Heat equation: Heat Kernel as $t\to0$ Consider heat flow on an infinite, 1D wire. The temperature T(x,t) obeys the diffusion equation,
$$
\frac{∂T}{∂t} = D \frac{∂^2T}{∂x^2}
$$
with initial condition $T(x,0) = δ(x)$.
The heat kernel is given by:
$$
T[x, t] =\frac{ 1}{\sqrt{4πDt}} \exp\left(-\frac{x^2}{4Dt}\right)
$$
I was asked to verify the heat kernel is a solution. It is easy to show that this satisfies the heat equation. However, to check the initial condition at $t=0$, I must take the limit as $t\to0$ (it shoots to infinity from the look of it). Could someone give me a hint on how to do this?
 A: It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since 
$$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$.
Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have
$$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$
Therefore
$$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) dx = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0)  \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$
In other words
$$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$
The only crucial passage is 
 $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z)  g(sz) dz$$
A quite mild condition that guarantees the passage is that $g$ is bounded as already required  (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that 
$$T[x,0^+] = \delta(x)\:.$$
The heat kernel is used to construct the solution of the heat equation 
$g=g(x,t)$ out of the initial condition $g(x)$:
$$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$
satisfies 
$$
\frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2}
$$
for $t>0$ with initial condition 
$$g(x,0) = g(x)\:.$$
The proof is immediate from (1).
A: Another way of getting at Valter Moretti's answer using a somewhat outdated (but altogether rigorous) notion of generalized functions is simply to witness that:
$$\int_{-\infty}^\infty T(x, t)\,\mathrm{d} t=1$$
i.e. $T$ is a normalized Gaussian function of $x$ and therefore that we can think of the equivalence class of sequences prototyped by the sequence $T\left(x, 1\right) ,\,T\left(x, \frac{1}{2}\right),\,T\left(x, \frac{1}{3}\right),\,\cdots$ as the generalized function $\delta(x)$ in the way defined in M. J. Lighthill, "An Introduction to Fourier Analysis and Generalized Functions". Here, we conceive of a generalized function not foremost as a member of the algebraic dual of the Schwartz space but rather as an equivalence class of sequences of functions, where the equivalence relationship is $f=\{f_n(x)\}_{n=0}^\infty\sim g=\{g_n(x)\}_{n=0}^\infty$ iff $\lim\limits_{n\to\infty}\langle f_n,\,h\rangle=\langle g_n,\,h\rangle\,\forall h\in\mathscr{S}$ where $\mathscr{S}$ is the Schwartz space. So we have, by the Lighthill definition of $\delta$, that $T(x,\,0^+)=\delta(x)$ and the rest of Valter's answer follows.
Now, I admit that this may seem a bit of a daft answer, because essentially all I am doing is saying in fancy "it's true because that is one possible definition of Dirac delta", but it does recall one approach to the introduction to the notion of generalized functions (the Lighthill / Temple approach) that is still sometimes used in introductory expositions of the idea. When this approach is discussed, the heat kernel is often explicitly singled out as Lighthill's "prototype" for the $\delta$ equivalence class. I sometimes find it helpful to think of generalized functions in this way to see certain results. So I answered, because your question evoked fond memories of my first grasp of the rigorous notion  of a generalized function through Lighthill's approach.
A: Take the integral of T with respect to x from - infinity to + infinity and show that it is equal to 1 at all times.
