Problem. Let $A = (a_{ij})$, $1\le i \le m$, $1\le j \le n$ and $\varepsilon \in (0,1/2) $. Let $\mathcal{N}$ be an $\varepsilon$-net of $S^{n-1}$ and $\mathcal{M}$ be an $\varepsilon$-net of $S^{m-1}$ ($S^{m-1}$ and $S^{n-1}$ are unit spheres). Prove that
\begin{align*}
\sup_{x \in \mathcal{N},\ y \in \mathcal{M}} \langle Ax,y\rangle \le \Vert A \Vert \le \dfrac{1}{1-2\varepsilon} \sup_{x \in \mathcal{N}, y \in \mathcal{M}} \langle Ax,y \rangle.
\end{align*}
My attempt: Let $x \in \mathcal{N}$ and $y \in \mathcal{M}$. Hence, there exist $x' \in S^{n-1}$, $y' \in S^{m-1}$ such that
\begin{align*}
& \Vert x – x' \Vert_2 \le \varepsilon,\\
& \Vert y – y' \Vert_2 \le \varepsilon.
\end{align*}
Hence, we have
\begin{align*}
\vert \langle Ax,y \rangle – \langle Ax',y'\rangle \vert
& = \vert \langle Ax,y\rangle – \langle Ax',y\rangle + \langle Ax',y\rangle – \langle Ax',y'\rangle\\
& = \vert \langle A(x-x'),y\rangle + \langle Ax',y-y'\rangle \vert\\
& = \vert \langle A(x-x'),y\rangle \vert + \vert \langle Ax',y-y'\rangle \vert \tag{triangle inequality}\\
& = \Vert A (x-x') \Vert_2 \cdot \Vert y \Vert + \Vert Ax'\Vert_2 \cdot \Vert y-y' \Vert_2 \tag{Cauchy-Schwarz inequality}\\
& \le \Vert A \Vert \cdot \Vert x-x' \Vert_2 \cdot \Vert y \Vert_2 + \Vert A \Vert \cdot \Vert x' \Vert \cdot y-y'\Vert \\
& \le \Vert A\Vert \varepsilon + \Vert A \Vert \varepsilon = 2\varepsilon \Vert A \Vert.
\end{align*}
Now I am stuck here. I would like to know if you can give me some ideas to deal with the problem.
Best Answer
Here is a sketch of one approach. First, show that given $\alpha \in (1,\infty )$, there are $x_{0}\in S^{n-1}$ and $y_{0}\in S^{m-1}$ such that $\|A\| \leq \alpha \langle A(x_{0}), y_{0}\rangle $. Second, given $x'\in \mathcal{N}$ and $y'\in\mathcal{M}$ such that $\|x' - x_{0}\| \leq \varepsilon$ and $\|y' - y_{0}\| \leq \varepsilon$, modify the work from the inequality in your answer to show that $\langle A(x_{0}), y_{0}\rangle \leq \langle A(x'), y'\rangle + 2\varepsilon \|A\|$. After that, use the previously obtained estimates to show that
$$(1-2\varepsilon\alpha )\|A\| \leq \sup_{x\in\mathcal{N}, y\in\mathcal{M}}\langle A(x), y\rangle .$$
The result follows from taking the supremum over all $\alpha \in (1,\infty )$.