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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 $x$.

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Fyi, $g(-nx)$ is not defined when $n\in\mathbb{N}$ and $x>0$ since $g:[0,\infty)\to\mathbb{R}$. Perhaps you need to change something? – J. Loreaux Sep 8 '12 at 15:56
I think there is a problem with signs. All $-n x$ should read $n x$, all$x\le0$ should read $x\ge0$, $x<0$ should read $x>0$. – Hagen von Eitzen Sep 8 '12 at 15:58
I want to show left side continuity of $f$ at zer0. – L.T Sep 8 '12 at 16:02
up vote 1 down vote accepted

Let $h(x)=\phi(x)g(x)$. Then $h\colon[0,\infty)\to\mathbb R$ is continuous and bounded by some $M$ and $h(x)=0$ for $x\ge2$. Given $\epsilon>0$, find $\delta$ such that $x<\delta$ implies $|h(x)-h(0)|<\frac\epsilon3$.

Then for $m\in \mathbb N$ $$\sum_{n=1}^\infty \frac1{2^n} h(nx)-h(0)=\sum_{n=1}^{m} \frac1{2^n} (h(nx)-h(0))+\sum_{n=m+1}^\infty \frac1{2^n} h(nx)-\frac1{2^m}h(0)$$ If $m<\frac\delta x$, then $$\left |\sum_{n=1}^{m} \frac1{2^n} (h(nx)-h(0))\right |<\sum_{n=1}^\infty \frac 1{2^n}\frac\epsilon3=\frac\epsilon3.$$ For the middle part, $$\left|\sum_{n=m+1}^\infty \frac1{2^n} h(nx)\right|<\frac M{2^m}$$ and finally $\left|\frac1{2^m}h(0)\right|\le \frac M{2^m}$. If $m>\log_2(\frac {3M}\epsilon)$, we find that $$\left|\sum_{n=1}^\infty \frac1{2^n} h(nx)-h(0)\right|<\epsilon$$ for all $x$ with $x<\min\{\delta,\frac Mm\}$.

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