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

Consider $\varphi \in S(R^n)$ (space of rapidly decreasing functions). Consider the heat kernel

$$K_t(x) = \displaystyle\frac{1}{{(4\pi t)}^{n/2}} \displaystyle e^{- \displaystyle\frac{|x|^2}{4t}}, t>0 , x \in R^n$$

I want to show that $$\displaystyle\lim_{t \rightarrow 0^{+}}\displaystyle\int_{R^n} e^{- \displaystyle\frac{|x|^2}{4t}} \varphi(x) \ dx = \varphi(0).$$

My try: we have $\displaystyle\int_{R^n} K_t(x) \ dx = 1$ and $\lim_{t \rightarrow 0^{+} }\displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}}=0$ for all $\epsilon >0$. Then

$$\displaystyle\lim_{t \rightarrow 0^{+} } |\displaystyle\int_{R^n} e^{- \displaystyle\frac{|x|^2}{4t}} \varphi(x) \ dx - \varphi(0)| = \displaystyle\lim_{t \rightarrow 0^{+} } | \displaystyle\int_{R^n} \displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}} \varphi(x) \ dx -\displaystyle\int_{R^n}\displaystyle\frac{1}{{(4\pi t)}^{n/2}} e^{- \displaystyle\frac{|x|^2}{4t}}\varphi(0) \ dx|$$

$$\leq \displaystyle\lim_{t \rightarrow 0^{+} } || \varphi - \varphi(0)||_{\infty}. \displaystyle\lim_{\epsilon \rightarrow 0^{+} } \displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}} \ dx$$

$$=\displaystyle\lim_{\epsilon \rightarrow 0^{+} } || \varphi - \varphi(0)||_{\infty} \displaystyle\lim_{t \rightarrow 0^{+} }\displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}} \ dx = 0$$

I dont know if my solution is correct (i am not sure about the last line). Someone can give me a hint to this exercise ?

• Why is this convergence true, $\lim_{t \rightarrow 0^{+} }\displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}}=0$. Does the sequence of functions, $\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}}$, converge uniformly to 0? – alpastor Dec 1 '18 at 2:29

For $\epsilon>0$, choose $\delta>0$ such that $|\varphi(x)-\varphi(0)|\leq\epsilon$ for $\|x\|\leq\delta$. We conclude from $(1)$ that
$$\left|\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}\varphi(x)-\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{- \frac{|x|^2}{4t}}\varphi(0) \right| \leq \\ \epsilon+\int_{|x|>\delta}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)|$$