relation between linear functional and kernel on nls

Question

Let $f$ be a linear functional on a normed linear space $X$. Prove that $f$ is discontinious iff $ker(f)$ is a proper dense subspace of $X$ \ I tried to get with discontinuity of $f$ but not getting the idea how to deal with kernel

Answer 1

Suppose that $f$ is discontinuous. Then $f\neq 0$, so $\ker f$ is proper. To prove that $\ker f$ is dense in $X$, let $x_0\in X$, let $\lambda:=f(x_0)$, and let $\left\{y_n\right\}\subset X$ be a sequence such that $\|y_n\|=1$ and $|f(y_n)|\to +\infty $ (it exists because $f$ is discontinuous). Now consider $$x_n:=x_0-\frac{y_n}{f(y_n)}\lambda $$ then $f(x_n)=0$, so that $x_n\in \ker f$ for all $n$, and $$\|x_n-x_0\|=\left|\frac{\lambda}{f(y_n)}\right|\to 0 $$ Therefore, $x_n\to x_0$, and $x_0\in \overline{\ker f}$.

Conversely, if $\ker f$ is proper and dense, it is non-closed, so since $\ker f$ is the preimage of the singleton $\left\{0\right\}$ (which is closed in $\mathbb{R}$), $f$ has to be discontinuous.

By the way, this is basically equivalent to the fact that a linear functional is continuous if and only if its kernel is closed.