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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!

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    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.2012-04-22

2 Answers 2

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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.

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    It was edited while ago, but thanks Abdelmajid.2012-04-22
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Hints:

As you said, use the Triangle inequality:

||a + b|| <= ||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 $

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    You mean 'scalar', not 'constant'.2012-04-22