I would need to approximate a step function by a differentiable function. Namely for any $\tau \in \mathbb{R}$, define the step function $$ F(x) = \begin{cases} 1 \mbox{ if $x < \tau$,} \\ 0 \mbox{ otherwise.} \end{cases} $$ Is it possible to find a family of functions $(F^{\delta})_{\delta \in [0,\infty)}$ which are at least twice differentiable and non-increasing with respect to $x$ and which converge point-wise monotonically to $F$ in the limit as $\delta \rightarrow \infty$?
It is not difficult some functions that approximate $F(x)$ but somehow I only find examples where the differentiability condition fails.
No problem. First take $$ g_0(x) = x^2(1-x)^2$$ On $[0,1]$, it's smooth and matches with $y=0$ to first order at the endpoints $x=0,1$. It's non-negative so it has some integral $\int_0^1 g_0=\frac1{30}=:c\ge 0$. Now $g_1(x):=g_0(x)/c$ satisfies $\int_0^1 g_1 = 1$. Then define $$ f(x)=\begin{cases}0 & x\le 0 \\ \int_0^xg_1(s)ds & x\in(0,1] \\ 1 &x>1\end{cases} = \begin{cases}0 & x\le 0 \\ x^3 (6 x^2 - 15 x + 10) & x\in(0,1] \\ 1 &x>1\end{cases}. $$ By construction $f$ is nondecreasing, and globally $C^2$. Since you actually want this backwards you can go ahead and set $$F^1(x) := f(\tau-x)$$ Then to get the approximation, define (aside- it pains me to consider $\delta\to \infty$) $$ F^{\delta}(x) := f(\delta(\tau - x))$$ It follows from the fact that $f$ is non-decreasing that for each $x$, $F^\delta(x)$ increases to $F(x)$. Quick graph on Desmos:
Of course, you can achieve any degree of matching by adjusting $g_0$ suitably.