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Let $\{f_k(x)\}_{k=0}^\infty$ be a Littlewood-Paley decompositon, that is, $$ f_k \in C_c^\infty $$ $$ \sum_{k=0}^\infty f_k (x) = 1,$$ $$ \text{supp} f_0 \subset \{ |x| \leq 2 \},$$ $$ \exists f \in C_c^\infty \; \text{such that}\; \text{supp} f \subset \{ 2^{-1} \leq |x| \leq 2 \} \; \text{satisfying}\; f_k (x) = f(x/2^k).$$ Then I hope to show that for each $x$, there are at most 3 nonzero terms in the summation. Here $C_c^\infty$ means $C^\infty$ functions with compact support and $\text{supp}$ means the support.

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The support of $f_0$ is contained in $|x|\le 2$. The support of $f_k$ is contained in $2^{k-1}\le |x|\le 2^{k+1}$. On the base 2 logarithmic scale these ranges become $$[-\infty, 1]\quad \text{ and }\ [k-1,k+1], \ k=1,2,\dots \tag1$$ It should not be hard to see that every number $x$ is covered by at most three of the intervals in (1).

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