Closed Ball is Convex Set

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

Let $\map { {B_r}^-} x$ be a closed ball in $X$ with radius $r \in \R_{>0}$ and center $x \in X$.

Then $\map { {B_r}^-} x$ is convex.

Proof
Let $y \in \map { {B_1}^-} {\mathbf 0}$.

From, it follows that:
 * $\norm {r y} = r \norm y$

It follows that:
 * $y \in \map { {B_1}^-} {\mathbf 0}$, $r y \in \map { {B_r}^-} {\mathbf 0}$

As $\norm {r y - \mathbf 0} = \norm {\paren {r y + x} - x}$, it follows that:
 * $r y \in \map { {B_r}^-} {\mathbf 0}$


 * $r y + x \in \map { {B_r}^-} {\mathbf x}$
 * $r y + x \in \map { {B_r}^-} {\mathbf x}$

It follows that:
 * $\map { {B_r}^-} {\mathbf x} = r \map { {B_1}^-} {\mathbf 0} + x$

From Closed Unit Ball is Convex Set, it follows that $\map { {B_1}^-} {\mathbf 0}$ is convex.

From Dilation of Convex Set in Vector Space is Convex, it follows that $r \map { {B_1}^-} {\mathbf 0}$ is convex.

From Translation of Convex Set in Vector Space is Convex, it follows that $r \map { {B_1}^-} {\mathbf 0} + x$ is convex.