Let $A\subseteq S^{d-1} = \{x\in \mathbb{R}^{d}: ||x||=1 \}$ be a subset of any unit sphere. Let $E \triangleq \{x\in \mathbb{R}: ||x||<1 \} \cup A$

Question

I promise I'll fix title in a few I'm very sorry mathjax is not my friend.

Let $A\subseteq S^{d-1}$ = {x $\in\mathbb R^d$: ||x||=1}$

be a subset of any unit sphere.

Define $$(E:={x\in\mathbb R^d:||x||<1}) \bigcup A$$ Is the set E convex?

So I know that a set E is convex if $tx+(1-t)y \in$E for every x,y $\in$E and every $t \in$[0,1].

Let x,y in $\mathbb R^d$

I need to use the triangle inequality to show $x,y \in A$ and $x \in B_1$(0) and $y\in B_1$(0) but I'm not sure how I'd go about showing $x \in B_1$(0) and $y\in B_1$(0)

Answer 1

Consider the segments contained in the closed unit ball where:

1) The two endpoints belong to $A$

2) Exactly one endpoint belongs to $A$

3) Neither endpoint belongs to $A$

The case 3) is immediate because the segment will be contained in $E$, which is convex. The other cases follows from the definition of convex set.

Answer 2

Let $x \cdot y = \sum_{i=1}^nx_iy_i$ denote the usual inner product on $\mathbb R^n$. The Cauchy- Schwarz-inequality says that $\lvert x \cdot y \rvert \le \lVert x \rVert \cdot \lvert y \rVert$, where equality holds if and only $x,y$ are linearly dependent.

It suffices to show that for $x,y \in E$ and $t \in (0,1)$ one has $tx + (1-t)y \in E$. We know

$$\lVert tx + (1-t)y \rVert^2 = (tx + (1-t)y)\cdot (tx + (1-t)y) = t^2\lVert x \rVert^2 + 2t(1-t) x \cdot y + (1-t)^2 \lVert y \rVert^2 \\ \le t^2\lVert x \rVert^2 + 2t(1-t) \lVert x \rVert \cdot \lVert y \rVert + (1-t)^2 \lVert y \rVert^2 = (t\lVert x \rVert +(1-t)\lVert y \rVert)^2 .$$ Thus, if at least one of $x, y$ has norm $< 1$, then $(t\lVert x \rVert +(1-t)\lvert y \rVert)^2 < (t + (1-t))^2 = 1$ which shows $tx + (1-t)y \in E$. If both $x,y$ have norm $1$, then $(t\lVert x \rVert +(1-t)\lvert y \rVert)^2 = 1$ and we see that $\lVert tx + (1-t)y \rVert^2 = 1$ if and only if $x \cdot y = \lVert x \rVert \cdot \lVert y \rVert $. This means that $x,y$ are linearly dependent and we conclude $y = rx$ for some $r \ne 0$ (note $x \ne 0$). Thus $r = r \lVert x \rVert^2 =x \cdot rx = x\cdot y = \lVert x \rVert \cdot \lVert y \rVert = 1$. Therefore $y = x$ and $tx + (1-t)y = x \in E$.