Define a convex set in a linear space $X$. Show that the open ball $ B(0, δ) $ is a convex set in a linear space $ \Bbb R^n $ over $ \Bbb R $.

Question

I know that a convex set in a linear space $X$ is defined thus:

Let $ X $ be a real linear space. A set $ E ⊂ X $ is said to be convex if and only if for each pair of points $ x, y ∈ E $, the line segment joining $ x $ and $ y $ lies in $ E $, i.e. if $ x, y ∈ E $, then $$ L [x, y] = \{z \in X\ :\ z = (1 - λ)x + λy, λ ∈ [0, 1]\} ⊂ E. $$

Now, the problem I have is this: How can I make use of the definition above to show that the open ball $ B(0, δ) $ is a convex set in a linear space $ \Bbb R^n $ over $ \Bbb R $.

Source: MTH 303 - Advanced Calculus/OAU - Harmattan Semester Examinations/2017 - 2018 Academic Session/Convex Set/Q1. (b)

Answer 1

If $\|x\|<\delta, \|y\|<\delta$ and $0\leq \lambda \leq 1$ then $\|\lambda x+(1-\lambda)y\| \leq \lambda \|x\|+(1-\lambda)\|y\| <\delta (\lambda +1-\lambda) =\delta$.

Answer 2

Let $x, y\in B(0, \delta)$, and let $\lambda\in[0,1]$.

If you can prove that $(1-\lambda)x+\lambda y \in B(0,\delta)$, then you have proven that $B(0,\delta)$ is convex. You can prove this easily using the triangle inequality.

Answer 3

Is $x,y\in B(0,\delta)$ and $\lambda\in[0,1]$, then$$\bigl\lVert(1-\lambda)x+\lambda y\bigr\rVert\leqslant(1-\lambda)\lVert x\rVert+\lambda\lVert y\rVert<(1-\lambda)\delta+\lambda\delta=\delta$$and therefore $(1-\lambda)x+\lambda y\in B(0,\delta)$.