Problem: Let $X$ be a normed space and $f$ be a non-zero linear functional on $X$. Then prove that $f$ is discontinuous $\iff$ kernel($f$) is dense in $X$.
I have proved that if $f$ is discontinuous, then kernel($f$) is dense in $X$. How to prove the other way?
So you want to prove that if $\ker(f)$ is dense, then $f$ is discontinuous, or equivalently, if $f$ is continuous, then $\ker(f)$ is not dense.
Suppose $f$ is continuous. Then $\ker(f)$, which is the preimage of the closed set $\{0\}$, is closed in $X$ by continuity. If it were dense, then $\ker(f)=X$, so $f$ would be the zero functional. Contradiction.
Suppose that $\mbox{kernel}(f)$ is dense in $X$.
Since $f$ is non-zero, there exists $x\in X$ such that $f(x)=1$.
Since $\overline{\mbox{kernel}(f)}=X$, there exists a sequence $(x_n)_{n\in\mathbb{N}}$ in $\mbox{kernel}(f)$ such that $\displaystyle \lim_{n\to\infty} x_n= x$.
It follows that, if $f$ were continuous, we would have $\displaystyle 1=f(x)=\lim_{n\to\infty} f(x_n)=\lim_{n\to\infty}0=0$.
So, $f$ is not continuous.
