Assume that $g : [0, \infty) \rightarrow \mathbb R$ is continuous and $\phi :\mathbb R \rightarrow \mathbb R$ is continuous with compact support with $0\leq \phi(x) \leq 1$, $\phi(x)=1$ for $ x \in [0,1]$ and $\phi(x)=0$ for $x\geq 2$.
I wish to prove that $ \lim_{x \rightarrow 0^-} \sum_{n=1}^\infty \frac{1}{2^n} \phi(-nx) g(-nx)=g(0). $
I try in the following way. Let $f(x)=\sum_{n=1}^\infty \frac{1}{2^n} \phi(-nx) g(-nx)$ for $x \leq 0$.
For $\varepsilon>0$ there exists $n_0 \in \mathbb N$ such that $\sum_{n\geq n_0} \frac{1}{2^n} \leq \frac{\varepsilon}{2|g(0)|}$. For $x<0$ there exists $m(x) \in \mathbb N$, $m(x)> n_0$ such that $\phi(-nx)=0$ for $n>m(x)$.
Then $ |f(x)-f(0)|\leq \sum_{n=1}^\infty \frac{1}{2^n} |\phi(-nx)g(-nx)-\phi(0)g(0|= \sum_{n=1}^{m(x)} +\sum_{n\geq m(x)} $ (because $\phi(0)=1$, $\sum_{n=1}^\infty \frac{1}{2^n}=1$). The second term is majorized by $\frac{\varepsilon}{2}$ , but I don't know what to do with the first one because $m(x)$ depend on $x$.