I need to prove that: If a nonzero linear functional $f$ on a Banach Space $X$ is discontinuous then the nullspace $N_f$ is dense in $X$.
To prove that $N_f$ is dense, it suffices to show that $\overline N_f = X$ which is equivalent to $(X \setminus N_f)^o=\emptyset$. (the interior of complement of $N_f$ is null set.)
Since $f$ is a linear functional and is discontinuous, it has to be unbounded. I don't know exactly how to utilize these observations.
Also on a related topic, I'm a little confused about how to exploit the a Linear Functional $f:X \to R$ or a Linear Operator $T:X \to Y$ being unbounded. Can I say that if a linear operator is unbounded then exists a sequence $<x_n>$ in $X$ s.t.
$||Tx_n|| > n^2||x_n||$ for each $n$ or $||Tx_n|| > n||x_n||$ ?
Best Answer
If $f$ is discontinuous, then you can find a sequence of non-zero vectors $(x_n)$ with $|f(x_n)|\ge n \Vert x_n\Vert$ for each $n$. Normalizing the $x_n$, and still calling them $x_n$, we obtain a sequence of norm one vectors $(x_n)$ such that $$\tag{1}|f(x_n)|\ge n,\quad\text{for each } n=1,2,\ldots.$$
Now suppose $x\notin {\text{Ker}(f)} $. Consider the sequence $$ z_n = x-\textstyle{f(x)\over f(x_n) } x_n. $$ One easily verifies that $z_n\in {\text{Ker}(f)}$ for each $n$. Moreover, from $(1)$, we have
$$\Vert z_n - x\Vert=\Bigl\Vert\textstyle{f(x)\over f(x_n) } x_n \Bigr\Vert =\Bigl|\textstyle{f(x)\over f(x_n) }\Bigr|\quad\buildrel{n\rightarrow\infty}\over\longrightarrow\quad0 .$$ From this it follows that $x\in\overline{\text{Ker}(f)}$. As $x$ was an arbitrary element not in $ {\text{Ker}(f)}$, it follows that $\overline{\text{Ker}(f)}=X$.
With regards to your last question, if $f$ is discontinuous and if $\alpha_n$ is any sequence of scalars, you can find a sequence of non-zero vectors $(x_n)$ with $|f(x_n)|\ge \alpha_n \Vert x_n\Vert$.