Dirac identity of Gaussian Heat Kernel

Question

I am having a real hard time trying to figure this problem out: $$u(\textbf{x},t)=\int_{\mathbb{R}^n}K(\textbf{x},\textbf{y},t)g(\textbf{y})d\textbf{y}$$ Where: $$ K(\textbf{x},\textbf{y},t):= \frac{e^{-\frac{|\textbf{x}-\textbf{y}|}{4t}^2}}{(4\pi t)^{n/2}} $$ show that $$ \lim_{x\to 0}u(\textbf{x},t)=g(\textbf{x}) $$

so far I have shown that $$ \int_{\mathbb{R}^n}K(\textbf{x},\textbf{y},t)d\textbf{y}=1 $$ But I can't seem to figure out how to deal with the limit since you have an infinity over infinity that can't seem to be solved using L'Hospital's rule. I get the general notion, namely that as t $\to 0$, the Gaussian Heat Kernel acts like the Dirac Delta function, but I can't seem to prove it. Thanks

Answer 1

HINT:

First, we have $\lim_{t\to 0^+}\frac{e^{-|\vec x-\vec y|^2/(4t)}}{(4\pi t)^{n/2}}=0$ for $|\vec x-\vec y|\ne 0$. So, we need to isolate the singularity point, $\vec x$, from the integration.

Proceeding, we write the integral as

$$\begin{align} u(\vec x,t)&=\int_{\mathbb{R}^n}K(\vec x,\vec y,t)g(\vec y)\,d\vec y\\\\ &=\int_{\mathbb{R}^n\setminus B(\vec x,\delta)}K(\vec x,\vec y,t)g(\vec y)\,d\vec y+\int_{B(\vec x,\delta)}K(\vec x,\vec y,t)(g(\vec y)-g(\vec x))\,d\vec y\\\\ &+g(\vec x)\int_{B(\vec x,\delta)}K(\vec x,\vec y,t)\,d\vec y\\\\ \end{align}$$

where $B(\vec x,\delta)$ is a ball of radius $\delta$ centered at $\vec x$.

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Moreover, we have

$$\begin{align} \lim_{t\to 0^+}\int_{B(\vec x,\delta)}K(\vec x,\vec y,t)\,d\vec y&=\lim_{t\to 0^+}\left(\frac{1}{(4\pi t)^{n/2}}\int_{S^{n-1}}\int_{0}^\delta e^{-r^2/4t}\,r^{n-1}\,dr\,\,dS^{n-1}\right)\\\\ &=\frac{2}{\Gamma(n/2)}\,\lim_{t\to 0^+}\int_0^{\delta/(2\sqrt t)}e^{-r^2}r^{n-1}\,dr\\\\ &=1 \end{align}$$

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Can you proceed to fill in the rest?

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Alternatively, translate coordinates by letting $\vec y-\vec x=\vec y_{\text{new}}$ so that $g(\vec y)=g(\vec x+\vec y_{\text{new}})$, scale the radial variable $r=|\vec y_{\text{new}}| \to r/(2\sqrt t)$, and simply apply the Dominated Convergence Theorem to bring the limit inside the integral. The result is $g(\vec x)\frac{2}{\Gamma(n/2)}\int_{\mathbb{R}^n}e^{-r^2}\,r^{n-1}\,dr=g(\vec x)$ as expected.