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

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.

## 1 Answer

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}