TITLE: If $f$ is not continuous then $\ker f$ is dense in $X$
QUESTION [5 upvotes]: Let $X$ be a normed space and $f:X\rightarrow \mathbb R$ a linear function.
I saw an old post with this problem, but there is not a complete proof.
For beginning I have to consider that  $\|f\|=\infty$.

REPLY [9 votes]: Note that $f$ is bounded if and only if it is continuous. So let $f$ be unbounded. This means for any $n \in \mathbb N$ we have an $x_n \in X$ so that $f(x_n) ≥ n \|x_n\|$. By rescaling set $\|x_n\|=1$.
Now let $z$ be in $X$. From the construction of the $x_n$ it follows that $z_n:= z - \frac{f(z)}{f(x_n)}x_n$ is a sequence that converges to $z$. But $f(z_n)=f(z)-\frac{f(z)}{f(x_n)}f(x_n)=0$, so $z_n$ lies in the kernel of $f$. But since $z_n \to z$ you have $z \in \overline{\ker(f)}$. Because $z$ was arbitrary the kernel is dense.