a Gaussian converges.

flyinginsectleopard
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  • Thank you very much. I just found in the , Measure and Integral text Wheeden and Zygmund that Theorem 9.5 implies that $\int_{|x|>\delta}K_t(x),dx\to 0$ when $t\to 0$ Although I still don't know why $|\varphi(x)-\varphi(0)|$ is bounded uniformly in $|x|>\delta$. – eraldcoil Jun 03 '22 at 20:10
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    I just saw why. Since $\varphi$ is a Schwartz function, it is bounded. – eraldcoil Jun 03 '22 at 20:16