Home > Functional Analysis > Functional is continuous iff kernel is closed

Functional is continuous iff kernel is closed

Let f be a linear functional on a topological vector space X. Assume f(x)\neq 0 for some x \in X. Then the following properties are equivalent:

  1. f is continuous
  2. \ker f=\{x : f(x)=0\} is closed
  3. \ker f is not dense in X
  4. f is bounded in some neighborhood of 0.

Proof: If f is continuous then \ker f=f^{-1}(0) and therefore is closed. If \ker f is closed and dense in X, then we would have f(x)=0,\ \forall x \in X, which contradicts our hypothesis.

Next, assume that \ker f is not dense in X. Then there is an open set which is outside of \ker f. 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 V (such that \lambda V \subset V,\ \forall \lambda with |\lambda|\leq 1)  and an element x \in X such that (x+V)\cap \ker f=\emptyset. Then f(V) is a balanced subset of \Bbb{K}, which is either bounded, either equal to the whole field \Bbb{K}. If f(V) is bounded, then we are done. Else there exists y \in V such that f(y)=-f(x) which means y+x \in \ker f in contradiction to (x +V) \cap \ker f=\emptyset.

For the last implication, if |f(x)|<M for all x \in V, then for every \varepsilon>0 considering W=(\varepsilon/M)V, we have |f(x)|<\varepsilon,\ \forall x \in W, which proves that f is continuous at the origin. Since X is a topological vector space, this means that f is indeed continuous.

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