# Closed Unit Ball is Convex Set

## Theorem

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

Let $\map {B_1^-} 0$ be the closed unit ball in $X$.

Then $\map {B_1^-} 0$ is convex.

## Proof

Let $x, y \in \map {B_1^-} 0$.

Let $\alpha \in \closedint 0 1$ be arbitrary.

Then:

 $\ds \norm {\paren {1 - \alpha} x + \alpha y}$ $\le$ $\ds \norm {\paren {1 - \alpha} x} + \norm {\alpha y}$ Norm Axiom $\text N 3$: Triangle Inequality $\ds$ $=$ $\ds \size {1 - \alpha} \norm x + \size \alpha \norm y$ Norm Axiom $\text N 2$: Positive Homogeneity $\ds$ $=$ $\ds \paren {1 - \alpha} \norm x + \alpha \norm y$ Definition of Convex Set (Vector Space): $0 \le \alpha \le 1$ $\ds$ $\le$ $\ds \paren {1 - \alpha} + \alpha$ $x, y \in \map {B_1^-} 0$ $\ds$ $=$ $\ds 1$

Therefore, $\paren {1 - \alpha}x + \alpha y \in \map {B_1^-} 0$.

By definition, $\map {B_1^-} 0$ is convex.

$\blacksquare$