Unique maximizer of inner product

Question

I am trying to prove that for each $x \neq 0 \in \mathbb{R}^2$, $x_0 := \frac{x}{|x|}$ is the unique maximizer of $x \cdot y$ among $|y| \leq 1$. The set $M:=\{y \in \mathbb{R}^2 : |y|\leq 1\}$ is bounded, closed and convex. The inner product is a convex function on $\mathbb{R}^2$ so the unique maximizer exists and is an element of the boundary of $M$. Therefore the maximizer has to be of the form $\frac{y}{|y|}$ for $y \in \mathbb{R}^2$. But why does it have to be $x_0$ and not any other element of the boundary of $M$? I've tried to prove this but somehow I'm missing the crucial argument even if it seems to me that the proof should be more or less straightforward.

Answer 1

Using the Cauchy-Schwarz inequality you obtain for any $y$ on the boundary of $M$ (you already explained why the maximum lies on the boundary of $M$)

$$ | x \cdot y| \leq |x| |y| = |x| = |x \cdot x_0|.$$