What function(s) can satisfy these conditions?

Question

For what function (or functions) is the following true:

1) $f(x)$ is positive for $x>0$

2) $\lim\limits_{x\to 0}{f(x)} = \infty$

3) $\lim\limits_{x\to\infty}f(x) = 0$

4) $\int_{0}^{\infty} {f(x)} dx = C$

5) $f(x)$ is symmetric over $y=x$

6) $f(x)$ isn't written in case structure

I am not sure I wrote the condition with an integral in the right way. If it is what you meant? — SBF, Aug 15 '11 at 16:51
If condition $4$ were $\int_0^{\infty}f(x),dx = C$, I think that taking any function defined on $[1,\infty)$ with $f(1)=1$, $f$ strictly decreasing, and $\int_1^{\infty}f(x),dx$ converges, and then extending to $(0,1)$ to ensure symmetry, will do. — Arturo Magidin, Aug 15 '11 at 16:59
Fixed. Was working on the equations without realizing they'd been formatted. The inverted f(x) was due to my use of an integral that contained a fraction as a reference to write mine. Thanks. — user474632, Aug 15 '11 at 17:03
@Arturo Magidin: I'm looking for a function that expresses these conditions as an inherent symmetry, not a constructed one. — user474632, Aug 15 '11 at 17:12
@bob: But what I'm pointing out is that, in general, there are lots of such functions, since you can construct them very easily from functions that satisfy relatively mild conditions. — Arturo Magidin, Aug 15 '11 at 17:16
What exactly do you mean by "$f(x)$ is symmetric over $y=x$"? I'd have thought you meant the graph of $f$ is invariant under flipping across the line $y=x$ but doesn't that just force $f(x)=x$? — Mike F, Aug 15 '11 at 17:19
@Mike: Why? $f(x) = \frac{1}{x}$ satisfies symmetry about the line $y=x$, since $(a,b)$ is in the graph of $f$ if and only if $(b,a)$ is in the graph of $f$. What it does force is $f(f(x))=x$ for all $x$; i.e., $f=f^{-1}$ on $(0,\infty)$. — Arturo Magidin, Aug 15 '11 at 17:22
@Jonas Teuwen: 6) requires that $\Delta dy/dx$ be capable of varying from one extreme to the other in a given function. — user474632, Aug 15 '11 at 17:36
I don't understand what you mean and I really doubt it that it matters because you can just use characteristic functions. — JT_NL, Aug 15 '11 at 17:38
@Arturo, Ilmari: Sorry, condition 4 was $\int_0^\infty \frac{1}{f(x)} dx = C$ when last I looked and this made me assume the values of the limits in 2,3 were switched. At least if $f$ was continuous this should have forced $f(x)=x$. — Mike F, Aug 15 '11 at 17:39
Just add a condition for $f$ to be smooth everywhere except $0$ and that should take out all of the "artificial" functions. I don't know why others are putting down bob's curiosity - it seems fairly natural to me to ask when there are "natural" functions satisfying so and so properties. — anon, Aug 15 '11 at 17:44
Actually, there is an infinite set of modular functions such that $f'(x)$(left)=$f'(x)$(right) at $f(x)=x$ — user474632, Aug 15 '11 at 18:01
@bob: "$f$ is defined by exactly one equation" is not a well-defined notion; condition (6) is either empty or incoherent. — Arturo Magidin, Aug 15 '11 at 18:09
@bob: You are still not giving a well-defined notion. What does it mean for a function to "contain conditional equations"? These are not well-defined concepts! — Arturo Magidin, Aug 15 '11 at 18:38
Sorry for the frustration, I'm not brilliant at terminology but I'm trying to be as clear as possible. — user474632, Aug 15 '11 at 18:41

score 9 · Accepted Answer · answered Aug 15 '11 at 18:18

A simple general recipe for these conditions is the implicit equation $$g(x)g(y) = 1,$$ with $g$ being an increasing function such that $g(0) = 0$ and $g(x) \to \infty$ as $x \to \infty$. Then $f$ is given by $$f(x) = g^{-1}\left(\frac 1 {g(x)}\right).$$

To get a finite integral for $\int _0^\infty f(x)$ under the given conditions, all you need is for $f(x)$ to decay faster than $1/x$ as $x \to \infty$. This is guaranteed as long as, informally, $g(x)$ grows "faster" towards $\infty$ as $x \to \infty$ than it decays to $0$ as $x \to 0$. (Formally, if $g(x)$ goes as $x^n$ as $x\to\infty$ and as $x^m$ as $x\to 0$, then you need $n > m$.)

