On the uniform convergence of series of functions with indexes in $\mathbb{Z}$

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Questions:

1. Is the definition below the common one for the uniform convergence of a series of functions of the form $\displaystyle\sum_{k=-\infty}^{\infty}f_k$ ?

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

3. 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.

Answer 1

1. 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.

2. 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.

3. 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.