Characteristic of a linear functional

Question

Let $(X,\|\;\|)$ be a vector space over $K\;(\Bbb R\text{ or }\Bbb C)$ and let $\;f:X\to K$ be a linear functional with $f\neq0$. I want to prove the following: $$ \text{If }\mathscr N=\{x\in X:f(x)=0\}\text{ is not dense in }X\Rightarrow\;f\text{ is bounded in some neighborhood of }0 $$

So my attempt goes like this:
Let's take $\;B_{r_0}(0)$, a neighborhood of $0$. Then, since $\;\;\overline{\mathscr N}\neq X$, there $\exists\;x_0\in X\setminus\overline{\mathscr N}$ which implies that $\;f(x_0)\neq0$ and thus $\;|f(x_0)|>0$

But got stucked here since I can't see a way to relate this to $f$ having to be bounded in $\;B_{r_0}(0)$.
Any hints or ideas would be appreciated.

Your very first step is wrong... You can not take $B_{r_o}$... YOu have to say there exists some $r_0$ — , Mar 09 '16 at 13:28
Every linear functional is such that $f(0)=0$. With this in mind, suppose that $f$ is not bounded in any neighborhood of $0$, then find a contradiction. — Larara, Mar 09 '16 at 13:31

score 1 · Accepted Answer · answered Mar 09 '16 at 14:25

If $f$ were unbounded on any neighbourhood of $0$, there were a sequence $(x_n)$ such that $\|x_n\| < 1/n$ and $f(x_n) > n$. We will show that $\mathscr N$ is dense: Let $x \in X$ and $\epsilon > 0$. Choose $n \in \mathbf N$ with $f(x_n) > |f(x)|$ and $1/n < \epsilon$. Define $$ y := x - \frac{f(x)}{f(x_n)}x_n $$ Then $$\|y-x\| \le \|x_n\| < \epsilon $$ and $$ f(y) = f(x) - \frac{f(x)}{f(x_n)}f(x_n) = 0 $$ That is $y \in \mathscr N \cap B_\epsilon(x)$ and therefore $\mathscr N$ is dense.

Characteristic of a linear functional

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