Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

[EDITED by Y. Choi - I have attempted to paraphrase the original question into something a bit terser and more precise; if this is not what the original poster intended, they should make corrections themselves.]

Let $G$ be a locally compact abelian (LCA) group and let $f\in L^1(G)$. Can we always find $g\in L^2(G)$ such that $h=f-g$ lies in $L^1(G)\cap B(G)$, where $B(G)$ is the Fourier-Stieltjes algebra of $G$?

($B(G)$ consists of all Fourier transforms of complex-valued regular Borel measures on $\Gamma$, the dual group of $G$.)

If there are counterexamples, are there counterexamples with $G={\mathbb R}^n$?

In the case $G={\mathbb R}^n$, as we know, the Calderon-Zygmund decomposition theorem asserts that every $f\in L^1({\mathbb R}^n)$ is the sum of its good part $g$ and bad part $b$. Since $g$ is bounded and belongs to $L^1({\mathbb R}^n)$, it is not hard to verify that $g$ belongs to $L^p({\mathbb R}^n)$ for every $p\ge 1$. But it is easy to see that there exists an $f$ such that the inversion formula of Fourier transform fails for $b$. That is to say, the Calderon-Zygmund decomposition is not the decomposition of $L^1({\mathbb R}^n)$ that I want.

share|improve this question
5  
What's the question? –  Stopple Feb 25 '11 at 17:36
    
Oh, but were it appropriate on this site to post a response in the form of a Letterman-style top ten list entitled "and the nominees for 'the following properties' are..." I can imagine some real gems, but, alas, I fear that I've already crossed a line. –  Ramsey Feb 26 '11 at 2:01
2  
Acky, if you have a question you want to ask, you should use the "edit" link, and add it to the essay you have written. Also, I suggest you use LaTeX notation for your mathematics. When you have done that, please flag the question for moderator attention. –  S. Carnahan Feb 26 '11 at 7:00
1  
Acky, your edit has not turned this into a question. –  Mariano Suárez-Alvarez Feb 26 '11 at 15:23
1  
I have just done some rewriting which hopefully recasts the question in slightly more well-defined form –  Yemon Choi Feb 26 '11 at 22:35

1 Answer 1

up vote 5 down vote accepted

I'm answering Yemon's version of the question.

The answer is trivially yes for discrete $G$ since $\ell^1(G) \subset \ell^2(G)$, so let me focus on the non-discrete case.

The first observation to make is that $B(G)$ is contained in the bounded (and uniformly continuous) functions of $G$. So the question asks in particular if every integrable function on $G$ is the sum of a bounded function and a square-integrable function.

This is clearly false for compact infinite $G$: For such $G$ we have the strict inclusions $L^\infty \subsetneqq L^2 \subsetneqq L^1$ so $L^\infty + L^2 \subset L^2$, and hence every function in $L^1 \smallsetminus L^2$ provides a counterexample to the question.

Since the question asks for a counterexample in $\mathbb{R}^{n}$, I'll give one for $\mathbb{R}$ which is easily adapted to the higher-dimensional case and with a little care should also gives a counterexample for any non-compact and non-discrete locally compact abelian group.

Take $f = \sum_{n=1}^{\infty} n \cdot [n,n+\frac{1}{n^{3}}]$. This is a function in $L^1 \smallsetminus L^2$. For a bounded function $h$ we have for all $n \geq \Vert h \Vert_{\infty}$ and all $x \in [n,n+\frac{1}{n^3}]$ that $|f(x) - h(x)| \geq n- \Vert h \Vert_{\infty}$, which implies that $g = f - h \notin L^2(\mathbb{R})$ by a straightforward estimate.

share|improve this answer
    
Thank you so much. Actually I realized that my question is silly yesterday since I have made a proof almost the same as yours. –  Acky Feb 27 '11 at 7:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.