$\ker f$ is either dense or closed when $f$ is a linear functional on a normed linear space 
Let $f$ be a linear functional on a normed linear space $X$. Prove that $\ker f$ is either dense or closed in $X$.

Two possibilities can occur, i.e either $f$ is bounded or unbounded. If it is bounded then $f$ is continuous and hence $\ker f=f^{-1}(0)$ which is closed 
How to show when $f$ is unbounded.
 A: Let $f$ be a non-trivial linear functional on $X$, and let $\mathcal{N}(f)$ be the null space of $f$. There exists $y \in X$ such that $f(y)=1$ because $f$ is non-trivial. Therefore,
$$
               x = f(x)y + (x-f(x)y) \in [\{y\}]+\mathcal{N}(f), \;\;\;(\dagger)
$$
where $[\{y\}]$ is the one-dimensional linear space spanned by $y$. Hence $X/\mathcal{N}(f)$ is one-dimensional, and the quotient map is
$$
                      x \mapsto f(x)y + \mathcal{N}(f).
$$
If $\mathcal{N}(f)^{c}$ denotes the norm closure of $\mathcal{N}(f)$, then it follows that either $\mathcal{N}(f)^{c}=\mathcal{N}(f)$ or $\mathcal{N}(f)^{c}=X$. If $\mathcal{N}(f)^{c}=\mathcal{N}(f)$, then $X/\mathcal{N}(f)$ is a normed space, and the quotient map from $X$ to $X/\mathcal{N}(f)$ is continuous in norm, and this map agrees with $f$ because $x\mapsto f(x)y$ under this quotient map, because of $(\dagger)$. So $f$ is continuous if $\mathcal{N}(f)$ is closed. Conversely, if $f$ is continuous, then $\mathcal{N}(f)=f^{-1}\{0\}$ is closed. If $f$ is not continuous then $\mathcal{N}(f)^{c}=X$, which means that $\mathcal{N}(f)$ is a dense subspace of $f$ in $X$.
A: Suppose $f$ is unbounded. Then there exists a sequence $\{x_n\}$ with $x_n\to0$ and $f(x_n)=1$ for all $n$. 
Now choose any $y\in X$. Then $y-f(y)x_n$ is in $\ker f$, since $$f(y-f(y)x_n)=f(y)-f(y)=0.$$ Thus $y=\lim_n y-f(y)x_n$, where the elements in the sequence $\{y-f(y)x_n\}$ are in $\ker f$. So $\ker f$ is dense in $X$. 
