# Show uniform convergence of function series

Let the function sequence $\left\{ f_n \right\}_{n\ge n_0}$ satisfy:

(1) $\forall_{x\in D}\forall_{n\in\mathbb{N}} \ f_n(x)\ge 0$

(2) $\forall_{x\in D}\forall_{n\in\mathbb{N}} \ f_n(x)\ge f_{n+1}(x)$

(3) $\sup_{x\in D}f_n(x)\rightarrow 0$

$D$ - domain

Show that function series $\displaystyle\sum_{n=n_0}^{+\infty}(-1)^{n+1}f_n(x)$ converges uniformly.

I don't know how to approach this. Hope the solution isn't too hard. From (2) and (3) and Leibniz theorem this series converge, but I don't know whether it will lead somewhere. Unfortunately rather Weierstrass M-test won't be helpful here.

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I don't know Abel's transform. It wasn't in school. Is there any simpler approach? –  xan Apr 25 '12 at 18:45

Essentially, such a series is alternating: for each $x\in D$, the series $\sum\limits_{n=1}^\infty{(-1)^{n+1}}f_n(x)$ is a convergent alternating series. From this, and 3) (note that 3) is saying that the sequence $(f_n)$ converges uniformly to $0$), it will follow that the series is uniformly Cauchy on $D$ and thus uniformly convergent on $D$.

Let $\alpha_n=\sup\limits_{x\in D}\,\{f_n(x)\}$.

Then for any positive integers $m$ and $n$ with $m\ge n$ and $x\in D$, using the fact that $\sum\limits_{n=1}^\infty{(-1)^{n+1}}f_n(x)$ is an alternating series of real numbers $$\tag{1} \Biggl|\,{(-1)^{n+1} } f_n(x)+{(-1)^{n+2} }f_{n+1}(x)+\cdots+{ (-1)^{m+1} }f_m(x)\,\Biggl|\ \le\ f_n(x)\le \alpha_n .$$ The term on the right hand side of $(1)$ is independent of $x$ and can be made as small as desired, since $\lim\limits_{n\rightarrow\infty }\alpha_n=0$. From this, it follows that the series $\sum\limits_{n=1}^\infty{(-1)^{n+1}}f_n(x)$ is uniformly Cauchy on $D$, and thus uniformly convergent on $D$.

Recall that a sequence $( g_n)$ of real-valued functions is uniformly Cauchy on a set $A$ if for any $\epsilon>0$, there exists a positive integer $N$ so that whenever $n$ and $m$ are positive integers with $m,n\ge N$ we have $\bigl|g_n(x)-g_m(x)\bigr|<\epsilon$ for all $x\in A$.

An easily proven result is that a sequence $( g_n)$ of real-valued functions is uniformly convergent on $A$ if and only if it is uniformly Cauchy on $A$. A proof of this result can be found here.

This result phrased for a series of functions would read as follows: a series $\sum\limits_{n=1}^\infty g_n$ of real valued functions converges uniformly on $A$ if and only if its sequence of partial sums is uniformly Cauchy on $A$.

We say a series of functions is uniformly Cauchy if its sequence of partial sums is. Note that $\sum\limits_{n=1}^\infty g_n$ is uniformly Cauchy on $A$ if and only if for any $\epsilon>0$, there is a positive integer $N$ so that for $m\ge n\ge N$, we have $\Bigl| \sum\limits_{k=n}^m g_k(x)\Bigr|<\epsilon$ for all $x\in A$.

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and why we have: $|(-1)^{n+1} f_n(x)+(-1)^{n+2} f_{n+1}(x)+\cdots+ (-1)^{m+1} f_m(x)|\le f_n(x)$ ? –  xan Apr 25 '12 at 19:16
@xan From the standard error estimate for a convergent alternating series. See here. You are using conditions 1) and 2). Fix an $x$. Then the $f_n(x)$ are nonnegative and decreasing. So $$f_n(x)-f_{n+1}(x)+\cdots +\epsilon f_m(x)\le f_n(x)$$ where $\epsilon$ is either $1$ or $-1$ as appropriate. –  David Mitra Apr 25 '12 at 19:23
David, your posts are always full of great tips. Thank you very much. –  xan Apr 25 '12 at 19:26