Mollifiers and density in L^p

Matt

Dec 19, 2011 13:11

I was going to ask if I got the big picture and then while typing I found an inconsistency. Now I'm going to ask the related question and maybe come back with this one.
So in that question they claim $J_\varepsilon$ is a mollifier. I looked up the definition and one of the properties that it should have is that $\lim_{\varepsilon \to 0} J_\varepsilon (x) = \delta(x)$, so I tried to verify this. Somehow I get $\infty$ for all $x$ in $\mathbb{R}$.

Jonas Teuwen

Isn't that a limit in distributional sense?

Matt

What does that mean?

t.b.

@JonasTeuwen yes it is, but it's not important here

Matt

I get $$ \lim_{\varepsilon \to 0} \frac{k}{\varepsilon} e^{- \frac{1}{1 - (\frac{x}{\varepsilon})^2}} = \infty$$ for all $x \in \mathbb{R}$

Which means I can't even compute a limit because I should get $0$ for $x \neq 0$.

Jonas Teuwen

I really dislike the fact that vector valued distributions are no longe the duals of nice spaces :(.

t.b.

Dec 19, 2011 13:15

@Matt wait. $J_\varepsilon(x) = 0$ for $|x| \gt \varepsilon$.

(that's because $J(x) = 0$ for $|x| \gt 1$.)

Matt

By definition. Doh. : S

Jonas Teuwen

8-). Do you know why we love mollifiers?

Matt

@tb Thank you.

Jonas Teuwen

Why state it that way with a distribution and not just $\|J_\varepsilon \ast f - f\| \to 0$?

Matt

@JonasTeuwen No.

Jonas Teuwen

Dec 19, 2011 13:19

Well, because what I wrote above. Your mollifier is nicely compactly supported and smooth.

So taking convolutions with $L^p$ functions gives us another of such a nice function.

So, then you can do cute calculations with that function and take limits to regain our original function.

Matt

@tb This morning I noticed that you had edited this question about a month after I'd asked it. Do you re-read old stuff at random to fix typos? Or did you go through my old stuff to see what I hadn't been doing properly?

@JonasTeuwen Yes I was thinking about this while trying to get the big picture of the question. Just going to go over it again now.

Jonas Teuwen

@Matt As that is a distributional limit it actually states the same as the convolution with the $\delta$ distribution in $x$ picks up the value of the original function in the point $x$. Roughly speaking.

A quite well-known Dutch (for Dutch standards) algebraist said about analysis that he never has an "Aha Erlebnis" when seeing analysis proofs. He says it is all about computing integrals by splitting them into pieces and calculate them separately... :'D.

t.b.

@Matt The answer is plain and simple: I was eradicating the typo "Lebesque" from this site, so I searched for it and edited all posts in which I found it. It was around the same time as Srivatsan and I were taking care of the various spellings of continuous...

I always wonder why people upvote such answers

Jonas Teuwen

I believe he proved a conjecture by Grothendieck in 2002. Frans Oort.

Matt

Yeah, embarrassing. I know how to spell it correctly. No idea how my finger slipped there. Thanks for correcting it.

t.b.

Dec 19, 2011 13:31

@JonasTeuwen Oh, I know him. I guess he's retired, now?

Jonas Teuwen

@tb Yes in 2000.

@tb Maybe the guy forgot the closure of that space?

But in that case the question is quite... strange.

t.b.

That's well possible, but you never know.

Matt

@JonasTeuwen I don't know either how I wandered off the right path. When I was in first year I was looking forward to not doing analysis anymore and only do discrete maths like e.g. algebra and graph theory. I wonder how I ended up the way I ended up. : P

t.b.

@JonasTeuwen Just to make you upset: $\|f(x)\|_p$ and $\|u(x)\|_{1,2}$

Jonas Teuwen

@Matt And how did you end up? :P.

@tb That second one is the Sobolev norm?

Or Triebel-Lizorkin or something like that...

t.b.

Dec 19, 2011 13:34

Just a random norm containing a function together with its argument

Jonas Teuwen

I would write that as (if I need the argument): $\|x \mapsto f(x)\|_{L^p(\mathbf R)}$.

t.b.

I know

Jonas Teuwen

Someone here needed a norm on something like $L^p(L^{p'}(L^q, \nu)), \mu)$

Matt

@JonasTeuwen I think I can't parse this sentence.

Jonas Teuwen

@Matt Which one?

Matt

Dec 19, 2011 13:37

But I think you can take a not necessarily continuous $f$ in $L^p$ and then mollify it to get a smooth function $f_\varepsilon$ by convoluting it with a mollifier.

@JonasTeuwen Click the up arrow next to your name.

@Matt where $f_\varepsilon$ is close to $f$.

Jonas Teuwen

@Matt Yes, and then it will converge in norm to $f$. I mean that "$(f \ast \delta)(x) = f(x)$".

So your identity is basically the same statement.

t.b.

