Results 1 to 13 of 13

Math Help - Convergence of Convolutions (Approximate Convolution Identity)

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    9

    Question Convergence of Convolutions (Approximate Convolution Identity)

    I'm trying to show that if a sequence of functions, (f_n) is uniformly convergent to a function f, and \phi_n (do people use LaTeX notation here?) is an approximation to the dirac delta (an approximate convolution identity), then f_n*\phi_n, i.e. the sequence of convolutions, is also convergent uniformly to f (for my purposes we can restrain ourselves to showing this on a compact set luckily!). This intuitively makes sense, but I'm having a doggone hard time showing it rigorously. Anyone have any ideas? Is this the right place for this post?
    Last edited by jon.s.beardsley; June 24th 2009 at 11:26 AM. Reason: noticed you could make things urgent!, plus new information
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jun 2009
    Posts
    9
    Let me make this simpler. Can we at least show that f_n*\phi_n is pointwise convergent to f?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by jon.s.beardsley View Post
    I'm trying to show that if a sequence of functions, (f_n) is uniformly convergent to a function f, and \phi_n (do people use LaTeX notation here?) is an approximation to the dirac delta (an approximate convolution identity), then f_n*\phi_n, i.e. the sequence of convolutions, is also convergent uniformly to f (for my purposes we can restrain ourselves to showing this on a compact set luckily!). This intuitively makes sense, but I'm having a doggone hard time showing it rigorously. Anyone have any ideas? Is this the right place for this post?
    Have you tried to write f_n\ast\phi_n-f=(f_n-f)\ast \phi_n + (f\ast\phi_n-f)? Then \|(f_n-f)\ast \phi_n\|_\infty \leq \|f_n-f\|_\infty \int\phi_n(t)dt=\|f_n-f\|_\infty tends to 0 (by uniform convergence of (f_n)_n to f), and \|f\ast \phi_n - f\|_\infty tends to 0 as well (usual result about convolution with an approximation of delta).

    nb: click on the formulas to see what I typed to get them.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2009
    Posts
    9
    So I'm assuming we can use \|_\infty because on a compact set the function is necessarily bounded? For f the only condition I have is that it is continuous (I could use locally integrable instead). Also, is there any way you might clarify the algebra in creating that first inequality you have? I understand rewriting the convolution (very clever!) but I have trouble seeing how we then immediately go to the inequality with the norms in it. Thanks so much.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by jon.s.beardsley View Post
    So I'm assuming we can use \|_\infty because on a compact set the function is necessarily bounded? For f the only condition I have is that it is continuous (I could use locally integrable instead). Also, is there any way you might clarify the algebra in creating that first inequality you have? I understand rewriting the convolution (very clever!) but I have trouble seeing how we then immediately go to the inequality with the norms in it. Thanks so much.
    The first inequality is: for every x, |((f_n-f)\ast \phi_n(x)|\leq \int |(f_n-f)(x-t)| |\phi_n(t)| dt \leq \int \|f_n-f\|_\infty |\phi_n(t)| dt = \|f_n-f\|_\infty \int\phi_n(t)dt (I bound |(f_n-f)(x-t)| by its maximum value, and |\phi_n(t)|=\phi_n(t) because \phi_n is positive (?)), hence we have an upper bound independent of x. This gives what I wrote.

    Since there is uniform convergence, by definition \|f_n-f\|_\infty is finite and converges to 0. The compacity is involved in the justification of the second term: why f\ast \phi_n converges uniformly to f.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2009
    Posts
    9
    Dearly sorry about the trouble but The first inequality is: for every x, [LaTeX Error: Image is too big (644x37, limit 600x220)]
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2009
    Posts
    9
    nevermind, i have the TeX in my e-mail. Thanks!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jun 2009
    Posts
    9
    How is it that \|f\ast\phi_n\|_\infty converges uniformly? As far as I understood, pointwise convergence on a compact set does not necessarily imply uniform convergence.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by jon.s.beardsley View Post
    How is it that \|f\ast\phi_n\|_\infty converges uniformly? As far as I understood, pointwise convergence on a compact set does not necessarily imply uniform convergence.
    I guess you mean f\ast\phi_n.

    Since this is the result in the case when f_n=f for all n, I supposed you already knew that f\ast\phi_n converges uniformly to f. (So that you were asked for a generalization of this fact)

    This is a usual property of approximations of delta, a quick one but not a trivial one. On a compact set, if f is continuous, \|f\ast\phi_n-f\|_\infty\to 0. The idea for the proof is to write (f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt, and to use the fact that \phi_n is 0 away from 0 (precisions depend on your own definition) so that the integral reduces to a segment around x of small width. Using the uniform continuity of f (that's were compacity is involved), we conclude. (Use an \varepsilon>0 for the proof) I guess you already met this proof in your lecture, didn't you?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jun 2009
    Posts
    9
    Thankyou for your help. I have not had any sort of lecture or class on this material. I am attempting to prove some properties of generalized functions for an undergraduate research thesis, and I'm having to figure much out my own (i.e. measure theory, distributions, a lot of functional analysis). I'm trying to show that a class of generalized functions form a sheaf (you know, those things Grothendieck liked) over locally compact spaces, and am starting with \mathbb{R}^n. Right now I'm working on a glueing property. Thanks again for all the help.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Jun 2009
    Posts
    9
    I cannot see how <br />
(f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt<br />
is true.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by jon.s.beardsley View Post
    I cannot see how <br />
(f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt<br />
is true.
    Because f\ast\phi_n(x)=\int f(t)\phi_n(x-t)dt( and =\int f(x-t)\phi_n(t)dt as well) and, since \int\phi_n(t)dt=1 (approximation of delta...), f(x)=f(x)\int \phi_n(t) dt=f(x)\int \phi_n(x-t) dt=\int f(x)\phi_n(x-t)dt (change of variable)
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Jun 2009
    Posts
    9
    All is clear now! Thankyou!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to evaluate these convolutions:
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 20th 2011, 01:34 PM
  2. solve by method of convolutions....
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: April 25th 2011, 10:16 PM
  3. Convolutions
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 21st 2011, 08:56 AM
  4. Inverse Fourier Transform using convolutions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 17th 2010, 03:18 PM
  5. Identity Element for Convolution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 22nd 2008, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum