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• CommentAuthordr
• CommentTimeJan 20th 2012 edited

Is this question http://math.stackexchange.com/q/100692/2987 appropriate on MO ?

I guess it isn't, but want to confirm.
• CommentAuthortheojf
• CommentTimeJan 23rd 2012

Hi Rajesh,

The wording of your question is quite convoluted. Probably this is because you are an expert in a language which is not mathematics (I see that you are a PhD student in Signal Processing), and you are having to produce a translation on the fly. Let me try to rewrite your question, to confirm that I understand it.

You consider the class of real-valued functions $f$ on the circle $S^1 = \mathbb R / \mathbb Z$ for which: (1) the set of points at which $f$ is not infinitely differentiable is countable and dense; (2) let $x$ be a point at which $f$ is $k$-times differentiable but not $(k+1)$-times, then both the left and right limits of the $(k+1)$th derivative of $f$ exist at $x$, and fail to agree. As far as I can tell, your notation is superfluous, and you never use the function you call $ck$. Oh, and you also request that $f$ be right-continuous at $0$.

Finally, you claim that this class of functions is a vector space, and you ask whether it is closed under convolution in $S^1$. Note that as written it definitely is not a vector space: the constant function $0$ fails condition $1$. But presumably you mean to weaken this?

It is not automatically clear to me that the convolution is even defined, but this is probably covered by your somewhat strict regularity requirements. Almost certainly the argument for closure under convolution is going to go something like this: by condition (2), the set of points at which any particular function fails to be $k$-times differentiable is discrete and hence finite; then the set of points at which the convolution of $f$ and $g$ fails to be regular is made out of those points at which $f$ and $g$ are each unregular, and you can do a direct analysis. This will be easier with the correct weakening in the preceding paragraph --- then almost certainly your class will be naturally filtered by easy conditions.

In order for this question to be suitable at MO, maybe most important is to improve the statement of the question. (Don't forget to look over http://mathoverflow.net/howtoask .) Currently you take three pages to say what could fit in less than one. I would also be interested to hear why you are interested in this question. But do give math.stackexchange a bit longer to provide an answer.

I worry that even with improvements in the statement, the question still will not be MO-appropriate. If my understanding of the question is correct, it is at an undergraduate-homework level. I tend to think graduate-level-in-math homework can be appropriate at MO, although there is disagreement as to this. And I understand that this question is not, in fact, undergraduate homework. But it's at that mathematical level.

I hope this helps,

Theo

• CommentAuthordr
• CommentTimeJan 24th 2012

Dear Theo, (I guess you are the same Theo I know from math.SE)

I'd like to clarify some of the issues you have raised.

1. you never use the function you call $ck$ ?

A. This is used in condition number 3 for the definition of the class of functions.

"3. \forall x \in (0,1), Cf(x) = ck(x)"

2. you claim that this class of functions is a vector space, "the constant function $0$ fails condition $1$"

I am claiming that the set $S_{pc}$ is a vector space and not the set $S_p$. The set $S_p$ is the class of periodic functions I have defined,
and hence they need to satisfy the conditions mentioned. Here I say "To get the closure property under addition, we can add the set of all constant functions $K$ to the set $Sp$ to form a new set $S{pc}$."
By this sentence I mean that the set $S_{pc}$ is the union of the two sets $S_p$ (defined class of functions) and $K$ the set of all constant functions.
Hence the constant function $0$ need not satisfy any condition in the definition of the class of functions defined earlier because it is not part of it.
"But presumably you mean to weaken this", I am not able to understand this comment clearly, but i guess you were talking about creating the set $S_{pc}$ as a union of two sets as mentioned above
and making it a vector space.

I also quote from the opening paragraph of my question "I started with a class of functions and made some modifications to show that it forms a vector space ....", Here note that I have
mentioned the phrase "...and made some modifications to show that it forms a vector space...", by this I meant the creation of the set $S_{pc}$.

4. "It is not automatically clear to me that the convolution is even defined,"

I request you to clarify what you mean by this statement. Do you mean to include the definition of (circular) convolution in the question ?

5. "This will be easier with the correct weakening in the preceding paragraph"

Request you to kindly clarify what you mean by 'weakening' and 'correct weakening' as I am new to this terminology.

Thank you very much for the comments and I request you to give clarifications of my above mentioned issues as it would be possible for me to improve the question.

I also request you to kindly participate in the thread at math.SE as well if it is possible.
• CommentAuthortheojf
• CommentTimeJan 30th 2012

Hi Rajesh,

(0. There are a couple "Theo"s on MO, and I'm not on M.SE, out of laziness. So probably you know a different "Theo"?)

1. I agree that you mention the function "ck" in the definition. My point was that you don't need to. As far as I can tell, your definition is: given f, define Cf as the counting the number of times f is differentiable at a given point. Now set ck = Cf. Now consider those functions for which ck = Cf has some property.

2. Probably I was skimming too fast. Your set S_pc is the union of your set S_p with the constant functions, and this certainly includes 0. It would then be better to never mention the set S_p, and define the set of functions that you want as being of some shape, or being constant.

Let me mention here that it's worth practicing and editing mathematical writing, because shorter writing is easier to skim.

1. I mean only that convolution involves integration, and not all functions are integrable. I'm not a real analyst, and so would like you to include one sentence for why your functions are integrable.

2. By "weakening" I meant that I thought the proceeding paragraph was incorrect, because the claim made was too strong.

Hope that helps, Theo

• CommentAuthordr
• CommentTimeJan 31st 2012

hi Theo,

Thanks for your reply. The Theo i've known from SE is Theo Buehler. I have added an answer to the question. It many not be in correct math language, but your comments and simplification of the definition of the class of functions was particularly helpful.
Here is a link to the answer.

http://math.stackexchange.com/a/104141/2987

Thanks again

Rajesh.D