Related to this question.
Questions:
Is the definition below the common one for the uniform convergence of a series of functions of the form $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ ?
Are the statements of the three propositions below true and, if so, are their presented proofs correct?
Is the definition below a particular case of a more general definition, for example that of the uniform convergence of a net of functions (see here)? In case the propositions below are true, are they particular cases of more general results?
Definition: Let $E$ be a set and $f_k:E\to\mathbb{C}$ ($k\in\mathbb{Z}$) be functions. Say that $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ converges uniformly (on $E$) to $f:E\to\mathbb{C}$ if $$ \forall\epsilon>0,\exists N,\forall m,n>N:\sup_{x\in E}\left|\sum_{k=-m}^nf_k(x)-f(x)\right|<\epsilon $$
Proposition: $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ converges uniformly to $f$ if, and only if, $\displaystyle\sum_{k=0}^{\infty}f_k$ and $\displaystyle\sum_{k=1}^{\infty}f_{-k}$ both converge uniformly and $\displaystyle\sum_{k=0}^{\infty}f_k+\displaystyle\sum_{k=1}^{\infty}f_{-k}=f$.
Proof: $[\Leftarrow]$ Let $\epsilon>0$. Then there are functions $g,h:E\to\mathbb{C}$ such that $g+h=f$ for which $$ \exists N_1,\forall n>N_1:\sup_{x\in E}\left|\sum_{k=0}^nf_k(x)-g(x)\right|<\frac{\epsilon}{2}\\ \exists N_2,\forall m>N_2:\sup_{x\in E}\left|\sum_{k=1}^mf_{-k}(x)-h(x)\right|<\frac{\epsilon}{2} $$ so that if $m,n>\max\{N_1,N_2\}$ then \begin{align} \sup_{x\in E}\left|\sum_{k=-m}^nf_k(x)-f(x)\right|&\leq\sup_{x\in E}\left|\sum_{k=0}^nf_k(x)-g(x)\right|+\sup_{x\in E}\left|\sum_{k=1}^mf_{-k}(x)-h(x)\right|\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon \end{align}
$[\Rightarrow]$ Let $\epsilon>0$. Then there exists $N$ such that $$ \sup_{x\in E}\left|\sum_{k=-m}^nf_k(x)-f(x)\right| $$ whenever $m,n>N$. Hence if $m>n>N$ then \begin{align} \sup_{x\in E}\left|\sum_{k=0}^mf_k(x)-\sum_{k=0}^nf_k(x)\right|&=\sup_{x\in E}\left|\sum_{k=n+1}^mf_k(x)\right|\\ &=\sup_{x\in E}\left|\left(\sum_{k=-n}^mf_k(x)-f(x)\right)-\left(\sum_{k=-n}^nf_k(x)-f(x)\right)\right|\\ &\leq\sup_{x\in E}\left|\sum_{k=-n}^mf_k(x)-f(x)\right|+\sup_{x\in E}\left|\sum_{k=-n}^nf_k(x)-f(x)\right|\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon \end{align} This shows that the series $\displaystyle\sum_{k=0}^{\infty}f_k(x)$ satisfies the Cauchy condition for uniform convergence and converges uniformly.
A similar argument shows that $\displaystyle\sum_{k=1}^{\infty}f_{-k}(x)$ also converges uniformly.
Finally, $$ \sum_{k=0}^{\infty}f_k+\sum_{k=1}^{\infty}f_{-k}=\sum_{k=-\infty}^{\infty}f_k=f $$
Proposition (Weierstrass M-test): Let $f_k:E\to\mathbb{C}$ ($k\in\mathbb{Z}$) be functions and $M_k\geq0$ be real numbers such that $|f_k(x)|\leq M_k$ for all $x$ and all $k$ and $\displaystyle\sum_{k=-\infty}^{\infty}M_k<\infty$. Then $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ converges uniformly on $E$.
Proof: Since $|f_k(x)|\leq M_k$ for all $x$ and all $k\geq0$ and $$ \sum_{k=0}^{\infty}M_k\leq\sum_{k=-\infty}^{\infty}M_k<\infty $$ the usual version of the Weierstrass M-Test shows that $\displaystyle\sum_{k=0}^{\infty}f_k$ converges uniformly.
Similarly, $\displaystyle\sum_{k=1}^{\infty}f_{-k}$ also converges uniformly.
By the above proposition, it follows that $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ converges uniformly.
Proposition: If $I\subseteq\mathbb{R}$, $f_k:I\to\mathbb{C}$ are continuous for all $k\in\mathbb{Z}$ and $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ converges uniformly to $f$ on $I$, then $f$ is continuous.
Proof: Then $\displaystyle\sum_{k=0}^{\infty}f_k$ and $\displaystyle\sum_{k=1}^{\infty}f_{-k}$ both converge uniformly, say to $g$ and $h$ respectively. So $g$ and $h$ are continuous. Since $f=g+h$, it follows that $f$ is continuous.