Show that $\frac{1}{n} \sum_{i=1}^n a_i \operatorname{erf} \left( \frac{b_i-x}{\sqrt{2}} \right) \to x$ for some sequence $\{a_n\}$ and $\{b_n\}$

Question

Consider the following function \begin{align} f_n(x)=\frac{1}{n} \sum_{i=1}^n a_i \operatorname{erf} \left( \frac{b_i-x}{\sqrt{2}} \right) \end{align} where $\operatorname{erf} $ is the error function.

Question: Can we show that there exist sequences $ \{a_k\}$ and $\{b_i\}$ such that \begin{align} \lim_{n \to \infty} f_n(x) \to x \end{align} for all x.

I attempted a proof of this by using Taylor series of ${\rm erf}$. However, that argument ended up being a dead end because it was not clear how to choose $a_i$ and $t_i$.

Answer 1

Let $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ be the derivative of your $\mathrm{erf}(x/\sqrt{2})$. Then we have $$\sqrt{2\pi}=\int_{-\infty}^\infty e^{-(x-t)^2/2}dt= \frac{1}{n}\sum_{k=-\infty}^\infty \int_{k/n}^{(k+1)/n}e^{-(x-t)^2/2}dt.$$ Let us compare this with the infinite integral sum of the same integral $$S_n=\frac{1}{n}\sum_{k=-\infty}^\infty e^{-(x-k/n)^2},$$ which is evidently uniformly convergent. The difference between the terms of the two series is at most $$\frac{1}{n}\max\left\{ \left|(d/dt)e^{-(x-t)^2/2}\right|:t\in[k/n,(k+1)/n]\right\}$$ $$\leq \frac{1}{n^2}e^{-(x-k/n)^2},$$ and the sum of these differences for $-\infty<k<\infty$ is $O(1/n)\to 0, $ uniformly in $x$.

So the difference between the integral and the integral sum tends to $0$, uniformly on the whole line. Thus $(\sqrt{2\pi})^{-1}S_n\to 1$ uniformly, and integrating from $0$ to $x$ we obtain the convergence of integrals uniform on compacts.

This shows how to choose your $b_k$ in practice: choose them equidistant with small distance, and $a_k$ independent of $k$.