6

Good evening guys!

I have to show that the unit sphere represented by unit sphere is convex.

A set is said to be convex when $sx + (1 - s)y \in M$, where $x, y \in M$ and $s \in (0,1)$

I've read on wikipedia that this can be proven over the triangle inequality, but I think it can be solved in another way? Would this be enough as proof:

For the unit sphere, we have to prove that $0 \leq sx + (1 - s)y \leq 1$ (because $||x||\leq 1$ therefore, $0 \leq x,y \leq 1$). Seeing as the maximal value that x and y can take are 1, the maximum the equation can achieve is 1 (when s=1,x=1 or s=0,y=1). The same can be shown for the minimum 0, therefore it is really between 0 and 1. Finished?

Many thanks in advance!

Clash
  • 1,401
  • 3
    you want to show the norm $||sx+(1−s)y|| <= 1$, not sx+(1−s)y, which is a Vector. – Long Apr 22 '12 at 20:07
  • It suffices to prove this in two-dimensions, i.e. a circular disk, since in higher dimensions you can slice the "sphere" (unit ball is a better term) so that two given points and the center are in a common plane. – hardmath Apr 22 '12 at 20:17

2 Answers2

10

Unfortunately you can not say from a vector whether it is between two reals or not. That's why you need to consider the norm of $sx+(1-s)y$. But as Wikipedia suggests, and basicly what you figured out in a different way, you see that if $x,y\in \bar{B_{X}}$ and $s\in[0,1]$, then by triangle inequality: \begin{align*} ||sx+(1-s)y||\leq ||sx||+||(1-s)y||=s||x||+(1-s)||y||\leq s+(1-s)=1 \end{align*} Hence $sx+(1-s)y\in \bar{B_{X}}$ and thus $\bar{B_{X}}$ is convex.

T. Eskin
  • 8,303
1

Hints:

As you said, use the Triangle inequality:

$||a + b|| &lt= ||a|| + ||b||$, with a = sx, b = (1-s)y

And remember s, 1-s are constant, and there is a formula:

$||\alpha * z|| = |\alpha| * ||z||$, z is a vector, and $\alpha$ is a constant

Also, $||x|| = 1, ||y|| = 1 $

Long
  • 1,630