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Given the unit sphere in $\mathbb{R}^n$ show that the extreme points of the unit ball are exactly the points on the unit sphere.

The geometry is obvious: Take any two points $||x||,||y||<1$ "truly" inside the ball, then their connection must be "truly" inside the ball as well.

My problem is using that approach as the basis of my proof, i.e. forming a convex combination $\lambda x+(1-\lambda)y$ and showing $||\lambda x+(1-\lambda)y||<1$ leads me to a big root term mess.

Is there a better way? Am I missing some algebraic trick to allow me to work with the root?

Edit: Not a duplicate of questions regarding function spaces. This is $\mathbb{R}^n$.

Edit2: Now that I see the answer for $\mathbb{R}^n$ the connection to the linked proof is clear though...

Ruben Kruepper
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Hint: $\|x+y\|=\|x\|+\|y\|$ implies that $x$ and $y$ are linearly dependent. This can be seen by squaring both sides and using the condition for equality in C-S inequality: $|\langle x , y \rangle|=\|x\|\|y\|$ implies that $x$ and $y$ are linearly dependent. But if two unit vectors are $x$ and $y$ are linearly dependent then either $x=y$ or $x=-y$. Can you proceed?

Kavi Rama Murthy
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  • The ball is convex and compact, so it is the convex hull of it's extreme points. Are you trying to argue that every point in the ball is a convex combination $x$ and $-x$ for $x$ on the sphere? If so how do I formalize that? – Ruben Kruepper Nov 12 '19 at 09:45
  • Assuming that $x$, $y$ and $\lambda x+(1-\lambda)y$ are all unit vectors with $0<\lambda <1$ we have to show that $x=y$. In my argument the case $x=-y$ easily leads to a contradiction so we get $x=y$. @RubenKruepper – Kavi Rama Murthy Nov 12 '19 at 09:49
  • @RubenKruepper Extreme points are always on the boundary so every extreme point has norm $1$. We have to show that a vector of norm $1$ cannot be written as a strict convex combination of two distinct unit vectors. – Kavi Rama Murthy Nov 12 '19 at 09:51
  • That clears it up, many thanks! – Ruben Kruepper Nov 12 '19 at 09:52