# Are there functions satisfying the following integral condition?

Can we find two functions $f$ and $g$ that are reasonably defined nontrivial(not everywhere zero, $f\neq g$, not linear polynomials) functions such that the following condition is satisfied?

$$f( \left(\int_{0}^{t} g(x) \ \text{d}x\right)) = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right))$$

P.S.: I migrated this question from here on Math.SE. I am sure this site hosts very knowledgeable mathematicians that keeping on migrating to another site is foolish. I felt a really good feeling for some time as nobody answered my question. But is usually the case that: "There is a general principle that a stupid man can ask such questions to which one hundred wise men would not be able to answer. In accordance with this principle I shall formulate some problems." Vladimir Arnold

Motivation: The equation that I wrote out was not random. At least, the symmetry I find in it and the absence of an iota of clue at proceeding with any method makes me fall in love with finding a solution. Part of the motivation was to find a function that in some way resembles the exponential function. The exponential map is invariant under differentiation. So, the natural curiosity to find a nontrivial map invariant under integration. For obvious reasons, such map does not exist because of the presence of the constant of integration in indefinite integrals. Hence, I added an extra condition that would make the would-be function more nontrivial and more appealing.

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Take, e.g., two (distinct, non-trivial) bump functions $F$ and $G$ s.t. $supp\: F\cap G\left(\mathbb{R}\right)=supp\: G\cap F\left(\mathbb{R}\right)=\emptyset$ . Then their derivatives $f=F^{\prime}$, and $g=G^{\prime}$ are clearly satisfying the required identity.

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@Ady, thanks. I found it hard to find sample functions such that $supp\: F\cap G\left(\mathbb{R}\right)=supp\: G\cap F\left(\mathbb{R}\right)=\emptyset$. Could you come up with explicit two functions? – Chulumba Feb 1 '11 at 15:07
Chulumba, it's not hard if you know how to construct bump functions. You can take $F(R) = G(R) = [0, 1]$, and say $supp(F) = [2, 3]$ and $supp(G) = [3, 4]$, as follows. Define f(x) = exp(-1/x)exp(1/(1-x)) for 1 > x > 0, then g(x) = f(x)/(1 + f(x)) for 1 > x > 0, then define a continuous function h(x) piecewise to be = 0 for 0 > x, = g(3x) for 1/3 > x > 0, = 1 for 2/3 > x > 1/3, g(3-3x) for 1 > x > 2/3, and 0 for x > 1. Finally, define F(x) = h(x-2), and G(x) = h(x-3). – Todd Trimble Feb 9 '11 at 17:00
@Todd, thanks. So,there are many types of counterexample functions. – Chulumba Feb 11 '11 at 20:22

I'm not sure I understand your quotation. It could equally be "A small child can find a rock so heavy that a hundred strong men could not lift it. In accordance with this principle, I shall go around pointing out heavy rocks."

Part of the skill in mathematics is knowing which problems from the millions available are likely to be solvable. You should probably try to give some explanation as to why you expect a solution - or at least, why you want one - so people have a reason to think about this particular question over any other.

(This is a comment, not an answer, but I don't have the power)

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Thanks for your comments. First, the quotation is not mine. It was the recently-late Arnold's way of joking when he told others about his conjectures. Second, the equation that I wrote out was not random. At least, the symmetry I find in it and the absence of an iota of clue at proceeding with any method makes me fall in love with finding a solution. Part of the motivation was to find a function that in some way resembles the exponential function. The exponential map is invariant under differentiation. (...continued..) – Chulumba Jan 29 '11 at 14:26
(...continued) So, the natural curiosity to find a nontrivial map invariant under integration. For obvious reasons, such map does not exist because of the presence of the constant of integration in indefinite integrals. Hence, I added an extra condition that would make the would-be function more nontrivial and more appealing. – Chulumba Jan 29 '11 at 14:34
Oh, ok! Well, the exponential map is also invariant under integration, and is also a solution here if you replace your lower limit of integration with minus infinity. – Lloyd Smith Jan 29 '11 at 21:22
That's an interesting observation. However, one requirement is that $f\neq g$ . So, the exponential map, even with the lower limit of integration replaced by minus infinity, fails one criterion. – Chulumba Jan 30 '11 at 6:13

For real analytic functions, just looking at the power series gives information. For example, it seems that if all the coefficients of the power series expansion are nonzero, then the only solution (class) is $f=g.$

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How, sir? I do not see how you determined equality to be the only solution class. – Chulumba Jan 30 '11 at 6:22
@Chulumba: This is pretty easy to prove by induction for the case when the constant terms of $f$ and/or $g$ are nonzero. I do not know how to do it for other cases. – S. Carnahan Jan 31 '11 at 3:03
@Scott: I suspect there are polynomial counterexamples for the "other cases", but have been too lazy to program this. – Igor Rivin Jan 31 '11 at 7:25
@Scott, thanks. I did not consider cases. So, that's one step ahead and makes the would-be solutions more interesting. @Igor, that's a good suggestion. I will try to find polynomial counterexamples. – Chulumba Feb 1 '11 at 15:15

Following Igor's comment(or his hunch), I started out with finding polynomial counterexamples of the type: $f(x)= ax^n$ and $g(x)= bx^m$ on $(0,\infty)$. It is easy to see that such polynomials satisfy the integral condition only if $m=n$, and $a=b$ if $n$ is even or $a=\pm b$ if $n$ is odd. Then a class of counterexamples on $(0,\infty)$ can be constructed by letting $a= -b$ with odd $n\geq3$ .

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Wow. Thanks. My question has been crashed. – Chulumba Feb 10 '11 at 2:13

The problem does not seem as far from the main stream as it might first appear. Pick any given function g and let G denote the integral of g. Let F denote the integral of f. Then the equation becomes F'(G(x))=g(F(x)), with initial condition F(0)=0. In other words, for any given g, the problem reduces to a functional differential equation.

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Thanks. The differential equation version of the problem was already discussed lightly on Math.SE. However, I doubt whether there is any simplification arising from this formulation. – Chulumba Jan 30 '11 at 6:24