Prove that $X_0$ is a closed subspace of $X$

Question

I encounter the following exercise in functional analysis:

Let $(X,\|\cdot\|_{X})$ and $(Y,\|\cdot\|_{Y})$ be Banach spaces and $\{T_n\}$ be a family of uniformly bounded linear maps form $X$ to $Y$, i.e. $\|T_n\| \leqslant C$ for some positive constant $C$ for all $n$.

Let $X_0 = \{x \in X \mid \lim_{n \to \infty} T_n(x) \text{ exists}$}. Show that $X_0$ is a closed subspace of $X$.

Here's my approach: Let $y$ be a limit point of $X_0$, then there exists a sequence $\{x_k\}$ in $X_0$ that converges to $y$ in the $\|\cdot\|_X$ norm. I want to show that $\{T_n(y) = \lim_{k \to \infty} T_n(x_k)\}$ is a Cauchy sequence in $Y$ so that $y \in X_0$. But I don't know how to go from there.

Answer 1

Assume $x_j\to x$, with all $x_j\in X_0$. Given $\epsilon$, let $j_0$ be so that $\|x_{j_0}-x\|<\frac{\epsilon}{3C}$. Now find $N$ so that $m, n>N$ implies $\|T_m x_{j_0}-T_n x_{j_0}\|<\frac{\epsilon}{3}$. Thus $$ \begin{aligned} \|T_mx-T_nx\|=&\|T_mx-T_mx_{j_0}+T_mx_{j_0}-T_nx_{j_0}+T_nx_{j_0}-T_nx\|\\ \leq & \|T_mx-T_mx_{j_0}\|+\|T_mx_{j_0}-T_nx_{j_0}\|+\|T_nx_{j_0}-T_nx\|\\ \leq & C\|x-x_{j_0}\|+\|T_mx_{j_0}-T_nx_{j_0}\|+C\|x_{j_0}-x\|\\ \leq & C\frac{\epsilon}{3C}+\frac{\epsilon}{3}+C\frac{\epsilon}{3C}=\epsilon. \end{aligned}$$