I'm reading a paper and I encountered the following:
Define $T$ to be a $1/2$-net of an $n$ dimensional unit sphere $S^{n-1}$ in the norm $\|\|_2$.
What is the definition of $1/2$ net of a sphere?
I'm not familiar with this terminology and here is my understanding from searching it: We must have $T \subseteq S$ and $T$ must be $n-1$ dimensional. For any vector $x \in S^{n-1}$, there exists $y \in T$ such that $\|x-y\|_2 \leq 1/2$. Intuitively, $T$ performs a discretization on the sphere. Is my understanding correct?
Given this definition on $T$, they write the following inequalities for the operator norm of matrix $A$:
$$\|A\|_2 = \sup_{x \in S^{n-1}, y \in \mathbb{R}^n-\{0\}} \frac{x^\intercal A y}{\|y\|_2} \leq 2 \sup_{x \in T, y \in > \mathbb{R}^n-\{0\}} \frac{x^\intercal A y}{\|y\|_2}$$
Why is the above inequality true?
Put $M=\sup_{x \in T, y \in \mathbb{R}^n-\{0\}} \frac{x^\intercal A y}{\|y\|_2}$. Let $x\in S^{n-1}$ and $y\in\Bbb R^n-\{0\}$. There exists $t\in T$ such that $\|t-x\|_2\le 1/2$. Then
$$\frac {x^{\perp}Ay}{\|y\|_2}= \frac {t^{\perp}Ay}{\|y\|_2}+\frac {(x-t)^{\perp}Ay}{\|y\|_2}\le M+\frac 12\|A\|_2$$
If follows $\|A\|_2\le M+\frac 12\|A\|_2$ and $\|A\|_2\le 2M$.