Thread:

1. 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?

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

3. 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 $f_n\ast\phi_n-f=(f_n-f)\ast \phi_n + (f\ast\phi_n-f)$? 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.

4. 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.

5. Originally Posted by jon.s.beardsley
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 $f$.

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

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

8. 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.

9. Originally Posted by jon.s.beardsley
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?

10. 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.

11. 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.

12. Originally Posted by jon.s.beardsley
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)

13. All is clear now! Thankyou!
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