Closed operator I've got a very straightforward question : if $T : B \rightarrow B$ is a linear continuous operator and $B$ is a Banach space, is $T$ a closed operator?
This is obviously true in finite dimension, but I'm not sure what can happen in infinite dimension. Maybe it can't get "too bad" if $B$ is a Banach.
 A: 1) If closed means the image of every closed set is closed: No. 
If this was true, any injective bounded operator with dense range would have to be surjective, whence invertible on a Banach space by the Banach isomorphism theorem. The compact diagonal operator $T(x_n)=(x_n/n)$ on $\ell^2$ is a natural counterexample.

2) If closed means the operator is closed, that is its graph is closed in $B\times B$: Yes.
Assume $(x_n,Tx_n)$ converges to $(x,y)$ in $X\times X$, that is $x_n\rightarrow x$ and $Tx_n\rightarrow y$ in $X$. By continuity of $T$, $Tx_n\rightarrow Tx$. By uniqueness of a limit in a metric space, $Tx=y$. So $(x,y)=(x,Tx)$ belongs to the graph of $T$.
A: Yes it is closed. Look at the definition of closedness, it is a weaker definition than continuity.
If $T$ is continuous, then for every sequence $x_{n}$ in the domain of $T$ converging to some $x$ in domain of $T$, the sequence $Tx_{n}$ has to converge and has to converge to $Tx$. The sequence $Tx_{n}$ has no choice!
And what is the definition of closed? If $x_{n}$ is sequence in domain of $T$ converging to some $x$ in domain of $T$, then we only require that if the sequence $Tx_{n}$ converges, then it should converge to the "correct" limit, that is $Tx$. But it may not converge at all.
In case of conitnuity, however, the poor sequence $Tx_{n}$ has to converge and converge to $Tx$. Thus, continuity implies closedness.
