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## 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)| 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.