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I wonder what differences are between the folowwing two versions of Abel's criteria for uniform convergence:

From Elementary classical analysis by Marsden and Hoffman:

Abel's Test. Let $A \subset R^n$ and $\phi_n: A \rightarrow R$ be a sequence of functions which are decreasing; that is, $\phi_{n+1}(x) \leq \phi_n(x)$ for each $x \in A$. Suppose there is a constant $M$ such that$|\phi_n(x)| \leq M$ for all $x \in A$ and all $n$. If $\displaystyle\sum_{n=1}^\infty f_n(x)$ converges uniformly on $A$, then so does $\displaystyle \sum_{n=1}^\infty \phi_n(x)f_n(x)$.

From Wikipedia:

Abel's uniform convergence test. Let $\{g_n\}$ be a uniformly bounded sequence of real-valued continuous functions on a set $E$ such that $g_{n+1}(x) \leq g_n(x)$ for all $x ∈ E$ and positive integers $n$, and let $\{f_n\}$ be a sequence of real-valued functions such that the series $\displaystyle\sum f_n(x)$ converges uniformly on $E$. Then $\displaystyle\sum f_n(x)g_n(x) $converges uniformly on $E$.

  1. Is the additional requirement of continuity for a sequence of functions in Wikipedia the only difference? If not, what else?
  2. Is this continuity unnecessary and can be ignored as in Marsden's? If yes, is Marsden's a more general version? Or do you have a different one?

Thanks and regards!

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Doesn't Marsden specify what $f_n$ is? The assumption of continuity in Wiki's version is superfluous, as well as the assumption that $f_n$ be real-valued, they could be complex-valued as well. I see no other difference except that assumption. There are many of results of this flavor, see e.g. Dirichlet's criterion and Dedekind's criterion for two of them. –  t.b. Aug 23 '11 at 16:34
    
@Theo: Thanks! Does Marsden? I cannot find it. –  Tim Aug 23 '11 at 16:40
    
I'm asking. I don't know. Probably he has a universal assumption what $f$ and $f_n$ is supposed to mean. Google doesn't let me look and I never held that book in my hands. –  t.b. Aug 23 '11 at 16:42
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1 Answer

I don't think that continuity is necessary.

Fix $\varepsilon>0$. The uniform convergence of the series implies that there exists $n$ such that $$ \left|\sum_{k=n}^m f_k(x)\right|<\varepsilon \ \ \ \text{ for all } x\in A, m>n. $$ The key to the proof is "summation by parts": we have $$ \sum_{k=n}^m\phi_k(x)f_k(x)=\phi_m(x)\sum_{k=n}^mf_k(x)-\sum_{k=n}^m(\phi_{k+1}(x)-\phi_k(x))\sum_{j=n}^kf_j(x). $$ Then

\begin{align} \left|\sum_{k=n}^m\phi_k(x)f_k(x)\right|&\leq|\phi_m(x)|\,\left|\sum_{k=n}^mf_k(x)\right|+\sum_{k=n}^m\,|\phi_{k+1}(x)-\phi_k(x)|\,\left|\sum_{j=n}^kf_j(x)\right|\\ &\leq M\varepsilon + \varepsilon \sum_{k=n}^m\,\phi_{k}(x)-\phi_{k+1}(x) =M\varepsilon + \varepsilon(\phi_n(x)-\phi_{m+1}(x))\\ &<3M\varepsilon. \end{align}

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