How can I prove that every point in the boundary of the unit disc in $\mathbb{R}^2$ is a extreme point?

Question

I know that a point in the circunference can't be in a segment with the edges inside de disk, but I'm having problems at proving it formally.

We defined extreme point of a set $S$ as a point $x_0 \in S$ such that $x_0 = \alpha x + (1-\alpha)y \Longleftrightarrow x_0 = y = x$ for $x,y \in S, \alpha \in [0,1]$

I tried with a proof by contradiction assuming there exists a segment with the edges inside of the disk that contains it but I don't know what steps to take to arrive to a contradiction.

Does someone have any ideas that could help?　Thanks.

Edit: Does this lead anywhere? (I don't think so) For any $x,y \in D, x_0 \in \partial D, || \alpha x + (1-\alpha)y|| \leq | \alpha |||x|| + |1-\alpha|||y|| = ||x_0|| = 1 \Longleftrightarrow ||x|| = ||y|| = 1$

Answer 1

As per what you've done, we can let $x, y \in D$ and $x_0 \in \partial D$ such that $x_0 = \lambda x + (1 - \lambda)y$ for some $\lambda \in (0, 1)$. As you've noticed, this can only happen if $\|x\| = \|y\| = 1$.

Let's first consider the special case of $\lambda = \frac{1}{2}$. In this particular case, we can fall back on some circle geometry. The line segment between $x$ and $y$ is a chord in the circle $\partial D$, and $x_0$ is the midpoint. By circle geometry, I would expect the line segment from $0$ (the centre) to $x_0$ (the midpoint) should be perpendicular to $y - x$ (the chord). Let's verify: $$x_0 \cdot (y - x) = \frac{1}{2}(y + x) \cdot (y - x) = \frac{1}{2}(y \cdot y - x \cdot x) = \frac{1}{2}(1^2 - 1^2) = 0,$$ as expected. Note also that $$x_0 + \frac{y - x}{2} = \frac{y + x}{2} + \frac{y - x}{2} = y,$$ so by Pythagoras's theorem, $$1 = \|y\|^2 = \|x_0\|^2 + \left\|\frac{y - x}{2}\right\|^2.$$ This implies that $\|x_0\| = 1$ if and only if $\left\|\frac{y - x}{2}\right\|^2 = 0$, i.e. $y = x$.

Now, the other cases, where $\lambda \neq \frac{1}{2}$, follow from the fact that every point in the line segment between $x$ and $y$ is either a convex combination of $x$ and $\frac{x + y}{2}$ or a convex combination of $y$ and $\frac{x + y}{2}$. As you already proved, in order for such a convex combination to be on the boundary circle, both endpoints must be on the boundary circle! But, we just proved that $\frac{x + y}{2}$ is not (unless $x = y$), and so the theorem is proven.