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 ?
 A: For the last line to be true, you need show that the limits can be interchanged. I will propose her another solution. Note
\begin{eqnarray}
 \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&       \nonumber \\
 \int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)|  &=&  \nonumber \\
 \int_{|x|\leq\delta}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)|       +\int_{|x|>\delta}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)|        \tag{1}\end{eqnarray}
For $\epsilon>0$, choose $\delta>0$ such that $|\varphi(x)-\varphi(0)|\leq\epsilon$ for $\|x\|\leq\delta$. We conclude from $(1)$ that  
$$\left|\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}\varphi(x)-\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{- \frac{|x|^2}{4t}}\varphi(0) \right| \leq \\ \epsilon+\int_{|x|>\delta}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)| $$
Now you can conclude.
