Bounded operators that are not closed. If a bounded operator, say $A:D(A)\to X$, have $D(A)=X$ then it is closed.
Can anybody construct an example of a bounded linear operator, without resorting and restricting to $D(A)=X$, that is not closed? 
Please give the examples including densely defined operators,
Thanks in advance.
 A: If $A : \mathcal{D}(A) \subseteq X\rightarrow X$ is a bounded linear operator on a Banach space $X$, then $A$ is closed iff $\mathcal{D}(A)$ is closed. So, for the case where $X$ is complete, the only example of a non-closed bounded operator $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is the restriction of a bounded operator to a non-closed subspace $\mathcal{D}(A)$ of $X$. Any non-closed subspace $\mathcal{D}(A)$ and any $A \in \mathcal{L}(X)$ will work to produce a non-closed $A : \mathcal{D}(A)\subset X\rightarrow X$.
For the case of an incomplete normed linear space $X$, and bounded densely-defined $A : \mathcal{D}(A)\subseteq X\rightarrow X$, some things can be said by densely and isometrically embedding $X$ in its completion $\hat{X}$, and then uniquely extending $A$ to a continuous linear operator $\hat{A} : \hat{X}\rightarrow \hat{X}$. The problem now looks like that of the previous paragraph, but is more difficult to characterize in terms of $X$ alone. However, if you start with a bounded $\hat{A} : \hat{X}\rightarrow\hat{X}$ on a Banach space $\hat{X}$, then an example is produced on some dense non-closed subspace $X$ providing (a) $X$ can be found such that $\hat{A}X \subseteq X$ and (b) a subspace $\mathcal{D}(A)\subset X$ can be found which is dense in $X$ but not closed in $X$ (all topologies are inherited from that of $\hat{X}$.)
A: A (probably to) simple example is $T:D\to X:x\mapsto 0$ for some unclosed dense $D\subsetneq\overline{D}=X$. There $T$ is bounded densely defined but not closed, however, closable.
A more sophisticated bounded nonclosable example would require a little more work since one needs to consider an incomplete space and a nice fitted operator since:
"A bounded operator $T:D\to Y$ with $D\subseteq X$ and $X$ not necessarily complete but $Y$ complete extends to a bounded operator $T_E:\overline{D}\to Y$ that is indeed closed."
