Is there a function that has these properties?
Points:
- $f(1)=\tfrac{1}{2}$
- $f(-1)=-\tfrac{1}{2}$
- $f(0)=0$
Bounds: $f$ is bounded between $(-1,1)$:
- $\forall x\in\mathbb{R}: -1 < f(x) < 1$
- $\lim_{x\to \infty} f(x)=1$
- $\lim_{x\to -\infty} f(x)=-1$
Slopes: $f$ is strictly increasing for all $x\in\mathbb{R}$ and at $x=0$, $f'(x)=1$:
$$\frac{d}{dx}f(x)=\begin{cases}0<f'<1&:x<0\\1 &:x=0\\0<f'<1&:x>0 \end{cases}$$
Concavity:
$$\frac{d^2}{dx^2}f(x) = \begin{cases}>0 &:x<0\\ 0 &:x=0\\ <0 &:x>0 \end{cases}$$
Smoothness: $f$ is infinitely differentiable and every derivative is continuous (or perhaps uniformly continuous):
$$\forall n\in\mathbb{N} : \forall x\in\mathbb{R} : \exists y\in\mathbb{R} : f^{(n)}(x) = y$$
I've tried $\tanh(x)$, $\frac{2}{\pi}\arctan(x)$, $\text{erf}(x)$, and others, but all of them were missing at least one of the properties listed above. Is there one that satisfies ALL those properties? If so, prove it or show an example.
Below is a graph (in red) of the type of function I'm looking for. The derivative is in green and the 2nd derivative is in blue. (The function shown is $\arctan x$ so it's not exactly what I'm looking for.)