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I have to prove or disprove that $$ \left\{x \in \Bbb R^n : \sum_{i=1}^n x_i^2 = 1\right\}\text{ is convex.} $$ I already know that it is not convex, so a disprove is the right way.

I know that I gotta use the following for a prove: $$ \text{A set }\Omega\subset\Bbb R^n\text{ is convex if }\alpha x+(1−\alpha)y\in\Omega,\forall x,y\in\Omega\text{ and }\forall\alpha\in[0,1]. $$ For a disprove I could take two points and show that the above condition is not met. But what confuses me is that the statement also depends on the value of $n$. If I choose e.g. the two points $\sqrt{\frac{7}{10}}$ and $\sqrt{\frac{3}{10}}$, how do I could use them to disprove it?

Daniele Tampieri
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floschen
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    $\sqrt{7/10}$ and $\sqrt{3/10}$ are not two points, they are the coordinates of one point that is in your set in the case $n=2$. – Robert Israel Nov 11 '19 at 02:36
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    In this case any two distinct points will do, and any $\alpha \in (0,1)$. But for convenience you might try an example where $\alpha x + (1-\alpha) y = 0$. – Robert Israel Nov 11 '19 at 02:37
  • Note that the unit ball is indeed convex, so I disagree with your second line. – Calvin Lin Nov 11 '19 at 02:38
  • what would be a point if $\sqrt{\frac{7}{10}}$ and $\sqrt{\frac{3}{10}}$ are coordinates? – floschen Nov 11 '19 at 02:55
  • @CalvinLin but this is a hollow unit sphere isn’t it? Someone pls help me out – floschen Nov 11 '19 at 03:41
  • Oh good point. For a concrete counterexample, you can show that $(1,0)$ and $(-1,0)$ lie on the set, but clearly no point in the line segment lies in the set. In general, Robert's comment works. – Calvin Lin Nov 11 '19 at 03:52

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