Linear functional on a Banach space is discontinuous then its nullspace is dense.

Solution 1:

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 $|f(x_n)|\ge \alpha_n \Vert x_n\Vert$.

Solution 2:

Hint: $\overline{N_f}$ is a subspace of $X$ containing $N_f$. What can you say about $\mathrm{codim}\, N_f$?

Ok. So I'll give the answer. As $N_f$ isn't closed (since $f$ is discontinuous [Do you know that?]), $\overline{N_f} \supsetneq N_f$. It follows that $\text{codim}\, \overline{N_f} < \text{codim}\, N_f = 1$ (since $\dim \text{Im}\, f = 1$). So $\text{codim}\,\overline{N_f} = 0$, i. e. $\overline{N_f} = X$.

HTH, AB