# Differentiability of some series

Assume that:

$\phi: \mathbb{R} \rightarrow \mathbb{R}$ is a function of class $C^\infty$ with compact support such that $\phi|_{[0,1]}=1$, $\operatorname{supp} \phi \subset [-2,2]$, $0 \leq \phi \leq 1$,

$g: [0, \infty) \rightarrow \mathbb{R}$ is of class $C^1$;

sequences $(a_n)$, $(b_n)$ of real numbers satisfy:

$b_n<0$, $b_n \rightarrow -\infty$,

$\sum_{n=1}^\infty |a_n| |b_n|^m< \infty$, for $m=0,1$,

$\sum_{n=1}^\infty a_n b_n^m=1$ for $m=0,1$.

Let $f: (\infty, 0) \rightarrow \mathbb{R}$, $f(x)=\sum_{n=1}^\infty a_n \phi(b_n x) g(b_n x)$ for $x<0$.

I know that for each subinterval $[c,d]\subset (-\infty,0)$ the sum is finite, whence $f$ is of class $C^1$ and

$f'(x)=\sum_{n=1}^\infty a_n [\phi(b_n x) g(b_nx)]' \textrm{ for } x<0$.

I wish to show that:

$\lim_{x\rightarrow 0^-} f(x)=\sum_{n=1}^\infty a_n \lim_{x\rightarrow 0^-} [\phi(b_n x) g(b_nx)$]$\textrm{ (the last expression } =g(0)$),

$\lim_{x\rightarrow 0^-} f'(x)=\sum_{n=1}^\infty a_n \lim_{x\rightarrow 0^-} [\phi(b_nx) g(b_nx)$]'$\textrm{(the last expression } =g'(0)$).

I think that it suffices to show that series $\sum_{n=1}^\infty a_n [\phi(b_nx) g(b_nx)]^{(m)}$, ($m=0,1$) are uniformly convergent on for example $(-1,0)$ but I don't know how to show this.

Thanks.

Edit.

I change one of assumptions. Now $f \in C^\infty[0,\infty)$ (instead of $f \in C^\infty(0,\infty)$ ).

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What is $b_x$? Is it $b_n$? – Davide Giraudo Sep 7 '12 at 8:37
When you say "lastt expression = $g(0)$", you mean "would be $g(0)$ if it were defined". – Hagen von Eitzen Sep 7 '12 at 13:31