Let $X=\ell^2:=\{(a_1, a_2, ...): \sum^{\infty}_{i=1} a_i<\infty\}$. Define $T_n:X\to\mathbb{R}$ to be $T_n(x):=(\log n)a_n$.
It is obvious that $||T_n||=\log n$, which is unbounded, violating the uniform boundedness theorem. However, I wonder which of the assumption(s) in the theorem is violated.
During the lecture, the lecturer said that it is because there exists $x\in X$ such that $||T_n x||$ is not bounded and gave an example that $x=(0,0,...,0,\frac{1}{\log n},0,...,0,0,0,...,0,\frac{1}{\log n},0,...)$ which means that there is sometimes an entry $\frac{1}{\log n}$, and the number of entries between the two non-zero terms are larger. However, I did not quite get the idea.
Let $x = (a_1,a_2,...)$ where
$$a_n=\begin{cases} 2^{-k/2} & \text{if $n = 2^{2^k}$}\\ 0 & \text{otherwise}. \end{cases}$$
Clearly $x\in \ell^2$, but for $n=2^{2^k}$ we have
$$T_n(x)=\log(2)\cdot 2^{k/2},$$
showing $||T_n(x)||$ is unbounded.