Show that for any $a,b \in \mathbb{R}\setminus\{0\}$ every point
$p=\begin{pmatrix} s\\ t \end{pmatrix}$ with property
$\frac{s^2}{a^2}+\frac{t^2}{b^2}=1$ is an extreme point of the convex
set $E(a,b)=\left\{\begin{pmatrix} x_1\\ x_2 \end{pmatrix} \in
\mathbb{R}^2 \, \Bigg\vert \, \frac{x_1^2}{a^2}+\frac{x_2^2}{b^2} \leq
1\right\}$
This is a task from an old exam but I like to know how it's solved and I'm very unsure about my solution:
So the extreme point of a strictly convex function on a convex set is always found at one of the extreme points of the set.
Luckily we already know that $E(a,b)$ is a convex set. Now what needs to be shown is that $\frac{s^2}{a^2}+\frac{t^2}{b^2}=1$ is strictly convex. It is strictly convex if its Hessian matrix is positive definite. So we need to calculate the Hessian matrix and show it's positive definite:
Here is the part I'm not sure about, the Hessian matrix should be $\begin{pmatrix}
\frac{2}{a^2}+\frac{t^2}{b^2}-1 & \frac{2s}{a^2}+\frac{2t}{b^2}-1\\
\frac{2s}{a^2}+\frac{2t}{b^2}-1 & \frac{s^2}{a^2}+\frac{2}{b^2}-1
\end{pmatrix}$ and it's positive definite if all its eigenvalues are greater than zero (because it's a symmetric matrix it's enough to check only this).
But here I have huge problem because it's too complicated to determine the eigenvalues and it also doesn't really seem to be greater than zero anyway because $a,b \in \mathbb{R}$…? : /
Maybe there is another better way of doing it?
Best Answer
For $(x,y)\in E$, $$ 0\le \frac{(x-s)^2}{a^2}+\frac{(y-t)^2}{b^2}=\underbrace{\frac{x^2}{a^2}+\frac{y^2}{b^2}}_{\le 1}+\underbrace{\frac{s^2}{a^2}+\frac{t^2}{b^2}}_{=1}-2\frac{xs}{a^2}-2\frac{yt}{b^2}$$ with equality only for $(x,y)=(s,t)$. Hence $$ \frac s{a^2}\cdot x+\frac t{b^2}\cdot y\le 1$$with equality only for $(x,y)=(s,t)$. We conclude that $E$ is contained in one of the closed half planes given by the line given with the equation $\frac s{a^2}\cdot x+\frac t{b^2}\cdot y=1$ and that $(s,t)$ is the only point of $E$ on that line.