## On a decomposition of L^1(G)

[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.

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What's the question? – Stopple Feb 25 2011 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 2011 at 2:01
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 2011 at 7:00
Acky, your edit has not turned this into a question. – Mariano Suárez-Alvarez Feb 26 2011 at 15:23
I have just done some rewriting which hopefully recasts the question in slightly more well-defined form – Yemon Choi Feb 26 2011 at 22:35

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.

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 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 2011 at 7:30