"Functional example" of Cauchy sequence which is not convergent We know that if a sequence $x_n$ is Cauchy in a metric space, it does not have to be convergent. For example, consider $x_n = \frac{1}{n}$ in metric space $X = (0,1]$.
Inspired by part (a) of this problem, I am wondering what would be an (perhaps "exotic") example where $T_n$ is a Cauchy sequence of bounded linear operator in $L(X,X)$ where $X$ is a normed vector space, but $T_n$ does not converge? 
 A: I assume that the norm on $L(X,X)$ is the following, if $A \in L(X,X)$ then the norm is
\begin{equation}
\| A \| := \sup_{\|x\|=1} \|Ax\|.
\end{equation}
Your question basically amounts to the question if the space $L(X,X)$ is complete. The answer is that this is true precisely when $X$ is complete. Let us show this as follows.
First let me give an example in the case where the normed space $X$ is not complete. In this case let $x_{n}$ be a cauchy sequence that does not converge. Let $e_{i}$ be a normalized basis for $X$. Then define, for each $n \in \mathbb{N}$ a linear operator by $T_{n}e_{1} = x_{n}$ and $T_{n} e_{i} = 0$, for $i \neq 1$. This sequence of operators is Cauchy,
\begin{equation}
 \|T_{n} - T_{m} \| = \|x_{n} - x_{m} \|,
\end{equation}
but does not converge. (If it did it, we would have $Te_{1} = \lim_{n \rightarrow \infty} x_{n}$, since
\begin{equation}
\|T e_{1} - x_{n} \| = \| T e_{1} - T_{n} e_{1}\| = \|(T - T_{n})e_{1} \| \leqslant \| T - T_{n} \|,
\end{equation}
 a contradiction.)
Suppose that $X$ is complete. Let $T_{n}$ be a Cauchy sequence in $L(X,X)$. Then given any $x \in X$, define a sequence $y_{n}^{x} := T_{n}x$. Claim: for any $x \in X$, this sequence is Cauchy. Without loss of generality, assume $\|x\| = 1$. Let $\epsilon > 0$. Then by assumption, there exists an $N>0$ such that for all $n,m >N$ one has $\|T_{n} - T_{m}\| < \epsilon$. We now compute
\begin{equation}
\|y_{n}^{x} - y_{m}^{x} \| = \|(T_{n}-T_{m})x\| \leqslant \|T_{n} - T_{m}\| < \epsilon.
\end{equation}
It follows that the sequence $y_{n}^{x}$ is Cauchy, for each $x \in X$. By completeness it follows that $\lim_{n \rightarrow \infty} y_{n}^{x}$ exists for each $x \in X$. This allows us to define $Tx = \lim_{n \rightarrow \infty} T_{n}x$.
The reverse triangle inequality
\begin{equation}
| \|T_{n}\| - \|T_{N} \| |  \leqslant \| T_{n} - T_{N} \| < \epsilon,
\end{equation}
tells us that the family of operators $T_{n}$ is bounded. The Principle of uniform boundedness can now be leveraged to prove that the operator defined as such is bounded.
I do not have time to complete the argument, but I think most ingredients are there now.
