In books, I always that a bounded linear operator on a Hilbert space is defined on all the Hilbert space, while an unbounded linear operator can not be defined on all the Hilbert space.
Nevertheless, I am not able to find a demostration for that.
Specifically, let $\mathcal{H}$ be a complex Hilbert space and let $A\in\mathcal{B}(\mathcal{H})$ be a bounded linear operator on $\mathcal{H}$.
I understood that the domain $D(A)$ of $A$ is the set:
$$
D(A):=\left\{\psi\in\mathcal{H}\,: A(\psi)\in\mathcal{H}\right\}
$$
I would like to know how to prove that the fact that $A$ is bounded implies that $D(A)=\mathcal{H}$, and the converse.
Thank You.
Best Answer
It doesn't. There's nothing to prevent you from defining a bounded linear operator on a proper subspace. There are also unbounded linear operators defined on the whole space (although you won't find explicit examples of this, because it requires some form of the Axiom of Choice).
However, a closed linear operator defined on the whole space is bounded (Closed Graph Theorem), and a closed linear operator that is bounded must have a closed domain.
EDIT: Here's the proof of this last statement. Suppose $x$ is in the closure of $\mathcal D(A)$ for a bounded closed operator $A$. Thus there is a sequence $x_n$ in $\mathcal D(A)$ with limit $x$. Since $A$ is bounded, $A x_n$ is a Cauchy sequence. A Hilbert space is complete, so $A x_n$ has a limit $y$. Thus the sequence of pairs $(x_n, A x_n)$ converges to $(x, y)$ in $\mathcal H \times \mathcal H$. But since $A$ is closed, i.e. its graph is closed, $(x,y)$ must be in the graph of $A$, which says $x \in \mathcal D(A)$.