Consider the block matrix $$ X = \begin{bmatrix} A &B \\ B^T &C \end{bmatrix} $$ where $A,B,C$ are all matrices of dimension $n\times n$.
I am interested in finding an upper bound for the operator norm of the $B$ sub-block, i.e. I want to find $\phi$ such that $$ \|B\|_{\text{op}} \le \phi. $$ What I have been able to show however is only a bound on the operator norm of $X$, i.e. I have $$ \|X\|_{\text{op}} \le \gamma. $$ I am wondering if there is a way to find $\phi$ based on $\gamma$?
An attempt:
it holds for any $y \in \mathbb{R}^{2n}$ that we can partition into two $n$-dimensional pieces $y=[y_1^T, y_2^T]^T$ that
\begin{align*} \|X\|^2_{\text{op}} &\ge \frac{\|X y\|^2_2}{\|y\|^2_2}. \end{align*} In particular if we take $y_1 = 0$, then \begin{align*} \|X\|^2_{\text{op}} &\ge \frac{\left \| \begin{bmatrix} By_2 \\ Cy_2\end{bmatrix}\right \|^2_2}{\|y_2\|^2_2}\\ &= \frac{\|By_2\|^2_2 + \|Cy_2\|^2_2}{\|y_2\|^2_2}\\ &\ge \frac{\|By_2\|^2_2}{\|y_2\|^2_2}. \end{align*} Since this holds for any $y_2 \in \mathbb{R}^n$, we can take a sup over $y_2$, so that $$ \|X\|_{\text{op}} \ge \sup_{y \in \mathbb{R}^n}\frac{\|By\|_2}{\|y\|_2} = \|B\|_{\text{op}} $$
