Let $A$ be a symmetric matrix in $\mathbb{R}^{n \times n}$. We can show that its operator norm can be written as $\lVert A \rVert = \max_{x \in S^{n-1}} \left|\langle Ax,x\rangle \right|$, where $\langle x,y\rangle = y^T x$ for $x,y \in \mathbb{R}^n$ ($S^{n-1}$ is the unit sphere on $\mathbb{R}^n$ i.e. $x \in S^{n-1}$ iff $x_1^2+ \ldots + x_n^2 = 1$).
Let $N$ be an $\epsilon$-net on $S^{n-1}$. This means that for any $x \in S^{n-1}$, we can find $x_0$ such that $\lVert x - x_0 \rVert_2 \leq \epsilon$.
I want to show that $\lVert A \rVert \leq \frac{1}{1-2\epsilon} \sup_{x \in N} \left| \langle Ax,x\rangle \right|$.
My idea is to let $x$ be such that $\lVert A \rVert = \langle Ax,x\rangle$, and choose $x_0 \in N$ such that $\lVert x-x_0 \rVert_2 \leq \epsilon$. Then $\left|\langle Ax,x\rangle\right| - \left|\langle Ax_0,x_0\rangle\right| \leq \left| \langle Ax,x\rangle - \langle Ax_0,x_0\rangle \right| = \left|\langle Ax, x-x_0\rangle + \langle A(x-x_0), x_0\rangle \right|$.
I would like to show $\left|\langle Ax,x-x_0\rangle\right| \leq \epsilon \lVert A \rVert$ and likewise for $\left|\langle A(x-x_0), x_0\rangle\right|$, but I only have control over $\max_{x \in S^{n-1}} \left|\langle Ax,x\rangle \right|$ and not $\max_{x,y \in S^{n-1}} \left|\langle Ax,y\rangle\right|$.
Thank you
Using the Cauchy Schwarz Inequality, you have $$|\langle Ax,x-x_0\rangle| \le \|Ax\| \cdot \|x-x_0\| \le \|A\|\|x\| \cdot \|x-x_0\| \le \|A\|\cdot 1 \cdot \epsilon = \epsilon\|A\|,$$ and similarly, $$|\langle A(x-x_0),x_0\rangle| \le \|A(x-x_0)\| \cdot \|x_0\| \le \|A\|\|x-x_0\| \cdot \|x_0\| \le \|A\|\cdot \epsilon \cdot 1 = \epsilon\|A\|.$$