I'm having some difficulty with a homework problem regarding Exercise #14 from Chapter 2 of Rudin's $\textit{Functional Analysis}$.
(a) Suppose $X,Y$ are topological vector spaces, $\{f_n\}$ is an equicontinuous sequence of linear maps from $X$ into $Y$, and $C$ is the set of all $x$ at which $\{f_n(x)\}$ is a Cauchy sequence in $Y$. Prove that $C$ is a closed subspace of $X$.
(b) Assume, in addition to the hypotheses in (a), $Y$ is an $F$-space and that $\{f_n(x)\}$ converges in some dense subset of $X$. Prove that then $f(x)= \lim_{n \rightarrow \infty} f_n(x)$ exists for every $x \in X$ and that $f$ is continuous.
Well, for (a), so far I just have that to every neighborhood $W$ of $0$ in $Y$, there corresponds a neighborhood $V$ of $0$ in $X$ such that $f_n(V) \subset W$, by definition of equicontinuity. Also, I denote $C= \{x: d(f_n(x),f_m(x))< \epsilon\}$, for some $N$ such that $m,n \geq N$ for each $\epsilon > 0$. This is where I got stuck. Isn't it already clear that $C$ is a subspace of $X$? I have not yet been able to show that it is closed. Perhaps it is some theorem I am missing? How about for (b)?
Actually, it seems that (b) might simply follow from Theorem 2.7(b) on pp. 45–46, where it is stated that if $L$ is the set of all $x \in X$ at which $f(x)= \lim_{n \rightarrow \infty} f_n(x)$ exists, if $L$ is of the second category in $X$, and if $Y$ is an $F$-space, then $L=X$ and $f: X \rightarrow Y$, give or take a condition. This might be wrong, however.
Let's recall the definition first. A sequence $(x_n)_{n\in\mathbb{N}}$ in a topological vector space $E$ is a Cauchy sequence if and only if for all neighbourhoods $V$ of $0$, there is an $N_V \in \mathbb{N}$ such that for all $n,m \geqslant N_V$, we have $x_n - x_m \in V$.
If $(x_n)$ and $(y_n)$ are two Cauchy sequences, then so is $(z_n)$, where $z_n = x_n + y_n$: Let $V$ an arbitrary neighbourhood of $0$. Choose a neighbourhood $U$ of $0$ with $U+U \subset V$. Then there are $K,L \in \mathbb{N}$ such that $n,m\geqslant K \Rightarrow x_n-x_m \in U$ and $n,m \geqslant L \Rightarrow y_n - y_m \in U$, since $(x_n)$ and $(y_n)$ are Cauchy sequences. Then, for $n,m \geqslant N_V = \max \{K,L\}$, we have $$z_n-z_m = (x_n-x_m) + (y_n-y_m) \in U+U\subset V.$$
$V$ was arbitrary, hence $(z_n)$ is a Cauchy sequence. I'll leave the proof that $(\lambda x_n)$ is a Cauchy sequence when $(x_n)$ is a Cauchy sequence and $\lambda\in\mathbb{K}$ as an exercise.
This shows that for every vector space $X$ and topological vector space $Y$, and every sequence of linear maps $f_n \colon X \to Y$, the space
$$C_{(f_n)} = \{x \in X : \bigl(f_n(x)\bigr) \text{ is a Cauchy sequence}\}$$
is a linear subspace of $X$.
Now we want to show that if $X$ is also a topological vector space and $(f_n)$ is equicontinuous, then $C_{(f_n)}$ is a closed subspace of $X$.
So let $x \in \overline{C_{(f_n)}}$. We must show that for every neighbourhood $V$ of $0$ in $Y$ there is an $N_V\in \mathbb{N}$ such that $f_n(x) - f_m(x) \in V$ for all $n,m\geqslant N_V$. For a nighbourhood $V$ of $0$, choose a balanced neighbourhood $U$ of $0$ with $U+U+U \subset V$. Choose, by equicontinuity, a balanced neighbourhood $W$ of $0$ in $X$ with $f_n(W) \subset U$ for all $n$. Since $x \in \overline{C_{(f_n)}}$, there is a $\xi \in (x + W)\cap C_{(f_n)}$. Since $\bigl(f_n(\xi)\bigr)$ is a Cauchy sequence, there is an $N_V$ with $n,m \geqslant N_V \Rightarrow f_n(\xi) - f_m(\xi) \in U$. I'll leave the short remainder to fill in.
For part b), note that if $Y$ is an $F$-space, then being a Cauchy sequence and being a convergent sequence are identical conditions. So by part a), the set $\{x\in X : \bigl(f_n(x)\bigr) \text{ is convergent}\}$ is a closed linear subspace of $X$. By the assumption, it contains a dense subset of $X$, hence ...
That $f(x) = \lim\limits_{n\to\infty} f_n(x)$ is then a linear map is a standard argument. To see that it is continuous, consider an arbitrary closed neighbourhood $V$ of $0$ in $Y$. By the equicontinuity of $(f_n)$, there is a neighbourhood $W$ of $0$ in $X$, such that ... . Then $f(W)\dotsc$.