Discontinuous Linear Map - Closed Operators

Closed Operators

Many naturally occurring linear discontinuous operators occur are closed, a class of operators which share some of the features of continuous operators. It makes sense to ask the analogous question about whether all linear operators on a given space are closed. The closed graph theorem asserts that all everywhere-defined closed operators on a complete domain are continuous, so in the context of discontinuous closed operators, one must allow for operators which are not defined everywhere. Among operators which are not everywhere-defined, one can consider densely-defined operators without loss of generality.

Thus let be a map with domain . The graph of an operator which is not everywhere-defined will admit a distinct closure . If the closure of the graph is itself the graph of some operator, is called closable, and is called the closure of .

So the right question to ask about linear operators that are densely-defined is whether they are closable. The answer is not necessarily; one can prove that every infinite-dimensional normed space admits a nonclosable linear operator. The proof requires the axiom of choice and so is in general nonconstructive, though again, if X is not complete, there are constructible examples.

In fact, an example of a linear operator whose graph has closure all of X×Y can be given. Such an operator is not closable. Let X be the space of polynomial functions from to R and Y the space of polynomial functions from to R. They are subspaces of C and C respectively, and so normed spaces. Define an operator T which takes the polynomial function x ↦ p(x) to on to the same function on . As a consequence of the Stone–Weierstrass theorem, the graph of this operator is dense in X×Y, so this provides a sort of maximally discontinuous linear map (confer nowhere continuous function). Note that X is not complete here, as must be the case when there is such a constructible map.