Open Ball is Convex Set
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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
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $3.1$: Norms