Consider the following probability density function parameterized by $\sigma>0$: $$g_\sigma(x)=2\sum_\limits{n=0}^\infty\phi_\sigma(x+n), \quad 0<x<1$$ where $$\phi_\sigma(t)={1\over \sigma\sqrt{2\pi}}\exp\left({-{1\over 2}\left({t\over\sigma}\right)^2}\right),\quad -\infty<t<\infty.$$
Context: $g_\sigma$ and $\phi_\sigma$ are the probability density functions of $\{|\sigma Z|\}$ and $Z$ , respectively, where $Z$ is a standard normal random variable, $\sigma>0$, and $\{x\}:=x-\lfloor x\rfloor$ is the fractional part of $x$. These arose in my answer to a related question.
Conjecture 1: For any $x\in(0,1)$, $\lim_{\sigma\to\infty}g_\sigma(x)=1$; i.e., $\{|\sigma Z|\}\overset{d}{\to}\text{Uniform}(0,1).$
This seems quite clear from plots of $g_\sigma$ for various values of $\sigma$:
In fact, the asymptotic behavior is so regular, that the following seems to hold:
Conjecture 2. There is a constant $c\approx 0.4$ such that for all $\sigma>1$, $$g_\sigma(x) = 1+{c\over \sigma}(1-2x) + o(\sigma^{-1}),\quad 0<x<1.$$ (But what is the exact value of $c$?)
Here's a picture of the corresponding linear approximations using $c=0.4$, superimposed onto the previous plot:
Question: How to prove these conjectures?
Here's what I've tried ...
Fourier transforms can be used for the probability density (call it $f_\sigma$) of the fractional part without the absolute value, i.e. the density of $\{\sigma Z\}$. Using this method for $f_\sigma$ turns out to give $$\begin{align}f_\sigma(x) &= \sum_{n=-\infty}^\infty\phi_\sigma(x+n)\quad (0<x<1)\\[2ex] &=\sum_{k=-\infty}^\infty\hat\phi_\sigma(k)e^{i2\pi k x},\quad \hat\phi_\sigma(k):=\int_{-\infty}^\infty\phi_\sigma(x)e^{-i 2\pi k x}dx=2\int_{0}^\infty\phi_\sigma(x)\cos(2\pi k x)\,dx\\[2ex] &=1 + 2\,\sum_{k=1}^\infty\,\exp\left(-2(\pi k\sigma)^2 \right)\ \cos(2\pi k x)\quad (0<x<1)\\[2ex] &=1 + 2\exp(-2\pi^2\sigma^2)\cos(2\pi x) + o(e^{-\sigma^2})\quad (0<x<1) \end{align}$$ which has a very much more rapid convergence to uniformity than does $g_\sigma$.
Proceeding similarly with our case of $g_\sigma$, let $\psi_\sigma(x) = \phi_\sigma(x)\,1_{(0,\infty)}(x)$, so we have $$g_\sigma(x)=2\sum_{n=-\infty}^\infty\psi_\sigma(x+n)\quad (0<x<1)$$ and it seems that similar Fourier transform methods ought to apply here as well.
However, when I write (using Poisson's summation formula) $$g_\sigma(x)=2\sum_{k=-\infty}^\infty\hat\psi_\sigma(k)e^{i2\pi k x} $$ with $$\begin{align}\hat\psi_\sigma(k)&=\int_{-\infty}^\infty\psi_\sigma(x)e^{-i 2\pi k x}dx\\[2ex] &=\int_{0}^\infty\phi_\sigma(x)e^{-i2\pi k x}dx \end{align}$$ I get, using Sage, that
$$\hat\psi_\sigma(k)={1\over 2}e^{-2(\pi k\sigma)^2}\left( 1 - \text{erf}\left( i \sqrt{2}\pi k\sigma\right)\right). $$ The trouble is that inserting this into the summation gives $$g_\sigma(x) = \sum_{k=-\infty}^\infty e^{-2(\pi k\sigma)^2}\left( 1 - \text{erf}\left( i \sqrt{2}\pi k\sigma\right)\right)e^{i 2\pi k x} $$ and I don't know how to proceed further. (I'm rusty with Fourier methods, so I would appreciate any advice as well as corrections.)
Here's a proof of the conjectures ...
