I've found on a book an example of closed operator which is not continuous.
It is the following:
$$
f:\mathbb{R} \rightarrow \mathbb{R} \quad f(t)=
\begin{cases}
\frac{1}{t} & \text{if $\, t \ne 0$} \\
0 & \text{if $\, t=0$}
\end{cases}
$$
Now, an equivalent definition of closedness says:
Let $(X,\Vert . \Vert_X)$ and $(Y, \Vert . \Vert_Y)$ be two normed vector spaces and $\, f:X \rightarrow Y$ a function between them; then $f$ is closed if: $\qquad x_n \rightarrow x \, \land \, f(x_n) \rightarrow y \quad \Rightarrow \quad f(x)=y$
Recalling the previous example, let $\, x_n=\frac{1}{n}$. We have: $$ x_n \rightarrow 0 \quad \land \quad f(x_n) = n \rightarrow +\infty \quad \text{BUT} \quad f(0)=0 \ne +\infty $$
My explanation is that it is not correct to say that $f(x_n) = n \rightarrow +\infty$ in $\mathbb{R}$. Or better: according to the definition of convergence in normed vector space, $f(x_n)$ does not converge in $\mathbb{R}$. Therefore, the definition of closed linear operator is fullfilled.
However, if we consider $f:\mathbb{R} \rightarrow \mathbb{R}^{*}$ the previous counterexample ($x_n = \frac{1}{n}$) is correct - since $+\infty \in \mathbb{R}^{*}$, so that $f(x_n) \rightarrow +\infty$ in $\mathbb{R}^{*}$. Therefore, $f$ is not closed in this last case.
Am I right?
If so, could you provide me a more significant example of closed but not continuous operator?
Thanks in advance.
