j********3
发帖数: 9 | 1 Let f(t) be a function and let F(w) denote its Fourier Transform (FT). Is
it possible for |w||F(w)| to be integrable yet f(t) is not differentiable?
Sorry if this is a stupid question, I'm a beginner trying to learn something
about Fourier Transforms.
Thank you very much. |
J*****n
发帖数: 4859 | 2 integrable不一定要differentiable |
B********e
发帖数: 10014 | 3 if 1. on 1-dimension space
2. squre-integrable instead of integrable
then f(t) must be differentiable,by sobolev imbedding thm
otherwise it need further investigation though i intend to say no.
in any case it's not trival, not a stupid question;)
something
【在 j********3 的大作中提到】
: Let f(t) be a function and let F(w) denote its Fourier Transform (FT). Is
: it possible for |w||F(w)| to be integrable yet f(t) is not differentiable?
: Sorry if this is a stupid question, I'm a beginner trying to learn something
: about Fourier Transforms.
: Thank you very much.
|
j********3
发帖数: 9 | 4 Thank you very much for the reply.
(1) Yes, the function that I have in mind is one-dimensional: f(t) is a
mapping from the R to C.
(2) I'm not familiar with anything Sobolev... I did some web search but couldn't understand many of the theorem statements. I will try to look at some textbooks. I did manage to find this related statement:
If |w|^s |F(w)| is in L2 and s > n + 1/2 then f(t) is n times continuously
differentiable.
How do I prove this statement? In terms of my original question, s= |
B********e
发帖数: 10014 | 5 you are right,not sufficient
couldn't understand many of the theorem statements. I will try to look at
some textbooks. I did manage to find this related statement:
【在 j********3 的大作中提到】
: Thank you very much for the reply.
: (1) Yes, the function that I have in mind is one-dimensional: f(t) is a
: mapping from the R to C.
: (2) I'm not familiar with anything Sobolev... I did some web search but couldn't understand many of the theorem statements. I will try to look at some textbooks. I did manage to find this related statement:
: If |w|^s |F(w)| is in L2 and s > n + 1/2 then f(t) is n times continuously
: differentiable.
: How do I prove this statement? In terms of my original question, s=
|
H****h
发帖数: 1037 | 6 我这样推理一下，不知道对不对。
既然F是f的FT，那么f也是F的FT。既然wF是可积的，假设g是wF的FT。
那么g恰好是f的微分。
something
【在 j********3 的大作中提到】
: Let f(t) be a function and let F(w) denote its Fourier Transform (FT). Is
: it possible for |w||F(w)| to be integrable yet f(t) is not differentiable?
: Sorry if this is a stupid question, I'm a beginner trying to learn something
: about Fourier Transforms.
: Thank you very much.
|
B********e
发帖数: 10014 | 7 应该是这样的，我想多了，呵呵
distribution意义下FT的唯一性保证了这个结论
还是康牛啊
【在 H****h 的大作中提到】
: 我这样推理一下，不知道对不对。
: 既然F是f的FT，那么f也是F的FT。既然wF是可积的，假设g是wF的FT。
: 那么g恰好是f的微分。
:
: something
|
j********3
发帖数: 9 | 8 Thanks again for the responses. I understand that if f(t) is differentiable, then the FT of f'(t) is iwF(w).
I am wondering if I can find a f(t) that is not differentiable yet |w||F(w)| is in L1.
It does not seem obvious to me that such f(t) does not exist?
Also, I still interested in knowing how to prove the statement in my previous post.
Thanks again for helping, I really appreciate it. |
H****h
发帖数: 1037 | 9 我前面的忽略了太多的细节。你可以这样证明。
现在已经知道f是F的FT。把f写成积分的形式。
然后求(f(x+h)-f(x))/h当h->0时候的极限。利用控制收敛定理。
当然，如果钻牛角尖的话，你可以改动f在几个点的取值，使得f不连续，但FT不变。
differentiable
)|
【在 j********3 的大作中提到】
: Thanks again for the responses. I understand that if f(t) is differentiable, then the FT of f'(t) is iwF(w).
: I am wondering if I can find a f(t) that is not differentiable yet |w||F(w)| is in L1.
: It does not seem obvious to me that such f(t) does not exist?
: Also, I still interested in knowing how to prove the statement in my previous post.
: Thanks again for helping, I really appreciate it.
|
