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I have to prove or disprove that {x ∈ $\mathbb{R}^n$: $\sum_{i=1}^n x_i^2 = 1$} is convex.

I don't even understand what kind of set this is, i.e. which values of $x$ are in this set.

I know that I gotta use the following for a prove:

A set $\Omega \subset \mathbb{R}^n$ is convex if $ \alpha x+(1−\alpha)y \in\Omega,∀x,y \in\Omega$ and $∀ \alpha ∈[0,1]$.

But maybe a disprove would be easier?

Mahesh Chandra Mukkamala
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floschen
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  • This set is a sphere and thus is not convex. In fact the connecting line between any two points is out of the set. Just pick any two point and compute – Severin Schraven Nov 10 '19 at 18:32
  • thank you, which would be two exemplary points of this set? – floschen Nov 10 '19 at 18:39
  • As a contrast, ${x\in \mathbb{R}^n: \sum_{i=1}^n x_i=1}$ is convex, and proving this would be a good exercise to aid your understanding. – RobPratt Nov 10 '19 at 18:43
  • which would be two exemplary points of your set Rob? my problem is that I don't fully understand the notation of these kind of sets – floschen Nov 10 '19 at 18:47
  • would just [0.3 ; 0,7] be points of the set? or [0,1 ; 0,3 ; 0,4 ; 0,5] ? do I get it correctly? – floschen Nov 10 '19 at 19:16
  • $(1,0,\dots,0)$ and $(-1,0,\dots,0)$ are two such points. For $n=3$, you can think of these as the north and south poles, respectively. – RobPratt Nov 11 '19 at 03:03
  • I'm confused, how do I work with points like "(1,0,…,0)" in an equation like this: αx+(1−α)y ? – floschen Nov 11 '19 at 03:08
  • @floschen may it would be easier to visualize in 2D. Just draw a circle and you can take any two points on the circle boundary then draw a line in between them. Apart from the two selected points, no other point lies on the boundary of the circle, which means your set is not convex. Is it clear? (Note that your set deals with the boundary only.) – Mahesh Chandra Mukkamala Nov 23 '19 at 23:34

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