I'm studying the following theorem:
Theorem 1. Let $C$ be a convex set and let $\textbf{y}$ be a point exterior to the closure of $C$. Then there is a vector $\textbf{a}$ such that $\displaystyle\textbf{a}^T\textbf{y}<\inf_{x\in C}\textbf{a}^T\textbf{y}$.
Proof. Let $$\delta=\inf_{x\in C}|\textbf{x}-\textbf{y}|>0.$$
There is an $\textbf{x}_0$ on the boundary of $C$ such that $|\textbf{x}_0-\textbf{y}|=\delta$. This follows because the continuous function $f(x)=|\textbf{x}-\textbf{y}|$ achieves its minimum over any closed and bounded set and it is clearly only necessary to consider $\textbf{x}$ in the intersection of the closure of $C$ and the $\color{red}{\boxed{\color{black}{\text{sphere of radius }2\delta\text{ centered at }\textbf{y}.}}}$
We shall show that setting $\textbf{a}=\textbf{x}_0-\textbf{y}$ satisfies the conditions of the theorem.
Let $x\in C$. For any $\alpha, 0 \le\alpha\le 1$, the point $\color{red}{\boxed{\color{black}{\textbf{x}_0+\alpha(\textbf{x}-\textbf{x}_0)\in \overline{C}}}}$ and thus $$|\textbf{x}_0+\alpha(\textbf{x}-\textbf{x}_0)-\textbf{y}|^2\ge|\textbf{x}_0-\textbf{y}|^2.$$ Expanding, $$2\alpha(\textbf{x}_0-\textbf{y})^T(\textbf{x}-\textbf{x}_0)+\alpha^2|\textbf{x}-\textbf{x}_0|^2\ge 0.$$ Thus, considering this as $\alpha \to 0+$, we obtain $$(\textbf{x}_0-\textbf{y})^T(\textbf{x}-\textbf{x}_0)\ge0$$ or, $$\begin{align} (\textbf{x}_0-\textbf{y})^T\textbf{x}\ge(\textbf{x}_0-\textbf{y})^T\textbf{x}_0 &= (\textbf{x}_0-\textbf{y})^T\textbf{y}+(\textbf{x}_0-\textbf{y})^T(\textbf{x}_0-\textbf{y})\\ &=(\textbf{x}_0-\textbf{y})^T\textbf{y}+\delta^2 \end{align}.$$ Setting $\textbf{a}=\textbf{x}_0-\textbf{y}$ proves the theorem. $\blacksquare$
I have highlighted the points I have questions about:
1st rectangle: Why did the author select sphere with radius $2\delta$? isn't sphere of radius $\delta$ enough?
2nd rectangle: I don't understand why $\textbf{x}_0 + \alpha(\textbf{x}-\textbf{x}_0) \in \overline{C}$ for any $0 \leq \alpha \leq 1$, because lets say that $\alpha = 1$. Then $\textbf{x}_0 + \alpha(\textbf{x}-\textbf{x}_0)$ becomes $\textbf{x}_0 + \textbf{x}-\textbf{x}_0 = \textbf{x} \in C$?!...what did I do wrong :) Can someone clarify my mistake :)
Thank you for the help :)
We need just a tiny bit more than $\delta$, to be sure we're dealing with a nonempty compact set (a priori, before we deduce that a minimum occurs).
What we have there is a convex combination of two points in $\bar{C}$. Since $C$ is convex, so is $\bar{C}$, hence closed under convex combinations of elements.