Im trying to work on the following exercise, Let $\phi$ be a function s.t. $\|\widehat{\phi}(\xi)\| \leq B \min (\| \xi\|^\delta, \| \xi\|^{-\epsilon}) $, where $\epsilon,\delta >0$. Then $$ \sum_{j\in Z} \| \widehat{\phi}(2^{-j}\xi)\| \leq C_n B $$ where $C_n$ is a constant depending on the dimension of the space the function is defined.
This appears in Grafakos proof of Litllewood-Paley theorem, and i'm just stuck on how to bound that series, i think the first approach is to consider splitting in the cases $\|2^{-j}\xi\| \leq 1$ and $\|2^{-j}\xi\| \geq 1$, for the signal of the expoents, which is the approach he uses in the theorem, but i don't know how to procede, any help would be appreciated.
Edit: The term in the series is not squared, i fixed it
I think i've solved it but i can't see the dimensional constant, so if anyone can correct me i'll also appreciate.
Basically if $2^{-j}\|\xi\| \leq 1$, which implies $2^{-j} \leq \|\xi\|^{-1}$, so there exists $N \in \mathbb{Z}$ s.t $j \geq N$, therefore $$ \sum_{j\in Z} \| \widehat{\phi}(2^{-j}\xi)\| = \sum_{-\infty}^{N} \| \widehat{\phi}(2^{-j}\xi)\| +\sum_{N}^{+\infty} \| \widehat{\phi}(2^{-j}\xi)\| $$
With some abbuse of notation the second term (analogously for the first) can be split into $$ \sum_{j =1}^{+\infty} \| \widehat{\phi}(2^{-j}\xi)\| + \sum_{j = N}^{0} \| \widehat{\phi}(2^{-j}\xi)\| $$ Now the second term in the last equation is a finite sum of bounded terms, and therefore is a constant, the first term is bounded by a convergent power series $$ \sum_{j =1}^{+\infty} \| \widehat{\phi}(2^{-j}\xi)\| \leq \sum_{j =1}^{+\infty} B (2^{-j}\|\xi\|)^\delta = B\|\xi\|^\delta \sum_{j =1}^{+\infty} 2^{-j\delta} = B C_1 $$ where $C_1 = \|\xi\|^\delta \sum_{j =1}^{+\infty} 2^{-j\delta}$, the same thing can be done in the other term, just noting that the expoent will also be negative, except for a finite sum.