Let $\{\phi_j\}_{j=0}^\infty$ be the Littlewood-Paley partition of unity, i.e., $$ \sum_{j=0}^\infty \phi_j(\tau) = 1 \; for \; \tau \geq 0; \;\;\;\;\phi_j \in C_0^\infty (\Bbb R), \;\phi_j \geq 0 \; for \; j \geq 0;$$ $$ \text{supp} \phi_j \subset [2^{j-1}, 2^{j+1}] \; for \; j \geq 1, \;\;\;\; \text{supp} \phi_0 \cap \Bbb R_{\geq 0} \subset [0,2].$$ Then how can I show that $$ \sup_{\tau \in [0,T]} \frac{\phi_j (\tau)}{\sqrt{1+\tau^2}} \leq 2^{-j}$$ for any $T >0$ ? (or can I find some Littlewood-Paley partition of unity $\{\phi_j\}$ satisfying this inequality?)

Here $C_0^\infty$ means the family of $C^\infty$ functions with compact support and $\text{supp}$ means the support.


1 Answer 1


We have $\phi_j(\tau)\leqslant 1$ and it's equal to $0$ for $\tau<2^{j-1}$. Hence we get that the supremum in the OP is smaller than $\frac 1{2^{j-1}}$.

  • $\begingroup$ Thank you for the comment! $\endgroup$
    – Leun Kim
    Apr 3, 2013 at 21:15

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