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:
\(\ds \map \phi {y_n}\) | \(=\) | \(\ds \map \phi x - \map \phi {\frac {\map \phi x} {\map \phi {x_n} } x_n}\) | Definition of Linear Functional | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x - \frac {\map \phi x} {\map \phi {x_n} } \map \phi {x_n}\) | Definition of Linear Functional | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x - \map \phi x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
So, we have:
- $y_n \in \map \ker \phi$
and so:
- $y_n \in U$
We also have:
\(\ds \norm {x - y_n}\) | \(=\) | \(\ds \norm {\frac {\map \phi x} {\map \phi {x_n} } x_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cmod {\map \phi x} } {\cmod {\map \phi {x_n} } } \norm {x_n}\) | positive homogeneity of the norm. | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cmod {\map \phi x} } {\cmod {\map \phi {x_n} } }\) | since $\norm {x_n} = 1$ |
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 $X$.
$\blacksquare$