# Bounding spectral norm of ACB with spectral norms of ACA, BCB?

Suppose real matrices $A$ and $B$ are square, symmetric, positive semidefinite. Furthermore, suppose for some real square matrix $C$ (not necessarily symmetric or PSD), it is the case that $\left\lVert ACA \right\rVert_2 \leq 1$ and $\left\lVert BCB \right\rVert_2 \leq 1$. Is it true that $\left\lVert ACB \right\rVert_2 \leq 1$? Here, $\left\lVert M \right\rVert_2 = \max_{\left\lVert x \right\rVert = \left\lVert y \right\rVert = 1} y^\top M x$ is the largest singular value of $M$.

I tried for a while to prove this, but could not do so -- then, I tried checking whether it was true for randomly generated PSD matrices $A, B$ and randomly generated matrix $C$ scaled appropriately so that $\left\lVert ACA \right\rVert_2 = \left\lVert BCB \right\rVert_2 = 1$, and it was always the case that $\left\lVert ACB \right\rVert_2 \leq 1$. So it seems true perhaps except for carefully constructed counterexamples, was just wondering if anyone knew how to show it or a counterexample. Thank you in advance!

Consider this counterexample: $$A = \begin{pmatrix}0&0\\0&1\end{pmatrix},B = \begin{pmatrix}1&0\\0&0\end{pmatrix},C = \begin{pmatrix}0&2\\2&0\end{pmatrix}.$$ Then, $$ACA = 0, BCB = 0$$ but $$ACB = \begin{pmatrix}0&2\\0&0\end{pmatrix}$$ which has spectral norm $2$. In words, $A$ keeps just the second coordinate, $B$ keeps just the first coordinate, and $C$ switches the coordinates and scales by $2$. Then, both $ACA$ and $BCB$ will throw away one and then the other coordinate, but $ACB$ will keep one coordinate, switch and scale, and then keep that coordinate again.