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From: TCL on 9 Jan 2010 18:28 Let S be a subset of {1,x,x^2,....}. Suppose that the linear span of S is dense in L^2(0,1). Does it follow that x^k is in S for every k>=1? -TCL
From: W^3 on 9 Jan 2010 22:06 In article <aa7cfd57-4fcc-4657-8a82-337e147c9820(a)l2g2000vbg.googlegroups.com>, TCL <tlim1(a)cox.net> wrote: > Let S be a subset of {1,x,x^2,....}. Suppose that the linear span of S > is dense in > L^2(0,1). Does it follow that x^k is in S for every k>=1? > -TCL No, let f be in C = C([0,1]). Then f(sqrt(x)) is in C, so there is a sequence of polynomials p_n(x) -> f(sqrt(x)) uniformly on [0,1], which implies p_n(x^2) -> f(x) uniformly on [0,1], hence in L^2. So the linear span of {1, x^2, x^4, ...} is dense in L^2. Much more can be said. The Muntz-Szasz theorem says that if 0 < p_1 < p_2 < ..., then the linear span of {1, x^(p_1), x^(p_2), ...} is dense in C (hence in L^2) iff sum 1/(p_n) = oo. A question I don't know the answer to: What is the Muntz-Szasz theorem for L^2? It is clearly a different criterion, because the linear span of {x, x^2, x^3, ...} is not dense in C but is dense in L^2.
From: David C. Ullrich on 10 Jan 2010 08:24 On Sat, 09 Jan 2010 19:06:10 -0800, W^3 <aderamey.addw(a)comcast.net> wrote: >In article ><aa7cfd57-4fcc-4657-8a82-337e147c9820(a)l2g2000vbg.googlegroups.com>, > TCL <tlim1(a)cox.net> wrote: > >> Let S be a subset of {1,x,x^2,....}. Suppose that the linear span of S >> is dense in >> L^2(0,1). Does it follow that x^k is in S for every k>=1? >> -TCL > >No, let f be in C = C([0,1]). Then f(sqrt(x)) is in C, so there is a >sequence of polynomials p_n(x) -> f(sqrt(x)) uniformly on [0,1], which >implies p_n(x^2) -> f(x) uniformly on [0,1], hence in L^2. So the >linear span of {1, x^2, x^4, ...} is dense in L^2. > >Much more can be said. The Muntz-Szasz theorem says that if 0 < p_1 < >p_2 < ..., then the linear span of {1, x^(p_1), x^(p_2), ...} is dense >in C (hence in L^2) iff sum 1/(p_n) = oo. > >A question I don't know the answer to: What is the Muntz-Szasz theorem >for L^2? It is clearly a different criterion, because the linear span >of {x, x^2, x^3, ...} is not dense in C but is dense in L^2. Thinking about it for a few minutes, my conjecture is that the answer is "the same, except it doesn 't matter whether or not 1 is included", which is to say the difference you mention is the only difference. There's one bit I haven't worked out in detail, and of course the details that I _have_ worked out I may have got wrong - you may enjoy trying to firm up what's below, if not I'll try to. First, let's say PW is the space of all holomorphic functions in the upper half-plane with bounded L^2 norm on horizontal lines. So PW is the "Paley-Wiener space", equal to the space of Fourier transforms of functions in L^2((0, infinity)). PW is often called H^(upper half-plane), but not today: Say H^2 of the upper half-plane is the image of the Hardy space H^2(unit disk) under the Cayley transform. The point being that we know exactly what the zero sets of H^2 functions are, and we need to know what the zero sets of PW functions are. It seems to me (also I seem to recall) that it's not hard to show that there exists a holomorphic function w(z) in the upper half-plane such that w has no zero and f(z) is in PW if and only if w(z) f(z) is in H^2. (w(z) is 1/sqrt(z+i) or something like that...) If that's so then (*) the zero sets of PW and H^2 are the same. Assuming that: If f is in L^2(0,1) let g(s) = exp(-s/2) f(exp(-s)) (s > 0). Then int_0^infinity |g|^2 = int_0^1 |f|^2 and int_0^1 t^k f(t) dt = int_0^infinity exp(-s(k+1/2)) g(s) ds = G(i(k+1/2)), where G is the Fourier transform of g. So there exists a non-trivial f in L^2(0,1) with int_0^1 t^k f(t) dt = 0 for all k in T if and only if there exists a non-trivial G in PW with G(i(k+1/2)) = 0 for all k in T.( Oops. Here T is the set of exponents in question: S = {t^k : k in T}.) Assuming (*) is correct, this holds if and only if sum_{k in T} 1/(k+1/2) < infinity, which is the same as sum_{k in T, k <> 0} 1/k < infinity. Surely (*) is either confirmed or denied in Garnett "Bounded Analytic Functions". I wouldn't be surprised if the answer to your question is somewhere in a very old monograph "Fourier Transforms in the Complex Domain" by Paley and Wiener. Alas it's much too cold to go to the office right now.
