Norm of symmetric positive semidefinite matrices If we have two symmetric, positive semidefinite matrices $A$ and $B$ which satisfy $\| A \|_2 \ge \| B \|_2$, how can we relate the $2$-norms of $CAC$ and $CBC$, where matrix $C$ is Toeplitz, symmetric and also positive semidefinite?

I have been researching a lot trying to find an answer to my question and didn't find any so I would appreciate it if anyone can help. Thank you for your help.
 A: Observe that the $2$-norm of a matrix is
$$\|A\|_2=\sup_{\|x\|=1}\sqrt{\langle x,A^*Ax\rangle}=\sup_{\|x\|=1}\sqrt{\|Ax\|^2}=\|A\|,$$
i.e. the 2-norm and the operator norm coincide. Knowing only $\|B\|\leq\|A\|$, I would not expect to be able to control $\|CBC\|$ in terms of $|\|CAC\|$. The following example shows that you need some type of extra condition to get a bound:
$$
A=\left(\matrix{
1&-1\\
-1&1
}\right),
C=\left(\matrix{
1&1\\
1&1
}\right).
$$
For these matrices, you get $CAC=0$.
Case 1: If we assume that $C$ is invertible, then
$$
\|CAC\|\geq \|C^{-1}\|^{-1}\|AC\|=\|C^{-1}\|^{-1}\|(AC)^*\|=\|C^{-1}\|^{-1}\|CA\| \geq \|C^{-1}\|^{-2}\|A\|,
$$
so
$$
\|CBC\|\leq \|C\|^2\|B\|\leq\|C\|^2\|A\|\leq\|C\|^2\|C^{-1}\|^2\|CAC\|.
$$
Case 2: Suppose $A$ is invertible, such that $I\leq \|A^{-1}\| A$. Observe that $B\leq\|B\|I\leq \|B\|\|A^{-1}\|A$, which implies $CBC\leq \|B\|\|A^{-1}\|CAC$, and thus
$$
\|CBC\|\leq \|B\|\|A^{-1}\|\|CAC\|.
$$
Note that this holds even without the condition $\|B\|\leq\|A\|$!
