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.
Best Answer
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.