From: David C. Ullrich on 12 Jan 2010 11:48 In article <s4kjk5d73obhqjveb60c0u4mc47t8s7pgs(a)4ax.com>, David C. Ullrich <ullrich(a)math.okstate.edu> wrote: > On Sat, 09 Jan 2010 19:06:10 -0800, W^3 <aderamey.addw(a)comcast.net> > wrote: > > >In article > ><aa7cfd57-4fcc-4657-8a82-337e147c9820(a)l2g2000vbg.googlegroups.com>, > > TCL <tlim1(a)cox.net> wrote: > > > >> Let S be a subset of {1,x,x^2,....}. Suppose that the linear span of S > >> is dense in > >> L^2(0,1). Does it follow that x^k is in S for every k>=1? > >> -TCL > > > >No, let f be in C = C([0,1]). Then f(sqrt(x)) is in C, so there is a > >sequence of polynomials p_n(x) -> f(sqrt(x)) uniformly on [0,1], which > >implies p_n(x^2) -> f(x) uniformly on [0,1], hence in L^2. So the > >linear span of {1, x^2, x^4, ...} is dense in L^2. > > > >Much more can be said. The Muntz-Szasz theorem says that if 0 < p_1 < > >p_2 < ..., then the linear span of {1, x^(p_1), x^(p_2), ...} is dense > >in C (hence in L^2) iff sum 1/(p_n) = oo. > > > >A question I don't know the answer to: What is the Muntz-Szasz theorem > >for L^2? It is clearly a different criterion, because the linear span > >of {x, x^2, x^3, ...} is not dense in C but is dense in L^2. > > Thinking about it for a few minutes, my conjecture is that the answer > is "the same, except it doesn 't matter whether or not 1 is included", > which is to say the difference you mention is the only difference. Sure enough, this is Szasz's Theorem, or Theorem XIV in Paley&Weiner. (That theorem is actually more general; if the p_j are as above, and if, as I've been tacitly assuming, the p_j tend to infinity, then the criterion is exactly sum 1/(1 + p_j) = infinity.) > There's one bit I haven't worked out in detail, and of course the > details that I _have_ worked out I may have got wrong - you > may enjoy trying to firm up what's below, if not I'll try to. > > First, let's say PW is the space of all holomorphic functions in the > upper half-plane with bounded L^2 norm on horizontal lines. > So PW is the "Paley-Wiener space", equal to the space of > Fourier transforms of functions in L^2((0, infinity)). > PW is often called H^(upper half-plane), but not today: > Say H^2 of the upper half-plane is the image of the Hardy > space H^2(unit disk) under the Cayley transform. > > The point being that we know exactly what the zero sets of H^2 > functions are, and we need to know what the zero sets of PW > functions are. It seems to me (also I seem to recall) that it's > not hard to show that there exists a holomorphic function > w(z) in the upper half-plane such that w has no zero and > f(z) is in PW if and only if w(z) f(z) is in H^2. (w(z) is > 1/sqrt(z+i) or something like that...) > > If that's so then > > (*) the zero sets of PW and H^2 are the same. > > Assuming that: > > If f is in L^2(0,1) let > > g(s) = exp(-s/2) f(exp(-s)) (s > 0). > > Then > > int_0^infinity |g|^2 = int_0^1 |f|^2 > > and > > int_0^1 t^k f(t) dt = int_0^infinity exp(-s(k+1/2)) g(s) ds > > = G(i(k+1/2)), > > where G is the Fourier transform of g. So there exists a non-trivial > f in L^2(0,1) with int_0^1 t^k f(t) dt = 0 for all k in T if and only > if there exists a non-trivial G in PW with G(i(k+1/2)) = 0 for > all k in T.( Oops. Here T is the set of exponents in question: > S = {t^k : k in T}.) > > Assuming (*) is correct, this holds if and only if > > sum_{k in T} 1/(k+1/2) < infinity, > > which is the same as > > sum_{k in T, k <> 0} 1/k < infinity. > > Surely (*) is either confirmed or denied in Garnett > "Bounded Analytic Functions". I wouldn't be surprised if the > answer to your question is somewhere in a very old monograph > "Fourier Transforms in the Complex Domain" by Paley and > Wiener. Alas it's much too cold to go to the office right now. -- David C. Ullrich
From: W^3 on 17 Jan 2010 21:11
