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

1. $h$ belongs to $L^1(G)$ and the inversion formula of Fourier transform holds for $h$; or, more strictly, $h$ satisfies that
2. $h$ belongs to $L^1(G)∩B(G)$, where $B(G)$ contains all functions of the form $\int (x,Y)d\mu$, where $\mu$ is a complex-valued regular measure on $\Gamma$. Here $\Gamma$ denotes the dual of $G$.

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.

[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.

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?

Suppose G is a locally compact abelian (LCA) group, then is there a decomposition of L^1(G) such that for every f belongs to L^1(G), we have f=g+h, where g belongs to L^2(G) and h satisfies the following properties: (i) h belongs to L^1(G) and the inversion formula of Fourier transform holds for h; or, more strictly, h satisfies that (ii) h belongs to L^1(G)∩B(G), where B(G) contains all functions of the form ∫(x,γ)dμ, where μ is a complex-valued regular measure on Γ. Here Γ denotes the dual of G.

In the case G=R^n, as we know, the Calderon-Zygmund decomposition theorem asserts that every f belongs to L^1(R^n) is the sum of its good part g and bad part b. Since g is bounded and belongs to L^1(R^n), it's not hard to verify that g belongs to L^p(R^n) for every p>=1. But it's easy to see that there exists a 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(R^n) which I want.

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

1. $h$ belongs to $L^1(G)$ and the inversion formula of Fourier transform holds for $h$; or, more strictly, $h$ satisfies that
2. $h$ belongs to $L^1(G)∩B(G)$, where $B(G)$ contains all functions of the form $\int (x,Y)d\mu$, where $\mu$ is a complex-valued regular measure on $\Gamma$. Here $\Gamma$ denotes the dual of $G$.

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.

Suppose G is a locally compact abelian (LCA) group, then is there a decomposition of L^1(G) such that for every f belongs to L^1(G), we have f=g+h, where g belongs to L^2(G) and h satisfies the following properties: (i) h belongs to L^1(G) and the inversion formula of Fourier transform holds for h; or, more strictly, h satisfies that (ii) h belongs to L^1(G)∩B(G), where B(G) contains all functions of the form ∫(x,γ)dμ, where μ is a complex-valued regular measure on Γ. Here Γ denotes the dual of G.

In the case G=R^n, as we know, the Calderon-Zygmund decomposition theorem asserts that every f belongs to L^1(R^n) is the sum of its good part g and bad part b. Since g is bounded and belongs to L^1(R^n), it's not hard to verify that g belongs to L^p(R^n) for every p>=1. But it's easy to see that there exists a 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(R^n) which I want.

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?

Suppose G is a locally compact abelian (LCA) group, then is there a decomposition of L^1(G) such that for every f belongs to L^1(G), we have f=g+h, where g belongs to L^2(G) and h satisfies the following properties: (i) h belongs to L^1(G) and the inversion formula of Fourier transform holds for h; or, more strictly, h satisfies that (ii) h belongs to L^1(G)∩B(G), where B(G) contains all functions of the form ∫(x,γ)dμ, where μ is a complex-valued regular measure on Γ. Here Γ denotes the dual of G.

In the case G=R^n, as we know, the Calderon-Zygmund decomposition theorem asserts that every f belongs to L^1(R^n) is the sum of its good part g and bad part b. Since g is bounded and belongs to L^1(R^n), it's not hard to verify that g belongs to L^p(R^n) for every p>=1. But it's easy to see that there exists a 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(R^n) which I want.