Discontinuous linear map: Difference between revisions

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== Closed operators ==

Many naturally occurring linear discontinuous operators occur are [[closed operator|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 <math>T</math> be a map <math>X\to Y</math> with ___domain <math>\operatorname{Dom}(T)</math>. The graph <math>\Gamma(T)</math> of an operator <math>T</math> which is not everywhere-defined will admit a distinct closure <math>\overline{\Gamma(T)}</math>. If the closure of the graph is itself the graph of some operator <math>\overline{T}</math>, <math>T</math> is called closable, and <math>\overline{T}</math> is called the closure of <math>T</math>.

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 function]]s from [0,1] to '''R''' and ''Y'' the space of polynomial functions from [2,3] to '''R'''. They are subspaces of ''C''([0,1]) and ''C''([2,3]) respectively, and so normed spaces. Define an operator ''T'' which takes the polynomial function ''x'' ↦ ''p''(''x'') on [0,1] to the same function on [2,3]. 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.