Here is the problem:
Let $A$ be the set of positive integers greater than 1. For each $L\in A$, we want to construct a smooth function $f_L$ with compact support such that
$$\frac{1}{m}=\sum_{j=0}^{\infty}\frac{1}{L^j}f_L\left(\frac{m}{L^j}\right)$$ for any nonzero integer $m$. Moreover, we require that $\frac{1}{L^j}\sum_{n\in \mathbb{Z}}\left|f_L(\frac{n}{L^j})\right|$ is bounded, uniformly in $j$ and $L$.
Here is what I have so far:
Let $\phi_L$ be a smooth function which is $0$ in $[-1,1]$ and $1$ in $[L,\infty] \cup [-\infty,-L]$. Let $$\eta_L(x)=\phi_L(Lx)-\phi_L(x)$$ Then we can see that $\eta_L$ is supported in $|x|\in \left[\frac{1}{L},L\right]$ and $\sum_{j=0}^{\infty}\eta_L\left(\frac{x}{L^j}\right)=1$ for any $x$ with $|x|\ge1$. Define $f_L(x)=\frac{\eta_L(x)}{x}$. Then $f_L$ satisfy all the required properties with the exception that $\frac{1}{L^j}\sum_{n\in \mathbb{Z}}\left|f_L(\frac{n}{L^j})\right|$ may not be bounded uniformly in $j$ and $L$. Can you prove that they are uniformly bounded? Or give a different construction for $f_L$? Thank you.