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?

Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?) — cammil, Mar 22 '19 at 18:10
@cammil a surjection (i.e. a function whose range is equal to its codomain). — , Mar 22 '19 at 18:19
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. — Ross Millikan, Mar 22 '19 at 20:03
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") — Apass.Jack, Mar 22 '19 at 22:58

eyeballfrog · Accepted Answer · 2019-03-23T16:30:58.513

21

$$ f(x) = x-e^{-x} $$ 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.

edited Mar 23 '19 at 16:30

answered Mar 22 '19 at 18:38

eyeballfrog

22,485

1

+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. – Calum Gilhooley Mar 22 '19 at 19:44
2

@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. – eyeballfrog Mar 22 '19 at 22:22
2

@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph! – Apass.Jack Mar 22 '19 at 22:51
I could’ve never come up with this. – Randall Mar 22 '19 at 23:55
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 $x^2-4xy+y^2=0$. – Calum Gilhooley Mar 23 '19 at 19:52

score 5 · Answer 2 · answered Mar 22 '19 at 18:33

5

How about

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

answered Mar 22 '19 at 18:33

paw88789

40,402

zhw. · Answer 3 · 2019-03-23T23:29:37.773

3

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

edited Mar 23 '19 at 23:29

answered Mar 22 '19 at 18:57

zhw.

105,693

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 $c>0$). – Marc van Leeuwen Mar 25 '19 at 17:57

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

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