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From: ArtflDodgr on 9 Jun 2010 14:37 In article <457408350.304655.1276088530123.JavaMail.root(a)gallium.mathforum.org>, Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > 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 According to Müntz's theorem, if 0 < n(1) < n(2) < ... is an increasing sequence of positive integers, then the linear span of {x^{n(k)}: k=1,2,...} is uniformly dense in C[0,1] if and only if the series sum[ n(k)^{-1} : k=1,2,... ] diverges. -- A.
From: Maury Barbato on 10 Jun 2010 07:14 ArtflDodgr wrote: > In article > <457408350.304655.1276088530123.JavaMail.root(a)gallium. > mathforum.org>, > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > > 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 > > According to Müntz's theorem, if 0 < n(1) < n(2) < > ... > is an increasing sequence of positive integers, then > the linear span > of {x^{n(k)}: k=1,2,...} is uniformly dense in > n C[0,1] if and only if > the series sum[ n(k)^{-1} : k=1,2,... ] diverges. > > -- > A. Ops, a very beatiful and elegant result!!! A little correction, we must have n(1)=0, that is, Muntz's Theorem is the following: Let if 0 = n(1) < n(2) < ... an increasing sequence of integers, then the linear span of {x^{n(k)}: k=1,2,...} is uniformly dense in C[0,1] if and only if the series sum[ n(k)^{-1} : k=1,2,... ] diverges. Thank you very very much for having acquainted me with this very remarkable result. Making some search on google, I found a lot of papers which generalize the original result of Muntz. The book "Polynomials and Polynomial Inequalities" by Borwein and Erdélyi is a great source about the subject. My Best Regards, Maury Barbato
From: W^3 on 11 Jun 2010 23:29 In article <457408350.304655.1276088530123.JavaMail.root(a)gallium.mathforum.org>, Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > 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])? Muntz-Szasz is a beautiful result, but we don't need it here. Let A = {c+x^kp(x) : p is a polynomial}. By Stone Weierstrass, A is uniformly dense in C([0,1]). So if f is in C([0,1]) with f(0) = 0, then there is a sequence c_j + x^kp_j(x) converging uniformly to f. Because f(0) = 0, c_j -> 0, hence x^kp_j(x) -> f uniformly. Thus {x^kp(x) : p is a polynomial} is dense in the continuous functions on [0,1] that vanish at 0, and these are dense in L^2.
From: W^3 on 12 Jun 2010 12:42 In article <aderamey.addw-C24D26.20292811062010(a)News.Individual.NET>, W^3 <aderamey.addw(a)comcast.net> wrote: > In article > <457408350.304655.1276088530123.JavaMail.root(a)gallium.mathforum.org>, > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > > 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])? > > Muntz-Szasz is a beautiful result, but we don't need it here. Let A = > {c+x^kp(x) : p is a polynomial}. By Stone Weierstrass, A is uniformly > dense in C([0,1]). So if f is in C([0,1]) with f(0) = 0, then there is > a sequence c_j + x^kp_j(x) converging uniformly to f. Because f(0) = > 0, c_j -> 0, hence x^kp_j(x) -> f uniformly. Thus {x^kp(x) : p is a > polynomial} is dense in the continuous functions on [0,1] that vanish > at 0, and these are dense in L^2. I meant of course A = {c+x^kp(x) : c in R, p is a polynomial}.
From: Maury Barbato on 14 Jun 2010 23:42
The World Wide Wade wrote: > In article > <aderamey.addw-C24D26.20292811062010(a)News.Individual.N > ET>, > W^3 <aderamey.addw(a)comcast.net> wrote: > > > In article > > > <457408350.304655.1276088530123.JavaMail.root(a)gallium. > mathforum.org>, > > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > > > > 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])? > > > > Muntz-Szasz is a beautiful result, but we don't > need it here. Let A = > > {c+x^kp(x) : p is a polynomial}. By Stone > Weierstrass, A is uniformly > > dense in C([0,1]). So if f is in C([0,1]) with f(0) > = 0, then there is > > a sequence c_j + x^kp_j(x) converging uniformly to > f. Because f(0) = > > 0, c_j -> 0, hence x^kp_j(x) -> f uniformly. Thus > {x^kp(x) : p is a > > polynomial} is dense in the continuous functions on > [0,1] that vanish > > at 0, and these are dense in L^2. > > I meant of course A = {c+x^kp(x) : c in R, p is a > polynomial}. I've read only now your very elegant and simple proof. Erdös would have said it comes from the Book! Thank you very much for your help! Best Regards, Maury Barbato |