@Matt The rule of thumb is that the convolution of two functions is as smooth as the smoother of the two.

Matt

Yes, I read that earlier today.

t.b.

So if you convolve an integrable function with a continuous function the result is continuous. If you convolve an integrable function with a $C^k$ function the result is $C^k$, and so on. The point is essentially that $(f \ast g)' = f' \ast g = f \ast g'$, so you can push the derivatives on the function which is differentiable.

Jonas Teuwen

@Matt Woah check out Rajesh's comments!

(but what is the derivative of a $L^1$ function? :-))

Matt

Dec 19, 2011 13:45

You take a representative of an equivalence class which is not necessarily continuous and not necessarily differentiable and try to take a derivative?

@JonasTeuwen Do I have to? I think I can't be bothered.

Jonas Teuwen

@Matt Of course not.

@Matt Well, in some cases you can take the weak derivative.

Density of the $C^\infty$ functions in our Sobolev spaces make that work.

@Matt So. You're doing analysis and forcing at the same time? That's peculiar.

@Matt In your topic why can you switch the derivative and the integral? I think you need to argue what that is allowed.

Matt

@JonasTeuwen I was badgered into it. I have to have what's called "complementary" courses. I didn't fancy the one's offered so I applied to get this set theory course credited as complementary course.

t.b.

@Matt So. You got this other question showing that $C_c$ is dense in $L^p$ with a rather detailed proof in the answer.

FreakEnum

I've a Q whose language I'm unable to understand (meaning) . May I the Q?

Jonas Teuwen

You could used the closedness of the derivative operator :-).

t.b.

Dec 19, 2011 13:50

@JonasTeuwen sounds like begging the question, doesn't it?

Matt

@tb, @JonasTeuwen Can I have a minute? You type and ask faster than I can type and answer.

Jonas Teuwen

Hmm, how do you mean? Taking weak derivatives is a closed operator isn't it?

t.b.

Yes. But proving that involves showing what you want to show.

@Matt sure.

@FreakEnum just ask

Jonas Teuwen

Oh, right in this case it does 8-).

FreakEnum

Q : There is a bacteria which has the probability of die 1/3 of its total number or it may tripled. Find out the probability //I am unable to understand what does author means by tripled death probability?

t.b.

Dec 19, 2011 13:54

So you have a probability that says that you end up with 2/3 of the bacteria you had or with 3 times the number of bacteria you had before.

FreakEnum

@tb still didn't understand that or part "or with 3 times the number of bacteria you had before."

t.b.

You start with x bacteria. With probability p you end up with (2/3) x bacteria and with probability 1-p you end up with 3x bacteria.

Matt

@JonasTeuwen What were you referring to?

QED

that seems unrealistic

FreakEnum

@tb so how do I solve this?

I've no Idea

Matt

Dec 19, 2011 14:01

@tb I haven't finished with this one. What were you going to say?

@FreakEnum Can you post the question verbatim?

FreakEnum

@Matt There is a bacteria which has the probability of die 1/3 of its total number or it may tripled. Find out the probability

t.b.

@Matt Well, if you know that for every $L^p$ function $f$ and $\varepsilon \gt 0$ you find a continuous function $g$ with compact support such that $\|f-g\|_p \lt \varepsilon$. Then the thing that remains to show is that you can find a smooth function with compact support which is arbitrarily close to $g$.

Matt

@tb That's what I was thinking about. The answer is to mollify $g$.

t.b.

Yes. Exactly. What you gain by replacing $f$ by $g$ is that now you have a function $g$ which is continuous and of compact support, hence it is uniformly continuous. Now if you take $\delta$ so small that $|x-y| \lt \delta$ implies $|f(x) - f(y)| \lt \varepsilon$ then $J_{\delta}(x) \ast g$ can't be too far away from $g$.

@FreakEnum I think there is some information that was lost in translation.

FreakEnum

@tb anyways I was thinking the same but how to calculate probability of getting tripled?

how to calculate probability of getting tripled?

t.b.

Dec 19, 2011 14:12

Well, could it be that the answer is just 1-p where p is the probability that one third dies?

FreakEnum

well the solution is:

It has to be 1/3 of dieing plus living which is given by 2/3 * p^3 as p is tripled so raised to 3
probability of die is 1/3
or
its get tripled
the probability of tripled is 2/3*p*p*p
that's 2/3*p^3

but I'm unable to understand the solution also

t.b.

I can't make head or tails of this, to be honest.

FreakEnum

here is another problem : If the area was hit by a virus and so the decrease in the population because of death was x/3 and the migration from other places increased a population by 2x then annually it had so many ppl. find our the population in the starting.

Can you please confirm the answer? : mine is "x"

Matt

@tb Did you mean this? I can't find one with $L^p$ in my old questions.

t.b.

Dec 19, 2011 14:29

@Matt Yes, that's the one I meant. Adapting it to $L^p$ is straightforward.

for ($1 \leq p \lt \infty$)

$Mathematics$ Mathematics

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