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From: Maury Barbato on 8 Jun 2010 05:16 Hello, let R be the field of real numbers and f:[0,1]->R a continuous function such that, for every positive integer n, we have int_{0 to 1} f(x)*(x^n) dx = 0. Is f(x)=0 for every x in [0,1]? Thank you very much for your attention. My Best Regards, Maury Barbato PS If we assume also that int_{0 to 1} f(x) dx = 0, then the answer is yes. In this case we have int_{0 to 1} f(x)*P(x) dx = 0 for every real polynomial P(x). For Weierstrass Theorem, there's a sequence of polynomials {P_n} which converges uniformly to f on [0,1]. So f*P_n converges uniformly to f^2, and we have int_{0 to 1} [f(x)]^2 = 0.
From: Robert Israel on 8 Jun 2010 14:58 On Tue, 08 Jun 2010 09:16:00 EDT, Maury Barbato wrote: > Hello, > let R be the field of real numbers and f:[0,1]->R > a continuous function such that, for every positive > integer n, we have > > int_{0 to 1} f(x)*(x^n) dx = 0. > > Is f(x)=0 for every x in [0,1]? > > Thank you very much for your attention. > My Best Regards, > Maury Barbato > > PS If we assume also that > > int_{0 to 1} f(x) dx = 0, > > then the answer is yes. In this case we have > > int_{0 to 1} f(x)*P(x) dx = 0 > > for every real polynomial P(x). For Weierstrass Theorem, > there's a sequence of polynomials {P_n} which converges > uniformly to f on [0,1]. So f*P_n converges uniformly to > f^2, and we have > > int_{0 to 1} [f(x)]^2 = 0. If int_0^1 f(x) dx = c, we have int_0^1 f(x) P(x) dx = c P(0) for every polynomial P. Now there are polynomials P(x) with P(0) = 1 and int_0^1 |P(x)| dx arbitrarily small. Since |int_0^1 f(x) P(x) dx| <= (max_{x in [0,1]} |f(x)|) int_0^1 |P(x)| dx we conclude that c = 0, bringing us back to your case. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: W^3 on 8 Jun 2010 15:06 In article <1743242637.298389.1276002990518.JavaMail.root(a)gallium.mathforum.org>, Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > Hello, > let R be the field of real numbers and f:[0,1]->R > a continuous function such that, for every positive > integer n, we have > > int_{0 to 1} f(x)*(x^n) dx = 0. > > Is f(x)=0 for every x in [0,1]? Yes. Note that int f*p = 0 for every polynomial p with p(0) = 0. But an easy argument shows the set of such polynomials is dense in L^2([0,1]). That is enough to give you f identically 0.
From: Maury Barbato on 9 Jun 2010 02:17 Robert Israel wrote: > On Tue, 08 Jun 2010 09:16:00 EDT, Maury Barbato > wrote: > > > Hello, > > let R be the field of real numbers and f:[0,1]->R > > a continuous function such that, for every positive > > integer n, we have > > > > int_{0 to 1} f(x)*(x^n) dx = 0. > > > > Is f(x)=0 for every x in [0,1]? > > > > Thank you very much for your attention. > > My Best Regards, > > Maury Barbato > > > > PS If we assume also that > > > > int_{0 to 1} f(x) dx = 0, > > > > then the answer is yes. In this case we have > > > > int_{0 to 1} f(x)*P(x) dx = 0 > > > > for every real polynomial P(x). For Weierstrass > Theorem, > > there's a sequence of polynomials {P_n} which > converges > > uniformly to f on [0,1]. So f*P_n converges > uniformly to > > f^2, and we have > > > > int_{0 to 1} [f(x)]^2 = 0. > > If int_0^1 f(x) dx = c, we have int_0^1 f(x) P(x) dx > = c P(0) for every > polynomial P. Now there are polynomials