Show that $$v_{t}(x) = (4 \pi kt)^{- \frac{1}{2}} \exp \left( -\frac{a x ^2}{4kt} \right)$$ converges to $\delta_{0}$ in $D'(\mathbb{R})$ when $t \to 0^{+}$.
Asumming that:
$$\int_{\mathbb{R}} \exp \left( -a x^2\right) dx = (\frac{\pi}{a})^{\frac{1}{2}}, (a > 0)$$
I've been working on this problem but I don't think that $v_t(x)$ converge to $\delta_{0}$. I think it converges to $\delta_{0} \, a^{-\frac{1}{2}}$.
My proof:
$$\lim_{t \to 0^{+}} \int_{\mathbb{R}}(4 \pi kt)^{- \frac{1}{2}} \exp \left( -\frac{a x^2}{4kt} \right) \phi(x)dx$$
Changing the variable $x = \sqrt{4kt}y \Rightarrow dx = \sqrt{4kt} dy$.
$\lim_{t \to 0^{+}} \int_{\mathbb{R}}(4 \pi kt)^{- \frac{1}{2}} \exp \left( -\frac{a x^2}{4kt} \right) \phi(x)dx = \lim_{t \to 0^{+}} \int_{\mathbb{R}} \frac{1}{\sqrt{\pi}} \exp \left( – ay^2 \right) \phi(\sqrt{4kt}y) dy \\ = \phi(0) \int_{\mathbb{R}} \frac{1}{\sqrt{\pi}} \exp \left( – ay^2 \right)dy = \phi(0) \frac{1}{\sqrt{a}} = \frac{\phi(0)}{\sqrt{a}} = \frac{\delta(\phi)}{\sqrt{a}}, \forall \phi \in D(\mathbb{R})$
We can conclude that $v_t(x)$ converge to $\delta_{0}$ in $D'(\mathbb{R})$.
I'm using the fact that Lebesgue integral of a continuous function is continuous.
Best Answer
Yes, your proof is correct. From the outset, the formula for $v_t$ suggested that the limit will involve $a$: the parameter $a$ affects the integral of $v_t$, because it rescales the function in "horizontal" dimension only. (Unlike $t$, which scales both horizontally and vertically, so that the integral of $v_t$ remains the same.)
The passage to the limit is justified by domination
$$\left|\exp \left( - ay^2 \right) \phi(\sqrt{4kt}y) \right| \le \exp \left( - ay^2 \right) \sup|\phi|$$