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There is a function:

$V(n) = \sum_{i=1}^\infty f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)$,

where $f$,$b$ and $a$ are $C^\infty$ smooth functions.

I know that we can prove that the function $V(n)$ is also smooth under some assumptions.

I SUGGEST THE NEXT SOLUTION:

We know that if

$\forall i$ $ f_i() \in C^{1} $

$s(n)=\sum_{i=1}^\infty f_i(n)< \infty$ (not neceesary uniformly)

and

$\sum_{i=1}^\infty f_i^{'}(n)=\sigma(n)$ uniformly

then:

$s(n) \in C^{1}$ and $s^{'}(n)=\sigma(n)$

So IN MY CASE we have $C^{\infty} $

$s(n)=\sum_{i=1}^\infty f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)< \infty$

and we need to show that:

$\sum_{i=1}^\infty h_i^{'}(n)$ converges uniformly,

where $h_i(n)=f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)$

MY QUESTION: Can somebody help me with any ideas how I can prove that $\sum_{i=1}^\infty h_i^{'}(n)$ converges uniformly.

I will really appreciate your help!

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If we don't place any further restrictions on $f,b_i$ and $a_i$, then it's not true that $\sum h_i(n)$ converges uniformly.

Let $\{U_i\}$ be a partition of unity on $\Bbb R$, and let $H_i$ be an antiderivative of $U_i$. We want to show that there exist smooth functions $f, b_i$ and $a_i$ such that $H_i(n)=f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)$. This is easily achieved: take $f\equiv 1$ and $a_i\equiv 0$ to be constant functions, and $b_i$ an antiderivative of $H_i$.

By definition, $\displaystyle\sum_{i=1}^\infty H_i(n) = \sum_{i=1}^\infty U_i(n) = 1$ for all $n$. However, this convergence cannot be uniform because the support of each $U_i$ is compact, hence the support of any partial sum is compact. In particular, for any $K$ there is always some number $n_0$ such that $\displaystyle \sum_{i=1}^K U_i(n_0) = 0$, and this is automatically an obstruction to uniform convergence.

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  • $\begingroup$ Thank you a lot for your answer! Can you please propose restrictions, which should be applied to $f_i$,$b_i$ and $a_i$ to achieve $\sum_{i=1}^\infty h_i^{'}(n)$ converges uniformly? $\endgroup$ – Caim Feb 18 at 13:16

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