Results 1 to 4 of 4

Math Help - Proof involving an integral of a product of 2 functions

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Proof involving an integral of a product of 2 functions

    Let n\in \mathbb{N}, consider \Psi _n (x)=\frac{\sin (nx)}{\pi x} for x\neq 0. Let \Psi _n (0)=\frac{n}{\pi} such that \Psi _n is continuous.
    Demonstrate that \lim _{n\to \infty} \int _{\mathbb{R}} \Psi _n (x) f(x)dx=f(0) for any functions f that are infinitely many times differentiable and with compact support.
    ------------------------------
    Believe it or not, this exercise is a problem I've in my electromagnetism course. We have not been introduced what a compact support is so I checked out in wikipedia and it states
    Quote Originally Posted by wikipedia
    the support of a function is the set of points where the function is not zero, or the closure of that set.
    which seems 2 different things to me.

    Do you have any tip, hint or idea about how to do the proof? I'm willing to put a lot of efforts to do it. I just don't know how to tackle it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by arbolis View Post
    Let n\in \mathbb{N}, consider \Psi _n (x)=\frac{\sin (nx)}{\pi x} for x\neq 0. Let \Psi _n (0)=\frac{n}{\pi} such that \Psi _n is continuous.
    Demonstrate that \lim _{n\to \infty} \int _{\mathbb{R}} \Psi _n (x) f(x)dx=f(0) for any functions f that are infinitely many times differentiable and with compact support.
    First, you can notice that the change of variable y=nx, together with letting n tend to infinity inside the integral gives a "formal" proof, provided you know \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}. Maybe this is enough for your electromagnetic course.

    If this were maths, you'd have to justify taking the limit inside the integral. One difficulty is that the limit is an improper integral, hence usual Lebesgue's theorem don't apply directly. Here's a suggestion. On [a,b] containing the support of f, write f(x)=f(0)+x\varphi(x) where \varphi is \mathcal{C}^\infty as well (but not necessarily compactly supported). Use linearity to split the integral; for the first part, change of variable, use the above classic integral, ok. For the second part, you can note this is a case of Riemann-Lebesgue lemma. It can be proved for step functions \varphi first and generalized by density, for instance. Or using an integration by part (derivating phi), if I remember well.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Laurent View Post
    First, you can notice that the change of variable y=nx, together with letting n tend to infinity inside the integral gives a "formal" proof, provided you know \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}. Maybe this is enough for your electromagnetic course.

    If this were maths, you'd have to justify taking the limit inside the integral. One difficulty is that the limit is an improper integral, hence usual Lebesgue's theorem don't apply directly. Here's a suggestion. On [a,b] containing the support of f, write f(x)=f(0)+x\varphi(x) where \varphi is \mathcal{C}^\infty as well (but not necessarily compactly supported). Use linearity to split the integral; for the first part, change of variable, use the above classic integral, ok. For the second part, you can note this is a case of Riemann-Lebesgue lemma. It can be proved for step functions \varphi first and generalized by density, for instance. Or using an integration by part (derivating phi), if I remember well.
    I hope the informal way is enough but I'm not really sure, my professor doesn't consider himself a physicist though he is one but I agree he looks like a mathematical physicist or so.
    So putting y=nx, \lim _{n\to \infty} converts into \lim _{y\to  \infty}.
    Putting the limit inside the integral, we obtain \int _{\mathbb{R}} \lim _{y \to \infty} \frac{\sin (y) f(x)}{\pi x} dx.
    Should I replace x by \frac{y}{n} to reach something involving \frac{\sin y}{y}?
    Also the integral is over \mathbb{R}, so \int _{\mathbb{R}} \frac{\sin x}{x} dx =\pi. Why did you took only the positive real numbers instead of the whole set?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by arbolis View Post
    I hope the informal way is enough but I'm not really sure, my professor doesn't consider himself a physicist though he is one but I agree he looks like a mathematical physicist or so.
    So putting y=nx, \lim _{n\to \infty} converts into \lim _{y\to  \infty}.
    Putting the limit inside the integral, we obtain \int _{\mathbb{R}} \lim _{y \to \infty} \frac{\sin (y) f(x)}{\pi x} dx.
    You shouldn't mix variables inside the integral. Putting y=nx, we have \int \frac{\sin(nx)}{\pi x}f(x)dx= \int \frac{\sin y}{\pi y}f\left(\frac{y}{n}\right)dy, and we still want n\to\infty (by the way, "y" is just a dummy variable). By continuity at 0, the "f" term converges to f(0).

    Also the integral is over \mathbb{R}, so \int _{\mathbb{R}} \frac{\sin x}{x} dx =\pi. Why did you took only the positive real numbers instead of the whole set?
    It was just because this is the way I remember the formula, but sure it is on the whole line here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof involving functions.
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: September 8th 2011, 03:11 PM
  2. Annoying Integral involving Bessel Functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 20th 2011, 01:42 PM
  3. [SOLVED] Proof involving closures of functions in metric spaces
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 2nd 2010, 06:23 AM
  4. Replies: 1
    Last Post: September 10th 2010, 10:35 PM
  5. Proof involving functions
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: October 27th 2007, 01:04 PM

Search Tags


/mathhelpforum @mathhelpforum