# About the property of Littlewood-Paley partition of unity.

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.

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 $\frac 1{2^{j-1}}$.

• Thank you for the comment! – Leun Kim Apr 3 '13 at 21:15