Confused about Domain of Unbounded Operators (Hilbert Spaces)

Question

I understand that a bounded (linear) operator $A$ on a Hilbert space $H$ satisfies a condition $||Av|| \le c ||v|| $ for some fixed real number $c$ and for all $v \in H$. So, the domain of a bounded operator is all of $H$.

1. Presumably an unbounded operator fails to satisfy this requirement in some way. On the other hand some references seem to take "unbounded" as "not necessarily bounded" for example, in nLab, https://ncatlab.org/nlab/show/unbounded+operator, "In particular every bounded operator $A: \mathcal{H} \to \mathcal{H}$ is an unbounded operator", So is an ubounded operator just an operator with a specified domain ?
2. From what I've read, an unbounded operator has a domain $D(A) \subset H$, often taken to be dense in $H$. It isn't clear to me whether or not an unbounded operator is bounded on its domain ? I suspect not since the closed graph theorem says a closed operator defined on the whole space is bounded which suggests the existence of non-closed operators defined for the whole space which are not bounded.

Answer 1

1. It seems that both conventions are used. In my world, the class of unbounded operators includes the class of bounded operators. I think that is a very common convention, although it would perhaps be more precise to refer to such operators as 'partially defined operators'. I was quite surprised when I read the tag description on this site, since I had never before seen the class of unbounded operators defined in a way such as to rule out bounded operators. On the other hand, it is hard to refute the soundness of such a convention.
2. Any unbounded, densely defined operator which is bounded on its domain extends uniquely to a bounded operator on the full Hilbert space. So any closed operator which is not defined on all of $H$ is necessarily not bounded on its domain. However, any unbounded operator A is bounded on its domain with respect to the graph norm $\lVert x \rVert_A = \lVert Ax \rVert + \lVert x \rVert$, and $\mathrm{Dom}(A)$ is complete with respect to the graph norm if and only if $A$ is closed. In this way, boundedness plays a crucial role in the theory of unbounded operators.

Answer 2

1) "In particular every bounded operator $A: \mathcal{H} \to \mathcal{H}$ is an unbounded operator". This is confusing at best. Seems to be saying that bounded operators are a subset of unbounded operators, but that is not the usual terminology AFAIK.

2)For example, multiplication by $n$ defined on vectors in $l^2(N)$ with finite numbers of non-zero components is unbounded.