Open Ball is Convex Set

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $V$ be a normed vector space with norm $\norm {\,\cdot\,}$ over $\R$ or $\C$.


An open ball in the metric induced by $\norm {\,\cdot\,}$ is a convex set.


Proof

Let $v \in V$ and $\epsilon \in \R_{>0}$.

Denote the open $\epsilon$-ball of $v$ as $\map {B_\epsilon} v$.

Let $x, y \in \map {B_\epsilon} v$.

Then $x + t \paren {y - x}$ lies on line segment joining $x$ and $y$ for all $t \in \closedint 0 1$.

The distance between $x + t \paren {y - x}$ and $v$ is:

\(\ds \norm {x + t \paren {y - x} - v}\) \(=\) \(\ds \norm {\paren {1 - t} \paren {x - v} + t \paren {y - v} }\)
\(\ds \) \(\le\) \(\ds \norm {\paren {1 - t} \paren {x - v} } + \norm {t \paren {y - v} }\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(=\) \(\ds \paren {1 - t} \norm {x - v} + t \norm {y - v}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(<\) \(\ds \paren {1 - t} \epsilon + t \epsilon\) as $x, y \in \map {B_\epsilon} v$
\(\ds \) \(=\) \(\ds \epsilon\)

Hence, $x + t \paren {y - x} \in \map {B_\epsilon} v$.

Thus, by definition, $\map {B_\epsilon} v$ is a convex set.

$\blacksquare$


Sources