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Thread: how to prove the Abel's test for uniform convergence?

  1. #1
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    how to prove the Abel's test for uniform convergence?

    The Abel's test for uniform convergence states: Assume$\displaystyle \sum f_n(x)$ converges uniformly on E. A uniformly bounded sequence of functions $\displaystyle \{g_n(x)\}$ satisfies $\displaystyle g_{n+1}(x)\leq g_n(x)$ for all n and x in E. Then $\displaystyle \sum f_n(x)g_n(x)$ converges uniformly.
    My tentative proof is as follows:
    Let $\displaystyle s_n(x)=\sum\limits_{k = 1}^n f_k(x)g_k(x), F_n(x)=\sum\limits_{k = 1}^n f_k(x), x\in E$. By the partial summation formula, we have $\displaystyle s_n(x)=F_n(x)g_1(x)+\sum\limits_{k = 1}^n (F_n(x)-F_k(x))(g_{k+1}(x)-g_k(x))$. So if n>m we can get $\displaystyle s_n(x)-s_m(x)=(F_n(x)-F_m(x))g_1(x)+\sum\limits_{k = m+1}^n (F_n(x)-F_k(x))(g_{k+1}(x)-g_k(x))$. If the uniform bound for $\displaystyle \{g_n(x)\}$ is M, the absolute value of the above difference satisfies $\displaystyle |s_n(x)-s_m(x)|\leq M|F_n(x)-F_m(x)|+2M\sum\limits_{k = m+1}^n |F_n(x)-F_k(x)|$. From the fact that $\displaystyle \sum f_n(x)$ converges uniformly, for any given $\displaystyle \epsilon>0$, there is an N such that for all m,n>N we have $\displaystyle |F_n(x)-F_m(x)|<\epsilon$. So the right side of the above inequality $\displaystyle \leq [1+2(n-m)]M\epsilon$. But the variable (n-m) makes the Cauchy condition inapplicable. Can you tell me how to proceed, or is there any other better proof? Thanks!
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  2. #2
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    the absolute difference should go as follows:
    For $\displaystyle N<m<n, |s_n(x)-s_m(x)|\leq M|F_n(x)-F_m(x)|+$$\displaystyle \sum\limits_{k = m+1}^n |F_n(x)-F_k(x)|[g_k(x)-g_{k+1}(x)]\leq \epsilon(M+\sum\limits_{k = m+1}^n [g_k(x)-g_{k+1}(x)])$$\displaystyle =\epsilon(M+g_{m+1}(x)-g_{n+1}(x))\leq 3M\epsilon$. Now the Cauchy condition can be applied.
    Another proof can be found in K. Knopp's "Theory and application of infinite series", P346-347. This proof is more complicated than the above one.
    Last edited by zzzhhh; Jan 27th 2010 at 01:03 PM.
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