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A finite subspace of L2

  1. Nov 10, 2009 #1
    Let S be a subspace of [tex]L^{2}(\left[0,1\right])[/tex] and suppose [tex]\left|f(x)\right|\leq K \left\| f \right\|[/tex] for all f in S.

    Show that the dimension of S is at most [tex]K^{2}[/tex]

    ---------

    The prof hinted us to use Bessel's inequality.

    Namely, let [tex]\left\{ u_1,\dots, u_m \right\}[/tex] be a set of orthonormal vectors in [tex]L^{2}(\left[0,1\right])[/tex]. Then [tex]\left\| f \right\|^2 \geq \sum_{k=1}^m \left| \left\langle f, u_k \right\rangle\right|^2[/tex]

    I just keep getting stuck, and getting things like

    [tex]\left\|f \right\|^2 =\int_0^1 \left|f(x)\right|^2 dx \leq \int_0^1 K^2 \left\|f \right\|^2 dx = K^2 \left\|f \right\|^2[/tex]

    and I can't figure out how to apply Bessel's inequality.

    I guess the goal is to assume m linear independent vectors, and show that m is less than K^2. Help, please?
     
  2. jcsd
  3. Nov 10, 2009 #2

    lanedance

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    haven't totally put it together yet, but I'm thinking start with assuming an orthonormal basis for S:{...,ui ...} of dimension m.

    Try expanding f in terms of the basis function & then knowing both the ui and f satsify the given inequality, hopefully you can find a contradiction if m>K2 as you say
     
    Last edited: Nov 10, 2009
  4. Nov 10, 2009 #3

    lanedance

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    considering a function which is a certain linear combination of an orthogonal basis for S should help. As S is a subspace, its closed under linear combinations, so the function is contained in S
     
    Last edited: Nov 11, 2009
  5. Nov 10, 2009 #4
    Still at a loss....

    So: Assume {u1, ..., um} is an orthonormal basis of S.
    I'm trying the function f= u1 + ... + um. Is this the right one to consider? It certainly is the nicest.

    Then ||f|| = sqrt(m).

    I still don't know where exactly to apply the Bessels inequality. If I do, all I keep getting is that
    m^2 < m^2 K
    which tells me nothing.

    More help? Thanks for the response.
     
    Last edited: Nov 11, 2009
  6. Nov 11, 2009 #5

    lanedance

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    must be missing something.. thought it would follow on, but when i went to work it completely, went in a circle...

    now i'm not even so sure i understand the question correctly, what about sinusoidal basis functions, couldn't you have an infinte orthogonal number of those with ||f|| =1, and maximum magnitude ~ sqrt(2)

    will pass it on to some of the other guys to have a look
     
    Last edited: Nov 12, 2009
  7. Nov 11, 2009 #6
    Ok, I'm started off with a different approach to see if I could understand this a bit better.... and it only leads me to believe that the problem must be wrong.
    ----

    Assume K=1 (since that is the smallest it can be). Then S is the set of functions on [0,1] such that |f(x)| < ||f|| for all x in [0,1]. For simplicity, let's just say that ||f|| =1.

    But the integral of |f(x)|^2 from 0 to 1 must be 1, and |f(x)| can never be less than 1. Thus, it must be the case that |f(x)|=1 for all x on the interval.

    So the modulus of f(x) is always 1, and thus [tex]f(x)=e^{ia(x)}[/tex] for some function a(x).

    Then the inner product of f with some other function g is
    [tex]\left\langle f,g\right\rangle=\int_0^1 e^{i(a(x)-b(x))}dx[/tex].

    But this integral does not necessarily have to have a modulus of 1 (indeed consider a(x)=2x and b(x)=x), and thus g is not a multiple of f. This leads me to believe that there must be more than one function in the basis for S, a contradiction to the dimension of S being less than or equal to K=1.
     
    Last edited: Nov 12, 2009
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