Open Ball with respect to Seminorm is Convex, Balanced and Absorbing
Theorem
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let $d_p$ be the pseudometric induced by $p$.
Let $B$ be the open unit ball in $\struct {X, d_p}$.
That is:
- $B = \set {x \in X : \map p x < 1}$
Then $B$ is convex, balanced and absorbing.
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Proof
Proof that $B$ is convex
Let $t \in \closedint 0 1$ and $x, y \in B$.
Then:
\(\ds \map p {t x + \paren {1 - t} y}\) | \(\le\) | \(\ds t \map p x + \paren {1 - t} \map p y\) | Seminorm Axiom $\text N 2$: Positive Homogeneity, Seminorm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds t + \paren {1 - t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
so:
- $t x + \paren {1 - t} y \in B$.
$\Box$
Proof that $B$ is balanced
Let $s \in \GF$ such that $\cmod s \le 1$.
Let $x \in B$.
Then, we have by Seminorm Axiom $\text N 2$: Positive Homogeneity we have:
- $\map p {s x} = \cmod s \map p x \le \map p x < 1$
so $s x \in B$.
So, we have:
- $s B \subseteq B$
for all $s \in \GF$ with $\cmod s \le 1$.
So $B$ is balanced.
$\Box$
Proof that $B$ is absorbing
From Characterization of Convex Absorbing Set in Vector Space, it is enough to show that:
- $\ds X = \bigcup_{n \mathop = 1}^\infty n B$
By Seminorm Axiom $\text N 2$: Positive Homogeneity, we have:
- $\map p {n x} < n$ if and only if $\map p x < 1$
for each $n \in \N$.
So, we have:
- $n B = \set {x \in X : \map p x < n}$
Clearly we have:
- $\ds \bigcup_{n \mathop = 1}^n n B \subseteq X$
Now let $x \in X$, then we have:
- $\ds \map p {\frac x {2 \map p x} } = \frac 1 2 < 1$
from Seminorm Axiom $\text N 2$: Positive Homogeneity, so:
- $\ds \frac x {2 \map p x} \in B$
Then we have:
- $x \in \paren {2 \map p x} B$
Taking $N \in \N$ with $N \ge \map p x$, we have $x \in N B$, and so:
- $\ds x \in \bigcup_{n \mathop = 1}^\infty n B$
So:
- $\ds X = \bigcup_{n \mathop = 1}^\infty n B$
$\blacksquare$
Sources
- 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.34$: Theorem