On the uniform convergence of series of functions with indexes in $\mathbb{Z}$ 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.
 A: 
  
*
  
*Is the definition below the common one for the uniform convergence of a series of functions of the form $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ ?
  

Yes, but there are situations (e.g. Fourier series) where one only considers the symmetric partial sums
$$\sum_{k = -n}^n f_k$$
and uses the same notation. In these cases, one says the series is uniformly convergent if the sequence of symmetric partial sums is uniformly convergent.


  
*Are the statements of the three propositions below true and, if so, are their presented proofs correct?
  

Yes. But you forgot a $< \frac{\epsilon}{2}$ at the beginning of the $[\Rightarrow]$ direction of the proof of the first proposition.


  
*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)?
  

Yes. A series $\sum_{k = -\infty}^{\infty} f_k$ is a special case of a net, the $\mathbb{N}\times \mathbb{N}$-indexed net
$$F_{(m,n)} := \sum_{k = -m}^n f_k,$$
where the index set is endowed with the product order (that is, we have $(m_1,n_1) \leqslant (m_2,n_2)$ if and only if $(m_1 \leqslant m_2)\land (n_1 \leqslant n_2)$).

In case the propositions below are true, are they particular cases of more general results?

The third proposition is definitely a particular case of the more general and immensely useful fact that the limit of a uniformly convergent net (or filter) of continuous functions is continuous.
The Weierstraß $M$-test is particular to series of functions. One can generalise it to arbitrary index sets, but since a sum of non-negative real numbers can only be finite if at most countably many of the numbers are non-zero, this isn't really a generalisation, it can however make the notation more convenient to allow arbitrary index sets. We can generalise the codomain; the same statement, with essentially the same proof works if the codomain is an arbitrary normed space. This is also useful.
I don't see a more general form for the first proposition (except for generalising the codomain), the splitting of the series into the part with negative indices and the part with non-negative indices (of course we could also split at any other integer, we're not tied to $0$) doesn't seem to have a more general close analogue.
