In Rudin's functional analysis, the remark of closed graph theorem says the statement:
If $\{x_n\}$ is a sequence in $X$ such that
$x=\displaystyle\lim_{n\rightarrow\infty} x_n$, $y=\displaystyle\lim_{n\rightarrow\infty}f(x_n)$ exist, then $y=f(x)$
is equivalent to the statement the graph $(x,f(x))$ is closed.
However, I am thinking that the first statement is equivalent to $f$ is continuous, that confuses me.
Consider the following counterexample: take $f:[0,1]\to \Bbb R$ to be given by $$ f(x)= \begin{cases} 1/x& x\neq 0\\ 0& x=0 \end{cases} $$ Then $f$ is discontinuous, but it has a closed graph.
Similarly, a closed operator $T:X\to Y$ fails to be continuous because of "vertical asymptotes", i.e. subspaces on which the norm of $\|Tx\|$ "diverges to $\infty$" (at least, what I've said is a vague interpretation of the closed graph theorem).
For example, let $X=\mathcal{C}^{\infty}[0,1]$ consist of all infinitely differentiable functions on $(0,1)$ that have, along with all orders of derivative, continuous extensions to $[0,1]$. Put the max norm on $X$. Then $Tf=f'$ is defined on $X$.
$T$ is not continuous because $f_n = \frac{1}{n}x^n$ tends to $0$ in norm, but $Tf_n = x^n$ does not converge to $0$ in norm because $\|x^n\|=1$ for all $n$; $T$ is definitely not bounded because $\|Tf_n\| \not\le C\|f_n\|$ for any some constant $C$ and all $n$.
However, $T$ is closed. To see that $T$ is closed, suppose $\{ f_n \} \subset X$ converges in $X$ to $f$ and suppose $\{ Tf_n \}$ converges in $X$ to $g\in X$. Then, $$ f_n(x)=f_n(0)+\int_{0}^{x}(Tf_n)(t)dt $$ gives the following in the limit $$ f(x) = f(0)+\int_{0}^{x}g(t)dt. $$ Therefore $Tf=g$.
If $X$ is a Banach space, and $T$ is a closed linear operator on $X$, then $T$ must be continuous. That's implied by the Closed Graph Theorem. If $X$ is not complete (such as this space) or if $T$ is not defined on the entire space $X$, then $T$ may be closed without being bounded.
