Prove that $F$ is continuous if and only if $F^{-1}(0)$ is a closed set . Let $X$ be a normed linear space and let $F$ be a linear functional defined on $X.$ Prove that $F$ is continuous if and only if $F^{-1}(0)$ is a closed set (in the topology defined by the norm).
Could anyone give me a hint for the solution?
 A: One direction is just the usual topological fact that the pre-image of a closed set by a continuous function is closed.
For the other direction, I give a sketch with some details left to fill in since you asked for a hint. Suppose for a contradiction that $F^{-1}(0)$ is closed but $F$ is discontinuous. Then there is a sequence $x_n$ such that $\|x_n\| = 1$ and $f(x_n) \geq n$ (Why?). Now pick $x_0 \in X \setminus F^{-1}(0)$. There is $\varepsilon > 0$ such that $B(x_0, \varepsilon) \cap F^{-1}(0) = \emptyset$ (Why?). Now show that for large enough $n$, $$x_0 - \frac{f(x_0)}{f(x_n)}x_n \in B(x_0,\varepsilon) \cap F^{-1}(0)$$ to get a contradiction.
A: If $\ker(F)$ is closed we can consider the quotient normed space $X/\ker(F)$ and let $q: X \rightarrow X/\ker(F)$ be the canonical map sending an $x\in X$ to $x+\ker(F)$. Then $q$ is norm decreasing, hence continuous. We also have a map $\tilde{F}: X/\ker(F) \rightarrow \mathbb{C}$ given by $\tilde{F}(x+\ker(F))=F(x)$ which is continuous since the domain is 1-dimensional. Then $F=\tilde{F}\circ q$, i.e. a composition of continuous maps. 
A: One way is obvious: $F^{-1}(0)$ is closed if $f$ is continuous. 
Suppose $F^{-1}(0)$ is closed. If possible let $F$ be discontinuous at $0$. Show that there exists $x _n \to 0$ such that $F(x_n) \to \infty$. Fix $y$ with $F(y) \neq 0$. [If there is no such $y$ the $F \equiv 0$ and we are done]. Consider the sequence $y-a_nx_n$ where $a_n=\frac {F(y)} {F(x_n)}$. This sequence is contained in $F^{-1}(0)$ and tends to $y$. This contradicts the hypothesis 
