Special smooth non-analytic function

Question

Is there a smooth function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f(x)=0 $ $\forall x\leq 0$, $f(x)=1 $ $\forall x\geq 1$, and it is mononically increasing?

Answer 1

There is a function $g\in C^\infty(\mathbb R)$ such that $g=0$ for $x\le 0,$ $g>0$ for $0<x<1,$ and $g=0, x\ge 1.$ Let $c = \int_{\mathbb R} g.$ Then

$$f(x) = \frac{1}{c}\int_{-\infty}^x g(t)\,dt$$

does what you want

Answer 2

Let$$g(x)=\begin{cases}0&\text{ if }x\leqslant0\\e^{-1/x}&\text{ otherwise,}\end{cases}$$which is smooth. Now, define $f(x)$ as$$\frac{g(t)}{g(t)+g(1-t)}$$and you'll have the function that you're after.

Answer 3

If $\lambda$ is a non-negative (and not identically zero) smooth function with support in $[0,1]$, then the function $$f:\mathbb{R} \to \mathbb{R}$$ given by $$f(x)=\frac{1}{\int_{-\infty}^{\infty}\lambda} \cdot\int_{0}^x \lambda$$ satisfies what you want. Indeed, $f'(x)=\frac{1}{\int_{-\infty}^{\infty}\lambda}\cdot \lambda(x) \geq 0$, and thus $f$ is monotonically increasing. It is then easy to check that $f$ satisfies all other requirements.