Are there functions satisfying the following integral condition? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:30:15Z http://mathoverflow.net/feeds/question/53697 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53697/are-there-functions-satisfying-the-following-integral-condition Are there functions satisfying the following integral condition? Chulumba 2011-01-29T07:56:08Z 2011-02-09T16:32:03Z <p>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?</p> <p>$$f( \left(\int_{0}^{t} g(x) \ \text{d}x\right)) = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right))$$</p> <p><strong>P.S.:</strong> I migrated this question from <a href="http://math.stackexchange.com/questions/18005/are-there-functions-satisfying-the-following-integral-condition" rel="nofollow">here</a> 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</p> <p><strong>Motivation:</strong> 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.</p> http://mathoverflow.net/questions/53697/are-there-functions-satisfying-the-following-integral-condition/53702#53702 Answer by Lloyd Smith for Are there functions satisfying the following integral condition? Lloyd Smith 2011-01-29T09:14:46Z 2011-01-29T09:14:46Z <p>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." </p> <p>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.</p> <p>(This is a comment, not an answer, but I don't have the power)</p> http://mathoverflow.net/questions/53697/are-there-functions-satisfying-the-following-integral-condition/53723#53723 Answer by Michael Renardy for Are there functions satisfying the following integral condition? Michael Renardy 2011-01-29T16:03:31Z 2011-01-29T16:03:31Z <p>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. </p> http://mathoverflow.net/questions/53697/are-there-functions-satisfying-the-following-integral-condition/53759#53759 Answer by Igor Rivin for Are there functions satisfying the following integral condition? Igor Rivin 2011-01-30T04:41:55Z 2011-01-30T04:41:55Z <p>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 $f=g.$</p> http://mathoverflow.net/questions/53697/are-there-functions-satisfying-the-following-integral-condition/53775#53775 Answer by Ady for Are there functions satisfying the following integral condition? Ady 2011-01-30T11:20:39Z 2011-01-30T11:20:39Z <p>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.</p> http://mathoverflow.net/questions/53697/are-there-functions-satisfying-the-following-integral-condition/54900#54900 Answer by To be cont'd for Are there functions satisfying the following integral condition? To be cont'd 2011-02-09T16:32:03Z 2011-02-09T16:32:03Z <p>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 $n\geq3$ .</p>