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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.

Deniz
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  • For example, self adjoint operators are closed (http://math.stackexchange.com/questions/260987/why-is-every-selfadjoint-operator-closed). 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
  • @PhoemueX Yes, it does. It seems there is no "general" argument of the kind I was looking for. – Deniz Sep 22 '14 at 13:39
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    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.) – Disintegrating By Parts Sep 22 '14 at 15:44
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    In short, there do exist everywhere defined, unbounded operators, but they're not good for anything. They are awful pathological axiom-of-choice monsters and they don't correspond in any reasonable way to any natural operations. Natural operations (multiplication, differentiation, etc) start out defined on a subspace of the Hilbert space, and there is no natural way to extend them to the entire space. The Zorn's lemma arguments basically choose an extension arbitrarily, which prevents it from having any particular meaning. – Nate Eldredge Apr 07 '18 at 17:24

1 Answers1

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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.

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