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Math Help - Dense subset

  1. #1
    Senior Member slevvio's Avatar
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    Dense subset

    Let X be the space of continuous functions f:[0,\pi] \rightarrow \mathbb{C} with norm || f|| = \int_0^\pi |f(x) |dx. I am trying to determine whether the linearly independent set \text{span}\{\sin nx\}_{n \in \mathbb{N}} is dense in X. Somebody gave me a hint that it is not dense and to look at the constant function 1. So that means I am trying to show that there is some open ball around the function 1, which contains no element of \text{span}\{\sin nx\}_{n \in \mathbb{N}}.

    This means that I need to show \int_0^\pi |1 - \sum_{i=0}^r \alpha_i \sin (n_i x) | dx \ge \epsilon for some \epsilon > 0.

    Has anyone got any hints regarding how to go about this? Thanks very much
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  2. #2
    Super Member Rebesques's Avatar
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    How about expressing cosx in terms of span{sinkx}?
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  3. #3
    Senior Member slevvio's Avatar
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    That is not possible since the set  \{f \mid f=\sum_{i=0}^r \alpha _i f_i, \text{ where }f_i = \sin n_i (\cdot) \text{ or } \cos n_i(\cdot) , r >0, \alpha_i \in \mathbb{C},  n_i\in\mathbb{N}\} is linearly independent.

    This means of course that |\cos x - \sum_{i=0}^r \alpha_i \sin n_i (x) |>0. But its not necessarily bigger than a constant \epsilon which I choose at the start
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  4. #4
    Super Member Rebesques's Avatar
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    Hmmm... What if x is near 0?
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  5. #5
    Senior Member slevvio's Avatar
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    at 0, the difference is 1, but how does the affect the integral between 0 and pi ? similarly at pi the difference is 1.
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  6. #6
    Super Member Rebesques's Avatar
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    Well... If you suppose that f=cosx lies in span{sinkx}, then, for every a>0 there is a finite sum S(a), such that
    ||f-S(a)||<a. But since f(0)=1 and all finite sums are zero at x=0, in a sufficiently small right interval about 0, we can have f-S(a)>a/pi.
    Integrating gives a>a.
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  7. #7
    Senior Member slevvio's Avatar
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    But I know in advance that cosx is not in that span, from linear independence.

    I am trying to prove the statement " \exists f \in C([0,\pi],\mathbb{C}) \text{ such that } \exists \epsilon > 0 \text{ such that }\forall g\in \text{span}\{\sin nx\}_{n \ge 0} , \int_0^{\pi} |f(x)-g(x)|\ge \epsilon "
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  8. #8
    Super Member Rebesques's Avatar
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    If you already know that cosx is not in the span, then the span is not dense... right? Why bother proving more? :S
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  9. #9
    Senior Member slevvio's Avatar
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    But it might be in the closure of the span? Is this is a standard result you are using?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    But it might be in the closure of the span? Is this is a standard result you are using?
    I have been trying to think of a clever continuous linear functional which is constant on \text{span}\{\sin(nx)\} but not constant on \mathcal{C}\left([0,1],\mathbb{C}\right). This of course would contradict density. Any ideas?
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  11. #11
    MHF Contributor
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    Quote Originally Posted by slevvio View Post
    Let X be the space of continuous functions f:[0,\pi] \rightarrow \mathbb{C} with norm \|f\| = \int_0^\pi |f(x) |dx. I am trying to determine whether the linearly independent set \text{span}\{\sin nx\}_{n \in \mathbb{N}} is dense in X. Somebody gave me a hint that it is not dense and to look at the constant function 1. So that means I am trying to show that there is some open ball around the function 1, which contains no element of \text{span}\{\sin nx\}_{n \in \mathbb{N}}.

    This means that I need to show \int_0^\pi |1 - \sum_{i=0}^r \alpha_i \sin (n_i x) | dx \ge \epsilon for some \epsilon > 0.

    Has anyone got any hints regarding how to go about this? Thanks very much
    The ingredients for a solution are already here in the comments of Rebesques and Drexel28. Define a continuous linear functional  \phi on X by \phi(f) = \int_0^\pi f(x)\cos x\,dx. Then \phi(\sin nx) = 0 for all n, but \phi(\cos x)\ne0. So the function \cos x is not in the kernel of  \phi, which is a closed subspace of X.

    [I have moved this thread, because it's really a problem in functional analysis rather than linear algebra. The solution depends on the fact that L^\infty[0,\pi] is the dual of L^1[0,\pi].]
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  12. #12
    Senior Member slevvio's Avatar
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    but \phi(\sin nx ) is not zero when n is even

    the map \theta(f) = \int_0^{\pi} f(x) sin(x) dx seems to work for  cos(2x). . Only sinx is not in the kernel here, but that could never be a sequence tending to cos(2x) anyway.
    Last edited by slevvio; January 10th 2011 at 10:10 AM.
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  13. #13
    Senior Member slevvio's Avatar
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    yes that map works ! Thanks all I see how it works
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  14. #14
    MHF Contributor
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    Angry

    Quote Originally Posted by slevvio View Post
    but \phi(\sin nx ) is not zero when n is even
    Dammit, you're right! And what's more, I'm changing my mind about whether this result is true.

    I now think that every continuous function on [0,\pi] can be approximated in the L^1-norm by a linear combination of the functions \sin kx. In fact, let  f be a continuous function on [0,\pi], and extend  f to an odd function  g on [-\pi,\pi], (so that g(-x) = -f(x)). Then the Fourier series of  g consists only of sine terms. Let s_n(g) denote the sum of the first n terms of that series. Then \|s_n(g) - g\|_2\to0 as n\to\infty, where \|\,.\,\|_2 is the L^2-norm, given by \|h\|_2 = \sqrt{\int_{-\pi}^\pi|h(x)|^2dx}. By the Cauchy–Schwarz inequality, if \|h\|_1 is the L^1-norm of h, given by \|h\|_1 = \int_{-\pi}^\pi|h(x)|\,dx, then

    \|h\|_1 = \int_{-\pi}^\pi|h(x)|\cdot1\,dx\leqslant\|h\|_2\|1\|_2 = \sqrt{2\pi}\|h\|_2.

    It follows that \|s_n(g)-g\|_1\leqslant\sqrt{2\pi}\|s_n(g)-g\|_2\to0, and if you restrict the interval of integration to [0,\pi], this tells you that \int_0^\pi|s_n(f)-f|\,dx\to0. In other words,  f can be approximated in the L^1[0,\pi]-norm by a linear combination of functions \sin kx.
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  15. #15
    Super Member Rebesques's Avatar
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    I'm not sure about this last argument Op, how can we restrict the interval to [0,pi]?


    Now, about Slev's question.
    How about using a little bit of Arzela-Ascoli to arrive at a contradiction?
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