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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?

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    $\begingroup$ $f(x) = -e^{-x}$? $\endgroup$ – Daniel Schepler Mar 22 at 17:58
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    $\begingroup$ @DanielSchepler I was just about to write the same, +1. $\endgroup$ – Michael Hoppe Mar 22 at 17:59
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    $\begingroup$ @cammil a surjection (i.e. a function whose range is equal to its codomain). $\endgroup$ – Jake Mar 22 at 18:19
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    $\begingroup$ If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice. $\endgroup$ – Ross Millikan Mar 22 at 20:03
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    $\begingroup$ A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave") $\endgroup$ – Apass.Jack Mar 22 at 22:58
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$$ f(x) = x-e^{-x} $$ 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: Plot

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.

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    $\begingroup$ +1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat. $\endgroup$ – Calum Gilhooley Mar 22 at 19:44
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    $\begingroup$ @CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^{-x}$), then fiddled with the parameters until both the function and its inverse came out looking nice. $\endgroup$ – eyeballfrog Mar 22 at 22:22
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    $\begingroup$ @eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph! $\endgroup$ – Apass.Jack Mar 22 at 22:51
  • $\begingroup$ I could’ve never come up with this. $\endgroup$ – Randall Mar 22 at 23:55
  • $\begingroup$ As in Ross Millikan's comment, $f$ is the lower branch of a hyperbola. Neatly, $f^{-1}$ is the upper branch. The lower branch passes through the points $(1,0)$, $(0,-1)$, $(4,1)$, $(-1,-4)$, $(15,4)$, $(-4,-15)$. (So the upper branch passes through the reflections of these six points in the line $x=y$.) This makes the hyperbola easy to plot (in GeoGebra, for instance). Its equation is $x^2-4xy+y^2=1$. The equation of its asymptotes is $x^2-4xy+y^2=0$. $\endgroup$ – Calum Gilhooley Mar 23 at 19:52
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How about

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

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

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  • $\begingroup$ I would have preferred to incorporate the $\pi x$ term into the integral, by adding a constant to the integrand. And using $\pi$ is maybe not the most obvious choice of a number larger than$~\frac\pi2$; the essential point is having the primitive of an everywhere decreasing function of $t$ that moreover is positively bounded away from $0$ (i.e., remains ${}>c$ for some constant $c>0$). $\endgroup$ – Marc van Leeuwen Mar 25 at 17:57

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