In article <dullrich-E82600.10483912012010(a)text.giganews.com>, "David C. Ullrich" <dullrich(a)sprynet.com> wrote: > In article <s4kjk5d73obhqjveb60c0u4mc47t8s7pgs(a)4ax.com>, > David C. Ullrich <ullrich(a)math.okstate.edu> wrote: > > > On Sat, 09 Jan 2010 19:06:10 -0800, W^3 <aderamey.addw(a)comcast.net> > > wrote: > > > > >In article > > ><aa7cfd57-4fcc-4657-8a82-337e147c9820(a)l2g2000vbg.googlegroups.com>, > > > TCL <tlim1(a)cox.net> wrote: > > > > > >> Let S be a subset of {1,x,x^2,....}. Suppose that the linear span of S > > >> is dense in > > >> L^2(0,1). Does it follow that x^k is in S for every k>=1? > > >> -TCL > > > > > >No, let f be in C = C([0,1]). Then f(sqrt(x)) is in C, so there is a > > >sequence of polynomials p_n(x) -> f(sqrt(x)) uniformly on [0,1], which > > >implies p_n(x^2) -> f(x) uniformly on [0,1], hence in L^2. So the > > >linear span of {1, x^2, x^4, ...} is dense in L^2. > > > > > >Much more can be said. The Muntz-Szasz theorem says that if 0 < p_1 < > > >p_2 < ..., then the linear span of {1, x^(p_1), x^(p_2), ...} is dense > > >in C (hence in L^2) iff sum 1/(p_n) = oo. > > > > > >A question I don't know the answer to: What is the Muntz-Szasz theorem > > >for L^2? It is clearly a different criterion, because the linear span > > >of {x, x^2, x^3, ...} is not dense in C but is dense in L^2. > > > > Thinking about it for a few minutes, my conjecture is that the answer > > is "the same, except it doesn 't matter whether or not 1 is included", > > which is to say the difference you mention is the only difference. > > Sure enough, this is Szasz's Theorem, or Theorem XIV in > Paley&Weiner. (That theorem is actually more general; if > the p_j are as above, and if, as I've been tacitly assuming, > the p_j tend to infinity, then the criterion is exactly > sum 1/(1 + p_j) = infinity.) > > > There's one bit I haven't worked out in detail, and of course the > > details that I _have_ worked out I may have got wrong - you > > may enjoy trying to firm up what's below, if not I'll try to. > > > > First, let's say PW is the space of all holomorphic functions in the > > upper half-plane with bounded L^2 norm on horizontal lines. > > So PW is the "Paley-Wiener space", equal to the space of > > Fourier transforms of functions in L^2((0, infinity)). > > PW is often called H^(upper half-plane), but not today: > > Say H^2 of the upper half-plane is the image of the Hardy > > space H^2(unit disk) under the Cayley transform. > > > > The point being that we know exactly what the zero sets of H^2 > > functions are, and we need to know what the zero sets of PW > > functions are. It seems to me (also I seem to recall) that it's > > not hard to show that there exists a holomorphic function > > w(z) in the upper half-plane such that w has no zero and > > f(z) is in PW if and only if w(z) f(z) is in H^2. (w(z) is > > 1/sqrt(z+i) or something like that...) > > > > If that's so then > > > > (*) the zero sets of PW and H^2 are the same. > > > > Assuming that: > > > > If f is in L^2(0,1) let > > > > g(s) = exp(-s/2) f(exp(-s)) (s > 0). > > > > Then > > > > int_0^infinity |g|^2 = int_0^1 |f|^2 > > > > and > > > > int_0^1 t^k f(t) dt = int_0^infinity exp(-s(k+1/2)) g(s) ds > > > > = G(i(k+1/2)), > > > > where G is the Fourier transform of g. So there exists a non-trivial > > f in L^2(0,1) with int_0^1 t^k f(t) dt = 0 for all k in T if and only > > if there exists a non-trivial G in PW with G(i(k+1/2)) = 0 for > > all k in T.( Oops. Here T is the set of exponents in question: > > S = {t^k : k in T}.) > > > > Assuming (*) is correct, this holds if and only if > > > > sum_{k in T} 1/(k+1/2) < infinity, > > > > which is the same as > > > > sum_{k in T, k <> 0} 1/k < infinity. > > > > Surely (*) is either confirmed or denied in Garnett > > "Bounded Analytic Functions". I wouldn't be surprised if the > > answer to your question is somewhere in a very old monograph > > "Fourier Transforms in the Complex Domain" by Paley and > > Wiener. Alas it's much too cold to go to the office right now. Thanks for that David. |