Frobenius norm bound

Question

Is there any way to bound the Frobenius norm of a product of square matrices $A,B$ and a vector $x$ in the following way:

$$ \|ABx\|≤ \|Ax\|\text{ and }\|B\|. $$

Answer 1

If $A$ is invertible, then $$\|ABx\|=\|ABA^{-1}Ax\|\leq\|ABA^{-1}\|\|Ax\|=\kappa(A)\|B\|\|Ax\|,$$ where $\kappa(A)=\|A\|\|A^{-1}\|$ is the condition number of $A$. So yes, $\|ABx\|$ can be bounded by "something" involving $\|Ax\|$ and $\|B\|$ if $A^{-1}$ exists.

Answer 2

The answer is no. Set $$ A = \begin{pmatrix} 0 & c \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$ where $c > 0$ and let $x = (1,0)^T$, then $Ax$ is the zero vector but $\|ABx\| = c$ can be arbitrarily large.