Prove that $\lim_{n\to \infty}\langle \operatorname{erfc}(-nx), \phi\rangle =\langle H_0, \phi\rangle $

Question

Define the error function $\operatorname{erf}(x)$ as:

\begin{equation} \operatorname{erf}(x):=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-y^2}dy \end{equation} and $\operatorname{erfc}(x):=1-\operatorname{erf}(x)$. Now, I would like to prove that the limit of the sequence of distributions which are defined from the sequence of functions: \begin{equation} f_n(x):=\frac{1}{2}\operatorname{erfc}(-nx) \end{equation} is the Heaviside distribution, i.e: $f_n \to H_0$ (in the way that convergence is defined in $\mathcal{D}'$).

I played around a bit with the integrals but I had no luck.

Thank you for your time!

Answer 1

Hints: You may notice the following facts:

$$ \text{erf}(-x)=-\text{erf}(x),\\ \lim_{x \rightarrow +\infty }\text{erf}(x)=1 $$

Therfore: $$ \lim_{x \rightarrow \infty }1-\text{erf}(x)=1-\text{sign}(x) $$

Can you take it from here?

Edit:

Another way to prove this would be to use the following representation of the $\delta$-distribution

$$\delta(x)=\lim_{n\rightarrow \infty}e^{-n x^2}\frac{\sqrt{n}}{\sqrt{2\pi}}$$

And the fact that $\int_0^x\delta(x')dx'=H(x)$