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

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 $||Tx_n|| > n||x_n||$ ?

• Condition that $X$ is banach space is not necessary here...
– user87543
Nov 9, 2015 at 17:57

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

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

• codim $N_f$ will be 1 since $im(f)$ is $R$ right? How does that help? Mar 22, 2012 at 14:11
• Yes, that's right. Now (as $N_f$ isn't closed) $\overline{N_f} \supsetneq N_f$, what follows for $\text{codim}\,\overline{N_f}$? Mar 22, 2012 at 14:12
• @swair: If a subspace has codimension 1, what are the subspaces which contain it? Mar 22, 2012 at 14:13
• I'm not getting it. What result to use here? Mar 22, 2012 at 14:53
• oh! right. $f$ is continuous iff $N_f$ is closed. Thanks! Mar 22, 2012 at 15:03