Why are “not bounded” operators not everywhere defined?

Let $X, Y$ be Banach spaces, $\mathcal{D}(T)$ a subspace of $X$, and $T\colon X\to Y$ a linear map. Such a $T$ is commonly called an unbounded linear operator, where unbounded just means that the domain $\mathcal{D}(T)$ is possibly a strict subspace of $X$.

I am confused by all the general questions flying around this definition and ask for some clarification. Is there a general argument why we consider such operators which are not defined on the whole space? In other words, I am interested in the following statement:

Statement. $\mathcal{D}(T)=X$ $\Rightarrow$ $T$ is bounded, or equivalently, $T$ is not bounded $\Rightarrow$ $\mathcal{D}(T)\neq X$.

In this context it might be interesting to look at the closed graph theorem: And operator defined on all of $X$ is bounded if and only if it is closed. Therefore it makes sense to ask whether there exist operators defined on all of $X$ which are not bounded.

There are many specific cases when this definition comes in handy. For example, differential operators are often first defined on a small class of functions (e.g. compactly supported, smooth functions) and can then be extended to larger domains. But here I am really considering any spaces and operators.

Of course, it would be interesting to know whether and how this changes when we restrict $X, Y$ to be Hilbert spaces.

• For example, self adjoint operators are closed (math.stackexchange.com/questions/260987/…). One often wants to consider self adjoint, but unbounded operators, so one has to accept the fact that they are not defined on the whole space. – PhoemueX Sep 22 '14 at 13:01
• This is an argument of the "specific" type: Certain operators cannot be defined on the whole space, therefore we define them only on a subspace. But my question is whether a not bounded (i.e. not continuous) operator can or cannot be defined on the whole space. – Deniz Sep 22 '14 at 13:07
• "where unbounded just means that the domain $\mathcal{D}(T)$ is possibly a strict subspace of $X$." No, unbounded means not (or "not necessarily") continuous. – Daniel Fischer Sep 22 '14 at 13:23
• Ok, thanks for the clarification. Does my answer solve your question? – PhoemueX Sep 22 '14 at 13:23
• John von Neumann initiated the study of closed operators when he was studying Quantum Mechanics, and found that a lot could be said, including the existence of adjoint, provided the operator was also densely-defined. Without a closed (or closable) graph, it's tough to say much about the operator. As pointed out by others, closed + everywhere-defined on a Banach space requires boundedness, which is out of the question for important classes of operators such as differential operators, including those of Quantum Mechanics; but these are often closable (definitely if symmetric on Hilbert space.) – DisintegratingByParts Sep 22 '14 at 15:44

You can always define an unbounded operator on the whole space $X$, as long as $X$ is infinite dimensional.
Simply take any unbounded linear functional $\varphi : X \to \Bbb{K}$ (with $\Bbb{K} \in \{\Bbb{R}, \Bbb{C}\}$) and some $x_0 \in X \setminus \{0\}$ and define $T : X \to X, x \mapsto \varphi(x) \cdot x_0$.
For the existence of $\varphi$, see On every infinite-dimensional Banach space there exists a discontinuous linear functional.