Is there a bijective, monotonically increasing, strictly concave function from the reals, to the reals?

Question

I can't come up with a single one.

The range should be the whole of the reals. The best I have is $\log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^{-x}$ only maps to half of the real line.

Any ideas?

Answer 1

$$ f(x) = x-e^{-x} $$ is such a function. Since $f''(x) = -e^{-x}$ is always negative, it is strictly concave, and it's not hard to show it hits every real.

Even better, $$ f(x) = 2x -\sqrt{1+3x^2} $$ has $f''(x) = -3(1+3x^2)^{-3/2} < 0$ everywhere and the explicit inverse $f^{-1}(x) = 2x+\sqrt{1+3x^2}$, clearly defined for all $x$.

EDIT: Since it was requested in the comments, here is a plot of this function and its inverse:

Note that even though the growth rate for positive $x$ is slow, the function is asymptotically linear (with slope $2-\sqrt{3}\approx 0.268$) and thus unbounded.

Answer 2

How about

$f(x)=\left\{\begin{array}{cc} \ln(x+1)& &x\ge 0\\1-e^{-x}& &x<0\end{array}\right.$

Answer 3

$f(x) = \pi x+ \int_0^x \arctan (-t)\,dt$ is an example. Many more examples like this one can be constructed.