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F integrable, find g that makes the integral as small as you wish

  1. Jan 6, 2008 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I'm reading a proof and they make the following statement as if obvious: "Let f be a 2 pi-periodic function Lebesgue integrable on [-pi,pi]. Then given e>0, there is a 2 pi-periodic function g:R-->R continuous on [-pi,pi] such that

    [tex]\frac{1}{2\pi}\int_{-\pi}^{\pi}|f-g|<\epsilon[/tex]

    (In other words, [tex]C_{2\pi}[/tex] is dense in [tex]L^1_{2\pi}[/tex].)

    Is the existence of such a g obvious?

    3. The attempt at a solution

    Let [tex]L^1_{\lambda}[/tex] denote the space of equivalence classes of integrable functions that are equal almost everywhere (that is to say [f]=[g] if f=g a.e.). I know that the space [tex]E^1_{\lambda}[/tex] of elementary function (i.e., function of the kind [tex]\phi=\sum_1^n a_k\mathbb{I}_{E_k}[/tex]) are dense in [tex]L^1_{\lambda}[/tex]. This means that I can find an elementary function [tex]\phi[/tex] such that

    [tex]\frac{1}{2\pi}||f-\phi||_1=\frac{1}{2\pi}\int_{-\pi}^{\pi}|f-\phi|<\epsilon/2[/tex].

    Moreover, elementary functions are of bounded variation, which guaranties that their Fourier series converges to them almost everywhere (Dirichlet). But that convergence need not be uniform, so I cannot find an N such that letting g denote the N-th partial sum of the Fourier series of [tex]\phi[/tex], we have, for all x, [tex]|\phi(x) - g(x)|<2\pi\epsilon/2[/tex], and hence

    [tex]\frac{1}{2\pi}||\phi-g||_1=\frac{1}{2\pi}\int_{-\pi}^{\pi}|\phi-g|<\epsilon/2[/tex]
     
    Last edited: Jan 6, 2008
  2. jcsd
  3. Jan 6, 2008 #2

    CompuChip

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    You could consider the set
    [tex]N = \{ x \in [0, 2\pi) : f(x) \text{ not continuous} \} [/tex]
    and show that for f to be integrable, N should be a null set (*). Then you can define a new function g by something like
    [tex]g(x) = \begin{cases} f(x) &\text{ if } x \not\in N \\ \lim_{y \to x} f(y) & \text{ if } x \in N \end{cases}[/tex]
    and show that it is continuous (**). Then you have the theorem that says that if f and g differ on a null set, their integrals are equal.
    The question is if (*) and (**) are possible and true :smile:
     
  4. Jan 6, 2008 #3

    quasar987

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    Unfortunately, (*) is not true. For instance the Dirichlet function [tex]\mathbb{I}_{\mathbb{Q}\cap [0,1]}[/tex] is discontinuous everywhere (which forms a set of measure 1), yet is Lebesgue integrable.
     
  5. Jan 6, 2008 #4

    CompuChip

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    Hmm, that's true: it differs from the continuous function [itex]x \mapsto 0[/itex] on a null set which is what I tried to use, but I did it wrong.

    OK, another idea then: It's been a while since I did measure theory, but I thought that if f is Lebesgue integrable, it can be written as a limit of step functions in standard representation:
    [tex]f = \lim_{n \to \infty} f_n; \qquad f_n = \sum_{A \in \mathcal{A}_n} (y_A)_n A[/tex]
    where [tex]A_1 \cap A_2 = \{ \}[/itex] for every [itex]A_1 \neq A_2 \in \mathcal{A}_n, \forall n[/itex].
    So perhaps an approach would be to make this continuous (by connecting the horizontal 'platforms' in a continuous way, e.g. by a linear curve of with [itex]\delta_n[/itex] in such a way that the total error introduces is below [itex]\epsilon[/itex]. I'm not sure how these new, continuous, functions would behave under the limit [itex]n \to \infty[/itex] though; but I hope there is a way to make it work out even in quasar987's case.
     
  6. Jan 6, 2008 #5

    quasar987

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    If you read my attempt at a solution, this is in the spirit of what I tried, except I tried to make the step functions (elementary functions) continuous by approximating them with a trigonometric polynomial.

    In your case, the method fails because the number of platform could easily be uncountable. Try to visualize what your method would look like in the case of the Dirichlet function :grumpy:
     
  7. Jan 6, 2008 #6
    Take a look at Rudin's Real and Complex Analysis. He proves that Cc(X) is dense in Lp(X) using Lusin's theorem. Cc is the set of continuous complex functions with compact support.
     
  8. Jan 6, 2008 #7

    quasar987

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    I am aware of this result, but I don't see how to make it work for us here. Remember that the function g must be not only 2 pi periodic but also continuous! :eek:
     
  9. Jan 6, 2008 #8

    Hurkyl

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    My instinct is smoothing; convolve your function with a continuous function highly concentrated at 0.
     
    Last edited: Jan 6, 2008
  10. Jan 6, 2008 #9
    I thought Lusin's theorem basically gave you the continuous g that you need. And rudin's proof provides an explicit construction of g.
     
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