Functional is continuous iff kernel is closed
Let be a linear functional on a topological vector space . Assume for some . Then the following properties are equivalent:
- is continuous
- is closed
- is not dense in
- is bounded in some neighborhood of .
Proof: If is continuous then and therefore is closed. If is closed and dense in , then we would have , which contradicts our hypothesis.
Next, assume that is not dense in . Then there is an open set which is outside of . This set can be written as a translation of a neighborhood of the origin (translation is a homeomorphism) and since multiplication by scalars is continuous, there exists a balanced neighborhood of the origin (such that with ) and an element such that . Then is a balanced subset of , which is either bounded, either equal to the whole field . If is bounded, then we are done. Else there exists such that which means in contradiction to .
For the last implication, if for all , then for every considering , we have , which proves that is continuous at the origin. Since is a topological vector space, this means that is indeed continuous.