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Consider the Hardy space $H^1$. We can define this space by means of an atomic decomposition: A function $a$ is called an $L^2$-atom for $H^1$ if it is supported in a ball $B$, one has $\|a\|_2 \leq |B|^{-1/2}$, and $a$ is mean value free. Now the Hardy space consists of all functions $f$ that can be written as $f = \sum_B \lambda_B a_B$ with $\sum |\lambda_B| < \infty$, and the norm is given as the infimum over $\sum_B |\lambda_B|$ for all such decompositions.

Now consider the following situation: We are given three balls $B$, $B'$, and $B''$ of the same size but different location and functions $b$ and $b'$ supported in $B$ and $B'$, respectively, with same mean value. Consider now the function $$f = b - 1_{B''} (b)_B - b' + 1_{B''} (b')_{B'}=: a + a'.$$ The support of $f$ is $B \cup B'$, but the atoms in this decomposition are supported in $B\cup B''$ and $B'\cup B''$, respectively, so the supports of the atoms are not contained in the support of the function that is decomposed. In this example it is of course trivial to find an atomic decomposition with atoms supported in the support of $f$ (e.g. by replacing $1_{B''}$ by $1_{B}$ in the definition of the atoms).

Question: Is it always possible to find an atomic decomposition such that the atoms are supported in the support of the function that is decomposed?

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It is true that such an atomic decomposition exists, but it is a deep fact shown in a paper by Chang, Krantz, and Stein.

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