The following is the Abel's Theorem. I proved it my own way.
Let $g_n(x)$ be a sequence of real-valued functions such that $g_{n+1}(x)\leq g_{n}(x),\forall \,x\in T$ and $n\in\Bbb{N}.$ If $\{g_{n}\}$ is uniformly bounded on $T$, and if $\sum f_n(x)$ converges uniformly, then $\sum f_n(x)g_n(x)$ converges uniformly on $T.$
MY TRIAL
Let \begin{align}F_n(x)=\sum^{n}_{i=1} f_n(x),\forall \,x\in T,\;n\in\Bbb{N}\end{align}
Let $N>M,$ then
\begin{align}
\left|S_N(x)-S_M(x) \right|
&=\left|\sum^{N}_{n=1} f_n(x)g_n(x)-\sum^{M}_{n=1} f_n(x)g_n(x)\right|\\&=\left|\sum^{N}_{n=M+1} f_n(x)g_n(x)\right|\\&=\left|\sum^{N}_{n=M+1} \left[F_n(x)-F_{n-1}(x)\right]g_n(x)\right|\\&=\left|\sum^{N}_{n=M+1}F_n(x)g_n(x)-\sum^{N}_{n=M+1}F_{n-1}(x)g_n(x)\right|\\&=\left|
\left[F_{M+1}(x)-F_{M}(x)\right]g_{M+1}(x)+\sum^{N}_{n=M+2}\left[F_n(x)-F_{n-1}(x)\right]g_n(x)\right|\\&\leq
\left|F_{M+1}(x)-F_{M}(x)\right| \left| g_{M+1}(x)\right|+\sum^{N}_{n=M+2}\left|F_n(x)-F_{n-1}(x)\right|\left|g_n(x)\right|\end{align}
Since $g_n$ is uniformly bounded, there exists $M>0$ such that
\begin{align} \left|g_n(x)\right|\leq M_{1},\forall \,x\in T\end{align}
Also, by uniform convergence of $\sum f_n(x)$, the sequence of partial sums is uniformly Cauchy. Let $\epsilon>0.$ Then, there exists $N_1=N(\epsilon)$ such that $\forall\,N>M\geq N_1,\;\forall\,x\in T$
\begin{align} \left|F_{N}(x)-F_{M}(x)\right|<\frac{\epsilon}{M_{1}+1}\end{align}
Hence, $\forall\,N>M\geq N_1,\;\forall\,x\in T$
\begin{align}
\left|S_N(x)-S_M(x) \right|& \leq
\left|F_{M+1}(x)-F_{M}(x)\right| M_{1}+M_{1}\sum^{N}_{n=M+2}\left|F_n(x)-F_{n-1}(x)\right|\\&< \frac{M_{1}\epsilon}{M_{1}+1}+\frac{M_{1}\epsilon}{M_{1}+1}<\epsilon\end{align}
And we are done!
Please, can you help check if my proof is correct? Constructive criticisms are welcome! Thanks!
Best Answer
To resolve some confusion, here is a concise proof of Abel's test for uniform convergence. The trick is to apply summation by parts with $f_n(x) = R_n(x) - R_{n+1}(x)$ where $R_n(x) = \sum_{j=n}^\infty f_j(x)$, rather than use partial sums $\sum_{j=1}^n f_j(x)$.
Summing by parts, we get
$$\sum_{j=n}^m f_j(x) g_j(x) = \sum_{j=n}^m (R_j(x)- R_{j+1}(x)) g_j(x) \\ = R_n(x) g_n(x) - R_{m+1}(x) g_m(x) + \sum_{j=n}^{m-1}R_{j+1}(x) ( g_{j+1}(x) - g_j(x)) $$
Using the monotonicity of the sequence $\{g_n\}$ where $|g_{j+1}(x) - g_j(x)| = g_{j}(x) - g_{j+1}(x)$, it follows that
$$\left|\sum_{j=n}^m f_j(x) g_j(x)\right| \leqslant |R_n(x)| |g_n(x)| + |R_{m+1}(x)| |g_j(x)| + \sum_{j=n}^{m-1}|R_{j+1}(x)| ( g_{j}(x) - g_{j+1}(x))$$
By hypothesis, $\{g_n\}$ is uniformly bounded and there exists $B > 0$ such that $|g_n(x)| \leqslant B$ for all $n$ and all $x \in T$. Since $\sum f_n$ is uniformly convergent there exists $n$ such that for all $j \geqslant n$ and all $x\in T$ we have $|R_j(x)| < \epsilon/(4B)$.
Thus,
$$\left|\sum_{j=n}^m f_j(x) g_j(x)\right| \leqslant \frac{\epsilon}{4B} \cdot B + \frac{\epsilon}{4B} \cdot B + \frac{\epsilon}{4B}\cdot \sum_{j=n}^{m-1}(g_j(x) - g_{j+1}(x)) \\ = \frac{\epsilon}{2} + \frac{\epsilon}{4B}(g_n(x) - g_m(x))\leqslant \frac{\epsilon}{2} + \frac{\epsilon}{4B}(|g_n(x)| + |g_m(x)|) \leqslant \epsilon$$
This implies uniform convergence by the Cauchy criterion.
Original Hint:
Summing by parts with $S_n(x) = \sum_{j=1}^n f_j(x)$, we get
$$\sum_{j=m+1}^n f_j(x) g_j(x) = g_{n+1}(x) S_n(x) - g_{m+1}(x) S_m(x) + \sum_{j=m+1}^n S_j(x)(g_j(x) - g_{j+1}(x)),$$
which implies
$$\left|\sum_{j=m+1}^n f_j(x) g_j(x)\right| \leqslant |S_n(x)||g_{n+1}(x) - g_{m+1}(x)| + |g_{m+1}(x)| |S_m(x) - S_n(x)| + \sum_{j=m+1}^n |S_j(x)|(g_j(x) - g_{j+1}(x))$$