You can recover Ilmari's example by taking $$g(x) = \begin{cases} x & \text{if }0 < x \le 1, \\ x^2 & \text{if }1 < x. \end{cases}$$

For a smooth example, let $g(x) = e^x - 1$. Then you get $$f(x) = -\ln\left(1 - e^{-x}\right),$$ for which WolframAlpha reports that $\int_0^\infty f(x) = \pi^2/6.$

To adjust the value of the integral of $f$ at a given value $C$, one can simply use the solutions $f_a(x)=a^{-1}f(ax)$ where $a$ is positive. The integral of $f_a$ being $\pi^2/(6a^2)$, it is $C$ for $a=\pi/\sqrt{6C}$. — Did, Aug 15 '11 at 21:33

Ilmari Karonen · Answer 2 · 2011-08-15T18:20:33.790

8

As a particular example of a function satisfying Arturo Magidin's conditions,

$$f(x) = \begin{cases} 1/\sqrt x & \text{if }0 < x \le 1 \\ 1/x^2 & \text{if }1 < x \end{cases}$$

ought to work. In particular, $\displaystyle \int_0^\infty f(x)\;dx = 3$ and $f(f(x)) = x$.

Addendum: If you don't want a piecewise defined function, you could write $f$ above as $f(x) = \exp\;g(\log x)$, where

$$g(u) = -\frac34 |u| - \frac54 u.$$

Of course, this is just a notational trick; $f$ still has a "kink" at $x=1$ due to the non-differentiability of $|u| = |\log x|$ there. However, it's a trick that points in a useful direction: if you also want the function to be everywhere differentiable, you can replace $|u| = \sqrt{u^2}$ in the definition of $g$ above with the hyperbola $\sqrt{1+u^2}$ to get

$$\tilde g(u) = -\frac 34 \sqrt{1 + u^2} - \frac 54 u,$$

and thus

$$\tilde f(x) = \exp \left( -\frac 34 \sqrt{1 + (\log x)^2} - \frac 54 \log x \right).$$

This function $\tilde f$ still satisfies $\tilde f(\tilde f(x)) = x$, and since $\tilde f(x) < f(x)$ for all $x$, we know that its integral from 0 to infinity must be less than 3. (Actually, it is about 2.19574343.)

edited Aug 15 '11 at 18:20

answered Aug 15 '11 at 17:20

Ilmari Karonen

26,117

This is correct, but I need f(x) to be represented by only one equation. Edited for reference. – user474632 Aug 15 '11 at 17:27
Is a piecewise definition consistent with condition (6) in the question. Disambiguating condition (6) is a tough one; all the rest are simple. – Michael Hardy Aug 15 '11 at 17:48
6

"One equation" is a poorly-defined condition, but you might ask for $f$ to be analytic. The first example that comes to mind (of an involution on $(0,\infty)$ that is analytic) is $f(x) = 1/x$, but of course that does not satisfy the integrability condition. – Robert Israel Aug 15 '11 at 18:14
@bob: I wonder why you have the restriction to "no piecewise functions"? I showed here how to "join up" a piecewise equation, and if you're allowing the construction $\sqrt{x^2}$, well... – J. M. ain't a mathematician Aug 16 '11 at 04:14
That's definitely the term my engrish was unable to translate (I accepted this answer because it is correct for both of my misspoken questions). If anyone knows how to meet these criteria with a non-piecewise function, I'd like to know (I believe its construction is not possible within the limitations of exponential form). – user474632 Aug 16 '11 at 10:38
"If anyone knows how to meet these criteria with a non-piecewise function" - what is your definition of "non-piecewise function" @bob? Again, if you allow the square root and squaring functions, then it is a simple matter to construct functions that are equivalent to explicitly piecewise functions. – J. M. ain't a mathematician Aug 16 '11 at 13:46
@bob: Anyway, what's wrong with my $\tilde f$ or Rahul Narain's $-\log(1-\exp(-x))$? – Ilmari Karonen Aug 16 '11 at 13:57
Nothing from your respective viewpoints. Half the problem was gauging correct terminology, so in that respect I was the one who was incorrect. – user474632 Aug 16 '11 at 15:48

What function(s) can satisfy these conditions?

2 Answers2