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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

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    I am not sure I wrote the condition with an integral in the right way. If it is what you meant?2011-08-15
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    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.2011-08-15
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    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.2011-08-15
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    @Arturo Magidin: I'm looking for a function that expresses these conditions as an inherent symmetry, not a constructed one.2011-08-15
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    @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.2011-08-15
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    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 $f(x)=x$?2011-08-15
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    @Mike: No, it forces $f(f(x)) = x$. Not the same thing.2011-08-15
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    @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 $(0,\infty)$.2011-08-15
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    6) is a nonsense condition, why do you want that?2011-08-15
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    @Jonas Teuwen: 6) requires that $\Delta dy/dx$ be capable of varying from one extreme to the other in a given function.2011-08-15
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    I don't understand what you mean and I really doubt it that it matters because you can just use characteristic functions.2011-08-15
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    @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 $f(x)=x$.2011-08-15
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    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.2011-08-15
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    Actually, there is an infinite set of modular functions such that $f'(x)$(left)=$f'(x)$(right) at $f(x)=x$2011-08-15
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    @bob: "$f$ is defined by exactly one equation" is not a well-defined notion; condition (6) is either empty or incoherent.2011-08-15
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    @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!2011-08-15
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    Sorry for the frustration, I'm not brilliant at terminology but I'm trying to be as clear as possible.2011-08-15

2 Answers 2

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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 $\int_0^\infty f(x) = \pi^2/6.$

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    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 $a=\pi/\sqrt{6C}$.2011-08-15
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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.)

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    This is correct, but I need f(x) to be represented by only one equation. Edited for reference.2011-08-15
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    Is a piecewise definition consistent with condition (6) in the question. Disambiguating condition (6) is a tough one; all the rest are simple.2011-08-15
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    "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.2011-08-15
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    @bob: I wonder why you have the restriction to "no piecewise functions"? I showed [here](http://math.stackexchange.com/questions/49963/54750#54750) how to "join up" a piecewise equation, and [if you're allowing the construction $\sqrt{x^2}$](http://math.stackexchange.com/questions/54685), well...2011-08-16
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    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).2011-08-16
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    "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.2011-08-16
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    @bob: Anyway, what's wrong with my $\tilde f$ or Rahul Narain's $-\log(1-\exp(-x))$?2011-08-16
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    Nothing from your respective viewpoints. Half the problem was gauging correct terminology, so in that respect I was the one who was incorrect.2011-08-16