I was wondering whether the following statement is true or not?
If $A$ is closed, then it follows from the closed-graph theorem that it is bounded iff $D(A)$ is closed.
I found this in a chapter of a book on operator theory; a counterexample can be the operator $A=d/dx: D(A)(\subset C[a,b])\to C[a,b]$ which is closed when the domain $D(A)$ is chosen to be $C^1[a,b]$. Under the sup-norm, space $D(A)=C^1[a,b]$ is closed, however, the operator is not bounded.