How to show $a \cdot \|x \| = \sup_{ \|s\| \leq a} s^T \cdot x$? In a paper I stumbled upon this where I was left a bit clueless on how to derive this equation:
$$a \cdot \|x \| = \sup_{ \|s\| \leq a} s^T \cdot x$$
where $a \in \mathbb{R}, (s,x) \in \mathbb{R}^n$ and $|| \cdot || $ is the L2-norm.

Any ideas how to make sense of it?
 A: I think that you need $a\geq 0$ - otherwise, the supremum is empty.  Recall that $s^Tx$ is really a way of writing the dot product $s\cdot x$.
Consider this, $s\cdot x=\|s\|\|x\|\cos(\theta)$, where $\theta$ is the angle between the vectors $s$ and $x$.  Since $\|s\|\leq a$ and cosine is at most $1$, we know that the RHS is at most $a\|x\|$.
On the other hand, if we let $s$ be a vector in the direction of $x$ and of length $a$, in other words, $s=\frac{a}{\|x\|}x$, then $s\cdot x=a\|x\|$.  Therefore the supremum is at last $a\|x\|$.
Therefore, the equality holds (when $a\geq 0$ so that the length of $\frac{a}{\|x\|}x$ is $a$; if $a<0$, then the length of $\frac{a}{\|x\|}x$ is $|a|=-a$).
A: We have that $s\cdot x \leq ||s|| \ ||x||$ by the Cauchy-Schwarz inequality, so for all $||s|| \leq a$, $s \cdot x \leq a \cdot || x ||$, and hence $$\sup_{||s||\leq a} \ s \cdot x \leq a \cdot || x||$$
On the other hand, consider using $s = \frac{a x}{||x||}$, so that $||s|| = a$, then $\ s \cdot x = a \cdot || x||$, so we have equality above.