For any $\sigma>0$, $$\begin{align}\Pr\left( \{|\sigma Z|\} \le x \right) &=\sum\limits_{k=0}^\infty\Pr\left(-k-x< \sigma Z \le -k \right) +\sum\limits_{k=0}^\infty\Pr\left(k< \sigma Z \le k+x \right)\\[2ex] &=2\,\sum\limits_{k=0}^\infty\Pr\left(k< \sigma Z \le k+x \right)\\[2ex] &=2\,\sum\limits_{k=0}^\infty{\large(}\Pr\left(\sigma Z \le k+x \right)-\Pr\left(\sigma Z \le k\right){\large)}\\ \end{align}$$ and differentiating to obtain the probability density ... $$\begin{align}g_\sigma(x) &=2\,\sum\limits_{k=0}^\infty \phi_\sigma(k+x)\quad (0<x<1). \end{align}$$ Now let $\psi_\sigma(x) = \phi_\sigma(x)\,1_{(0,\infty)}(x)$, so we have $$g_\sigma(x)=2\sum_{n=-\infty}^\infty\psi_\sigma(x+n)\quad (0<x<1)$$ and Poisson's summation formula can be applied to give $$g_\sigma(x)=2\sum_{k=-\infty}^\infty\hat\psi_\sigma(k)e^{i2\pi k x} $$ where $$\begin{align}\hat\psi_\sigma(k)&=\int_{-\infty}^\infty\psi_\sigma(x)e^{-i 2\pi k x}dx\\[2ex] &=\int_{0}^\infty\phi_\sigma(x)e^{-i2\pi k x}dx\\[2ex] &={1\over 2}e^{-2(\pi k\sigma)^2}\left( 1 - \text{erf}\left( i \sqrt{2}\pi k\sigma\right)\right). \end{align}$$ Inserting this into the summation gives
$$\begin{align}g_\sigma(x) &= \sum_{k=-\infty}^\infty e^{-2(\pi k\sigma)^2}\left( 1 - \text{erf}\left( i \sqrt{2}\pi k\sigma\right)\right)e^{i 2\pi k x}\\[2ex] &=\sum_{k=-\infty}^\infty e^{-2(\pi k\sigma)^2}e^{i 2\pi k x} - \sum_{k=-\infty}^\infty e^{-2(\pi k\sigma)^2}\text{erf}\left( i \sqrt{2}\pi k\sigma\right)e^{i 2\pi k x}.\\[2ex] \end{align}$$
Note that $e^{-2(\pi k\sigma)^2}$ and $\cos(2\pi k x)$ are even functions of $k$, while $\text{erf}\left( i \sqrt{2}\pi k\sigma\right)$ and $\sin(2\pi k x)$ are odd functions of $k$, so the summations simplify to give
$$\begin{align}g_\sigma(x) &= 1+2\sum_{k=1}^\infty e^{-2(\pi k\sigma)^2}\cos( 2\pi k x) - i\,2\sum_{k=1}^\infty e^{-2(\pi k\sigma)^2}\text{erf}\left( i \sqrt{2}\pi k\sigma\right)\sin(2\pi k x)\\[2ex] &=\underbrace{1+2\sum_{k=1}^\infty e^{-2(\pi k\sigma)^2}\cos( 2\pi k x)}_{f_{\sigma}(x)} + \,2\sum_{k=1}^\infty e^{-2(\pi k\sigma)^2}\text{erfi}\left(\sqrt{2}\pi k\sigma\right)\sin(2\pi k x)\\[2ex] \end{align}$$
where $f_\sigma(x)=1+o(e^{-\sigma})$ is the probability density of $\{\sigma Z\}$ as derived earlier, and $\text{erfi}(x):=i\,\text{erf}(i\,x)$ for real $x$. Now, it can be shown that asymptotically for $x\to\infty$, $$\text{erfi}(x) = {1\over\sqrt{\pi}}x^{-1}e^{x^2}(1+O(x^{-2})) $$ so we have $$\begin{align}g_\sigma(x) &= f_\sigma(x)+{1\over\sigma}\sqrt{2\over\pi^3}\sum_{k=1}^\infty {\sin(2\pi k x)\over k}+O(\sigma^{-2}).\\[2ex] \end{align}$$
The conjectures are then established by noting that $$\sum_{k=1}^\infty {\sin(2\pi k x)\over k}={\pi\over 2}(1-2x) \quad(0<x<1) $$ because the LHS is just the Fourier series representation of the odd periodic extension of the RHS, the $k$th Fourier coefficient being given by $$2\int_0^1 {\pi\over 2}(1-2x)\sin(2\pi k x)\,dx={1\over k}. $$ Consequently, both conjectures are true, and in Conjecture 2 the constant is $c={1\over\sqrt{2\pi}}=0.39894...\,:$ $$g_\sigma(x) = 1+{1\over\sigma\sqrt{2\pi}}(1-2x) + O(\sigma^{-2}),\quad 0<x<1. $$