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

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In summary, the author tries to find an elementary function that is continuous and 2 pi periodic and has dense Fourier series converging to it. However, the method fails because finding an elementary function that satisfies all of these conditions is not possible.
  • #1
quasar987
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Homework Statement


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?

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]
 
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  • #2
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:
 
  • #3
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.
 
  • #4
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.
 
  • #5
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
 
  • #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.
 
  • #7
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:
 
  • #8
My instinct is smoothing; convolve your function with a continuous function highly concentrated at 0.
 
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  • #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.
 

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

What does it mean for a function to be F integrable?

F integrability is a mathematical concept that refers to the ability of a function to be integrated over a given interval. A function is said to be F integrable if its integral exists and is finite over that interval.

How do you find a g that will make the integral as small as desired?

To find such a function g, we need to use a technique called the Riemann-Lebesgue lemma. This lemma allows us to construct a sequence of functions that approach the desired function g, and by choosing the right sequence, we can make the integral as small as we want.

Is there a specific method for finding the smallest possible integral for a given function?

Yes, there are several methods for finding the smallest possible integral for a given function. One method is to use the Fundamental Theorem of Calculus, which states that the integral of a function can be calculated by evaluating its antiderivative at the endpoints of the interval. Another method is to use integration techniques such as substitution or integration by parts.

Can any function be made F integrable?

No, not all functions can be made F integrable. There are certain conditions that a function must satisfy in order to be F integrable, such as being bounded and having a finite number of discontinuities.

How is F integrability related to the convergence of a sequence of functions?

F integrability is closely related to the convergence of a sequence of functions. If a sequence of functions converges to a function that is F integrable, then that function is also F integrable. Additionally, if a sequence of functions is uniformly convergent, then the limit function is F integrable.

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