# Convergence of Convolutions (Approximate Convolution Identity)

• Jun 24th 2009, 10:19 AM
jon.s.beardsley
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?
• Jun 24th 2009, 11:03 AM
jon.s.beardsley
Let me make this simpler. Can we at least show that f_n*\phi_n is pointwise convergent to f?
• Jun 24th 2009, 11:10 AM
Laurent
Quote:

Originally Posted by jon.s.beardsley
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 $\displaystyle f_n\ast\phi_n-f=(f_n-f)\ast \phi_n + (f\ast\phi_n-f)$? Then $\displaystyle \|(f_n-f)\ast \phi_n\|_\infty \leq \|f_n-f\|_\infty \int\phi_n(t)dt=\|f_n-f\|_\infty$ tends to 0 (by uniform convergence of $\displaystyle (f_n)_n$ to $\displaystyle f$), and $\displaystyle \|f\ast \phi_n - f\|_\infty$ 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.
• Jun 24th 2009, 11:33 AM
jon.s.beardsley
So I'm assuming we can use $\displaystyle \|_\infty$ because on a compact set the function is necessarily bounded? For $\displaystyle f$ 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.
• Jun 24th 2009, 11:41 AM
Laurent
Quote:

Originally Posted by jon.s.beardsley
So I'm assuming we can use $\displaystyle \|_\infty$ because on a compact set the function is necessarily bounded? For $\displaystyle f$ 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 $\displaystyle x$, $\displaystyle |((f_n-f)\ast \phi_n(x)|\leq \int |(f_n-f)(x-t)| |\phi_n(t)| dt$ $\displaystyle \leq \int \|f_n-f\|_\infty |\phi_n(t)| dt = \|f_n-f\|_\infty \int\phi_n(t)dt$ (I bound $\displaystyle |(f_n-f)(x-t)|$ by its maximum value, and $\displaystyle |\phi_n(t)|=\phi_n(t)$ because $\displaystyle \phi_n$ is positive (?)), hence we have an upper bound independent of $\displaystyle x$. This gives what I wrote.

Since there is uniform convergence, by definition $\displaystyle \|f_n-f\|_\infty$ is finite and converges to 0. The compacity is involved in the justification of the second term: why $\displaystyle f\ast \phi_n$ converges uniformly to $\displaystyle f$.
• Jun 24th 2009, 11:42 AM
jon.s.beardsley
Dearly sorry about the trouble but The first inequality is: for every $\displaystyle x$, [LaTeX Error: Image is too big (644x37, limit 600x220)]
• Jun 24th 2009, 11:46 AM
jon.s.beardsley
nevermind, i have the TeX in my e-mail. Thanks!
• Jun 24th 2009, 11:57 AM
jon.s.beardsley
How is it that $\displaystyle \|f\ast\phi_n\|_\infty$ converges uniformly? As far as I understood, pointwise convergence on a compact set does not necessarily imply uniform convergence.
• Jun 24th 2009, 12:14 PM
Laurent
Quote:

Originally Posted by jon.s.beardsley
How is it that $\displaystyle \|f\ast\phi_n\|_\infty$ converges uniformly? As far as I understood, pointwise convergence on a compact set does not necessarily imply uniform convergence.

I guess you mean $\displaystyle f\ast\phi_n$.

Since this is the result in the case when $\displaystyle f_n=f$ for all $\displaystyle n$, I supposed you already knew that $\displaystyle f\ast\phi_n$ converges uniformly to $\displaystyle f$. (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 $\displaystyle f$ is continuous, $\displaystyle \|f\ast\phi_n-f\|_\infty\to 0$. The idea for the proof is to write $\displaystyle (f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt$, and to use the fact that $\displaystyle \phi_n$ is 0 away from 0 (precisions depend on your own definition) so that the integral reduces to a segment around $\displaystyle x$ of small width. Using the uniform continuity of $\displaystyle f$ (that's were compacity is involved), we conclude. (Use an $\displaystyle \varepsilon>0$ for the proof) I guess you already met this proof in your lecture, didn't you?
• Jun 24th 2009, 12:20 PM
jon.s.beardsley
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 $\displaystyle \mathbb{R}^n$. Right now I'm working on a glueing property. Thanks again for all the help.
• Jun 24th 2009, 01:10 PM
jon.s.beardsley
I cannot see how $\displaystyle (f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt$ is true.
• Jun 24th 2009, 01:43 PM
Laurent
Quote:

Originally Posted by jon.s.beardsley
I cannot see how $\displaystyle (f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt$ is true.

Because $\displaystyle f\ast\phi_n(x)=\int f(t)\phi_n(x-t)dt$( and $\displaystyle =\int f(x-t)\phi_n(t)dt$ as well) and, since $\displaystyle \int\phi_n(t)dt=1$ (approximation of delta...), $\displaystyle f(x)=f(x)\int \phi_n(t) dt=f(x)\int \phi_n(x-t) dt=\int f(x)\phi_n(x-t)dt$ (change of variable)
• Jun 26th 2009, 05:22 AM
jon.s.beardsley
All is clear now! Thankyou!