On a decomposition of L^1(G) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:33:21Z http://mathoverflow.net/feeds/question/56651 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56651/on-a-decomposition-of-l1g On a decomposition of L^1(G) Acky 2011-02-25T17:16:34Z 2011-02-28T13:06:41Z <p>[<strong>EDITED</strong> 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.]</p> <p>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 $G$?</p> <p>($B(G)$ consists of all Fourier transforms of complex-valued regular Borel measures on $\Gamma$, the dual group of $G$.)</p> <p>If there are counterexamples, are there counterexamples with $G={\mathbb R}^n$?</p> <p>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.</p> http://mathoverflow.net/questions/56651/on-a-decomposition-of-l1g/56791#56791 Answer by Theo Buehler for On a decomposition of L^1(G) Theo Buehler 2011-02-27T05:27:41Z 2011-02-28T13:06:41Z <p>I'm answering Yemon's version of the question.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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 <s>should</s> also gives a counterexample for any <s>non-compact and</s> non-discrete locally compact abelian group.</p> <p>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.</p>