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.

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.

$\blacksquare$

1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Exercise $1$
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2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $3.1$: Norms