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[Edit, Will Jagy: The OP takes an entire paragraph to describe a certain decomposition. The question is whether this decomposition existsEDITED by Y. As the second paragraph gives an example in Choi - I have attempted to paraphrase the negative, one might regard original question into something a bit terser and more precise; if this as fairly open ended.

(1) does is not what the decomposition never exist, (2) does it always exist, (3) if neither extremeoriginal poster intended, when does it exist?

End editthey should make corrections themselves.]

Suppose

Let $G$ is be a locally compact abelian (LCA) group . Is there a decomposition of $L^1(G)$ such that for every and let $f\in L^1(G)$, . Can we have $f=g+h$, where $g$ belongs to always find $g\in L^2(G)$ and $h$ satisfies the following properties?

• $h$ belongs to $L^1(G)$ and the inversion formula of Fourier transform holds for $h$;or, more strictly, $h$ satisfies such that $h$ belongs to h=f-g$lies in$L^1(G)∩B(G)$, L^1(G)\cap B(G)$, where $B(G)$ contains all functions of is the formFourier-Stieltjes algebra of $\int G$?

(x,Y)d\mu$, where$\mu$is a$B(G)$consists of all Fourier transforms of complex-valued regular measure Borel measures on$\Gamma$. Here$\Gamma$denotes \Gamma$, the dual group of $G$.G$.) If there are counterexamples, are there counterexamples with$G={\mathbb R}^n$? • Post Reopened by Qiaochu Yuan, Daniel Litt, Gil Kalai, Andres Caicedo, François G. Dorais 6 Tried to clean up and added some TeX. It may benefit from a second pass. [Edit, Will Jagy: The OP takes an entire paragraph to describe a certain decomposition. The question is whether this decomposition exists. As the second paragraph gives an example in the negative, one might regard this as fairly open ended, . (1) does the decomposition never exist, (2) does it always exist, (3) if neither extreme, when does it exist? End edit] Suppose G$G$is a locally compact abelian (LCA) group, then is . Is there a decomposition of L^1(G)$L^1(G)$such that for every f belongs to L^1(G)$f\in L^1(G)$, we have f=g+h,$f=g+h$, where g$g$belongs to L^2(G)$L^2(G)$and h$h$satisfies the following properties: (i) h ? 1.$h$belongs to L^1(G)$L^1(G)$and the inversion formula of Fourier transform holds for h;$h$; or, more strictly, h$h$satisfies that(ii) h 2.$h$belongs to L^1(G)∩B(G),$L^1(G)∩B(G)$, where B(G)$B(G)$contains all functions of the form (x,γ)dμ,$\int (x,Y)d\mu$, where μ$\mu$is a complex-valued regular measure on Γ.$\Gamma$. Here Γ$\Gamma$denotes the dual of G.$G$. In the case G=R^n$G={\mathbb R}^n$, as we know, the Calderon-Zygmund decomposition theorem asserts that every f belongs to L^1(R^n)$f\in L^1({\mathbb R}^n)$is the sum of its good part g$g$and bad part b.$b$. Since g$g$is bounded and belongs to L^1(R^n)$L^1({\mathbb R}^n)$, it's it is not hard to verify that g$g$belongs to L^p(R^n)$L^p({\mathbb R}^n)$for every p>=1$p\ge 1$. But it's it is easy to see that there exists a f an$f$such that the inversion formula of Fourier transform fails for b.$b$. That is to say, the Calderon-Zygmund decomposition is not the decomposition of L^1(R^n) which$L^1({\mathbb R}^n)\$ that I want.

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Post Closed as "not a real question" by Andres Caicedo, S. Carnahan
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