Closed Unit Ball is Convex Set
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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.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Exercise $1$
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed spaces
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $3.1$: Norms