Convergence of Convolutions (Approximate Convolution Identity)

jon.s.beardsley · #1 $Old$ June 24th, 2009, 11:19 AM

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?

jon.s.beardsley · #2 $Old$ June 24th, 2009, 12:03 PM

Let me make this simpler. Can we at least show that f_n*\phi_n is pointwise convergent to f?

Laurent · #3 $Old$ June 24th, 2009, 12:10 PM

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.

jon.s.beardsley · #4 $Old$ June 24th, 2009, 12:33 PM

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.

Laurent · #5 $Old$ June 24th, 2009, 12:41 PM

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.

jon.s.beardsley · #6 $Old$ June 24th, 2009, 12:42 PM

Dearly sorry about the trouble but The first inequality is: for every $x$ , [LaTeX Error: Image is too big (644x37, limit 600x220)]

jon.s.beardsley · #7 $Old$ June 24th, 2009, 12:46 PM

nevermind, i have the TeX in my e-mail. Thanks!

jon.s.beardsley · #8 $Old$ June 24th, 2009, 12:57 PM

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.

Laurent · #9 $Old$ June 24th, 2009, 01:14 PM

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?

jon.s.beardsley · #10 $Old$ June 24th, 2009, 01:20 PM

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.

jon.s.beardsley · #11 $Old$ June 24th, 2009, 02:10 PM

I cannot see how $(f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt$ is true.

Laurent · #12 $Old$ June 24th, 2009, 02:43 PM

Quote:

Originally Posted by jon.s.beardsley $View Post$

I cannot see how $(f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt$ 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)

jon.s.beardsley · #13 $Old$ June 26th, 2009, 06:22 AM

All is clear now! Thankyou!