P(x) with > P(0) = 1 and > int_0^1 |P(x)| dx arbitrarily small. I write down the simple proof of this statement here for my own future memory. Given eps > 0, let n be a positive integer such that 2*n > 1/eps, and set g(x) = 1 - n*x (0<= x <= 1/n) g(x) = 0 (1/n <= x <= 1). Now, for Weierstrass Theorem, there's a sequence of polynomials {Q_n(x)} which converge uniformly to g on [0,1]. So Q_n(0) -> 1. So, if we set P_n(x) = Q_n(x) + 1 - Q_n(0), we obtain a sequence of polynomials {P_n(x)} such that P_n(0)=1 and P_n -> g uniformly to g on [0,1]. Let m be such that |P_m(x)-g(x)| < eps on [0,1]. We have eps > int_0^1 |g(x)-P_m(x)| dx >= >= int_0^1 (|P_m(x)|-|g(x)|) dx = int_0^1 |P_m(x)| dx + - 1/(2*n), and we conclude that int_0^1 |P_m(x)| dx < 2*eps. > Since > |int_0^1 f(x) P(x) dx| <= (max_{x in [0,1]} |f(x)|) > int_0^1 |P(x)| dx > we conclude that c = 0, bringing us back to your > case. > > -- > Robert Israel > israel(a)math.MyUniversitysInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada Thank you very much, prof. Israel! Friendly Regards, Maury Barbato
From: Maury Barbato on 9 Jun 2010 05:01
The World Wide Wade wrote: > In article > <1743242637.298389.1276002990518.JavaMail.root(a)gallium > .mathforum.org>, > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > > Hello, > > let R be the field of real numbers and f:[0,1]->R > > a continuous function such that, for every positive > > integer n, we have > > > > int_{0 to 1} f(x)*(x^n) dx = 0. > > > > Is f(x)=0 for every x in [0,1]? > > Yes. Note that int f*p = 0 for every polynomial p > with p(0) = 0. But > an easy argument shows the set of such polynomials is > dense in > L^2([0,1]). That is enough to give you f identically > 0. This is a proof of your statement. Set M = max_{x in [0,1]} |f(x)|, and g_n(x) = f(1/n) + f(1/n)*(n*x - 1) if 0 <= x <= 1/n, g_n(x) = f(x) if 1/n <= x <= 1. For every n, g_n is a continuous fucntion with g_n(0) = 0. So, using Weierstrass Theorem we can find a polynomial P_n such that P_n(0) = 0 and |g_n(x) - P_n(x)| < 1/n for every x in [0,1]. We have (int_0^1 [f(x) - P_n(x)]^2 dx)^(1/2) <= <= (int_0^1 [f(x) - g_n(x)]^2 dx)^(1/2) + + (int_0^1 [g_n(x) - P_n(x)]^2 dx )^(1/2) <= <= [(2*M)^2)]/n + 1/(n^2). Since the set C([0,1]) is dense in L^2([0,1]) (please, correct me if I'm wrong: I don't know almost anything about Lebesgue theory of Integration ...), we can conlude that the set of polynomials {P(x) : P(0) = 0} is dense in L^2([0,1]). Coming back to my original problem, I'm now realizing that a simple reasoning prove the following more strong result. Theorem. Let f:[0,1]->R a continuous function and k a non negative integer such that for every integer n >= k we have int_0^1 f(x)*(x^n) dx = 0. Then f(x) = 0 for every x in [0,1]. Proof. Set g(x) = f(x)*(x^k). Then int_0^1 g(x)*(x^n) dx = 0, for every non negative integer. So (see the PS to my original post), we have g(x) = 0 for every x in [0,1]. This implies f(x) = 0 in (0,1], and for the continuity of f, we have f(x) = 0 in [0,1]. QED Anyhow, a question arises now in my mind. Let k be a non negative integer and let a_n(P) denote the coefficient of x^n in the polynomial P(x). Then set S = {P(x) is a real polynomial, and a_0(P)=...=a_k(P)=0}. Is S dense in L^2([0,1])? Thank you very much for your help. My Best Regards, Maury Barbato |