What can you say about the limit of a series as a linear transformation? I have a question.
Let $V$ be the space of convergent real infinite sequences.
I know that the following function: $lim: V \rightarrow \textbf{R}, lim(a_n) = lim_{n\rightarrow \infty} a_n$ is a linear transformation (Because $lim(a_n + b_n) = lim(a_n) + lim(b_n), and\ lim(ca_n) = clim(a_n)$). Is there anything special about this from a linear algebra perspective? The only thing I could find is that it's a functional. Is there anything special you can say about this from an abstract algebra persepctive?
Thanks!
 A: Good question! This is related to the notion of a Banach limit although that's not quite the same (there the intent is to try to assign a meaningful limit to a sequence which doesn't converge in the usual sense). Here are some other properties of $\lim$ as a linear functional on $V$:

*

*For any $k$, $\lim$ does not depend on the first $k$ terms of a sequence; explicitly, for any $k$, if $\{ a_i \}, \{ b_i \}$ are two sequences such that $a_i = b_i$ for $i \le k$, then $\lim a_i = \lim b_i$. This condition rules out the other obvious linear functionals on $V$, which are given by taking linear combinations of the projections $\{ a_i \} \to a_k$.


*$\lim$ is not only linear, it is multiplicative: we have $\lim a_n b_n = (\lim a_n)(\lim b_n)$. We also have that the limit of the constant sequence with constant value $1$ is $1$; altogether this implies that $\lim$ is a ring homomorphism, or more precisely an $\mathbb{R}$-algebra homomorphism.

Proposition: These properties, together with linearity, uniquely determine $\lim$. That is, $\lim$ is the unique map $V \to \mathbb{R}$ which is linear, multiplicative (including that it sends $1$ to $1$), and for all $k$ does not depend on the first $k$ terms of a sequence.

Proof. Probably a fairly concrete argument is possible here but here's an abstract one. $V$ can be identified with the $\mathbb{R}$-algebra of continuous real-valued functions on the one-point compactification $\mathbb{N}_{\infty} = \mathbb{N} \cup \{ \infty \}$ of $\mathbb{N}$, where the function corresponding to a convergent sequence $\{ a_i \}$ takes the value $a_i$ at $i \in \mathbb{N}$ and takes the value $\lim a_i$ at $\infty$.
Since $\mathbb{N}_{\infty}$ is compact Hausdorff, a standard exercise (which is done e.g. here and which I believe is an exercise in Atiyah-MacDonald) implies that every $\mathbb{R}$-algebra homomorphism $V \to \mathbb{R}$ is an evaluation homomorphism at some point of $\mathbb{N}_{\infty}$. So any such homomorphism must either be one of the projections $\{ a_i \} \to a_k$ or $\lim$, and the condition that it does not depend on the first $k$ terms of a sequence uniquely picks out $\lim$. $\Box$
