TITLE: Looking for function $f$ such that $f'<0$ and $(xf)'>0$
QUESTION [7 upvotes]: I'm looking for a function $f(x)$ with the following properties for $x\ge 0$:
$$0\le f(x)\le 1$$
$$f(0)=1$$
$$f'(x)\le 0$$
$$f(x)+xf'(x)\ge 0$$
$$\lim_{x\to\infty} xf(x)=L$$
where $L$ is a positive constant. Essentially I want $xf(x)$ to initially approximate $x$ and then level off at $L$. Candidates include:
$$f(x)={1\over 1+x/L}$$
$$f(x)={\tanh (x/L) \over x/L}$$
but neither of these have an extra parameter that allows me to control how quickly $xf(x)$ approaches $L$ while keeping the initial slope of $xf(x)$ equal to $1$. This is what I would like, ideally.
The motivation for this: $f(x)$ can be thought of as an efficiency with $x$ the input and $xf(x)$ the output. The system is perfectly efficient at zero input and decreases in efficiency with increasing input, but you never get less output from more input.
Thanks for your responses.

REPLY [2 votes]: $$f(x)=\frac{1}{2x}\left(1-\frac{1}{(x+1)^2}\right)=\frac{x+2}{2(1+x)^2} $$
fulfills the given constraints: $f$ is bounded between $0$ and $1$ on $\mathbb{R}^+$, $f(0)=1$, $f$ is decreasing, $xf$ is increasing and $L=\frac{1}{2}$. A parametric family is given by:

$$ f_a(x) = \frac{a^3}{2x}\left(\frac{1}{a^2}-\frac{1}{(x+a)^2}\right)=\frac{a(2a+x)}{2(x+a)^2} $$

with $\color{red}{L=\frac{a}{2}}$ for any $a>0$.