Are these two Abel's criteria for uniform convergence different?

Solution 1:

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}