Prove that the foundamental solution of the heat equation is an approximated identity for convolution. I have some problem to prove that $$\Gamma_D(x,t):= \left(\frac{1}{4\pi Dt}\right)^{\frac{n}{2}}e^{-\frac{|x^2|}{4Dt}}$$
(where $D>0$ is constant and $n$ is the dimension of $\mathbb{R}^n$) satisfies
$$\lim_{t\to 0^{+}}\int_{\mathbb{R}^{n}}\Gamma_D(x,t)\phi(x)dx=\phi(0)$$
for every $\phi$ continous that for $a,b\geq0$ satisfies $|\phi(x)|\leq a e^{b|x|^2}$.
I've tried to split the integral using that for every $t\neq0$ the integral of $\Gamma_D$ is $1$.  So as $\phi$ is coninous in $0$ there is a ball $B(0,R_{\epsilon})$ such that $|\phi(x)-\phi(0)|<\epsilon$ so $$\phi(0)=\int_{\mathbb{R}^{n}}\Gamma_D(x,t)[\phi(0)+\phi(x)-\phi(x)]dx,$$ so it suffices to show that $$\int_{\mathbb{R}^{n}}\Gamma_D(x,t)[\phi(0)-\phi(x)]\to 0$$ as $t\to 0^{+}$.
By taking the complement of the ball $B^c$ on has
$$\int_{\mathbb{R}^{n}}\Gamma_D(x,t)(\phi(0)-\phi(x))dx\leq\epsilon+\int_{B^c}|\Gamma_D(x,t)||\phi(0)-\phi(x)|dx.$$
How can I show that the second integral is zero as $t\to0^{+}$?   Any suggestion, solution, advice will be welcome
 A: Since $|\phi(x)| \leqslant ae^{b|x|^2}$, we have $|\phi(x) - \phi(0)| \leqslant  a + ae^{b|x|^2}$ and
$$\left|\int_{B^c}\Gamma_D(x,t)[\phi(0)-\phi(x)]\,dx\right| \leqslant \frac{1}{(4\pi D)^{n/2}} \frac{1}{t^{n/2}}\left(a\int_{B^c}e^{-\frac{|x|^2}{4Dt}}\,dx+ a\int_{B^c}e^{-\left(\frac{1}{4Dt}-b\right)|x|^2}\,dx\right)$$
For $0 < t < \frac{1}{8Db}$ we have $-(\frac{1}{4Dt}-b) <  -\frac{1}{8Dt}$. In that case  the second integral on the RHS converges, and we have the bound
$$\tag{*}\left|\int_{B^c}\Gamma_D(x,t)[\phi(0)-\phi(x)]\,dx\right| \leqslant \frac{a}{(4\pi D)^{n/2}} \frac{1}{t^{n/2}}\int_{B^c}e^{-\frac{|x|^2}{4Dt}}\,dx+ \frac{a}{(4\pi D)^{n/2}} \frac{1}{t^{n/2}}\int_{B^c}e^{-\frac{|x|^2}{8Dt}}\,dx$$
We can now show that the RHS of (*) converges to $0$ as $t \to 0+$. Both terms on the RHS are similar and upon changing to polar coordinates with $r =|x|$, they assume the form
$$\tag{**}C t^{-n/2}\int_{R_\epsilon}^\infty r^{n-1}e^{-\alpha r^2/t} \, dr,$$
where integration over the angular coordinates has been absorbed into the constant $C$.
For any integer $m$, we have $e^{\alpha r^2/t} > \frac{(\alpha r^2/t)^m}{m! }$ and
$$t^{-n/2}r^{n-1}e^{-\alpha r^2/t}  < m!\alpha^{-m}t^{m-n/2}r^{-2m+n-1}$$
If we choose $m > n/2$, then we have $-2m+n-1 <-1$ and the integral on the RHS of (**) is convergent.  We also have $m - n/2 > 0$, and consequently
$$\lim_{t \to 0+} C t^{-n/2}\int_{R_\epsilon}^\infty r^{n-1}e^{-\alpha r^2/t} \, dr \leqslant \lim_{t \to 0+} t^{m-n/2}\left(Cm!\alpha^{-m}\right)\int_{R_\epsilon}^\infty r^{-2m+n-1} \, dr = 0$$
