Hyperplane in Normed Vector Space generated by Unbounded Linear Functional is Everywhere Dense

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.

Let $\phi : X \to \Bbb F$ be a linear functional that is not bounded.

Let $U$ be a hyperplane in $X$ given by:


 * $U = \map \ker \phi$

Then:


 * $U$ is everywhere dense in $X$.

Proof
Since $\phi$ is not bounded, for each $n \in \N$ we can pick $v_n \ne 0$ such that:


 * $\cmod {\map \phi {v_n} } \ge n \norm {v_n}$

Set:


 * $\ds x_n = \frac {v_n} {\norm {v_n} }$

for each $n \in \N$.

Then for each $n \in \N$, we have:


 * $\cmod {\map \phi {x_n} } \ge n$

from linearity with:


 * $\norm {x_n} = 1$

Fix $x \in X$ and $\epsilon > 0$.

We aim to show that there exists $y \in U$ such that:


 * $\norm {x - y} < \epsilon$

It suffices to find a sequence $\sequence {y_n}_{n \mathop \in \N}$ in $U$ such that:


 * $y_n \to x$

Then, from the definition of convergent sequence, we can find $N \in \N$ such that:


 * $\norm {x - y_n} < \epsilon$

For each $n \in \N$, let:


 * $\ds y_n = x - \frac {\map \phi x} {\map \phi {x_n} } x_n$

Then we have:

So, we have:


 * $y_n \in \map \ker \phi$

and so:


 * $y_n \in U$

We also have:

Since:


 * $\cmod {\map \phi {x_n} } \ge n$

We have:


 * $\ds \frac {\cmod {\map \phi x} } {\cmod {\map \phi {x_n} } } \le \frac {\cmod {\map \phi x} } n$

So, we have:


 * $\ds \norm {x - y_n} \le \frac {\cmod {\map \phi x} } n$

We therefore have:


 * $\ds \norm {x - y_n} \to 0$ as $n \to \infty$.

From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have:


 * $\ds y_n \to x$

So $\sequence {y_n}_{n \mathop \in \N}$ is the desired sequence converging to $x$.

Since $x$ was arbitrary, we have that:


 * $U$ is everywhere dense in $X$.