# $\|Ax - By\| \leq c \|x-y\|$ for some constant $c$

I am studying the sub-multiplicative property of matrix norm. For the same linear transformation, it can be derived that $$\|Ax - Ay\| \leq \|A\|\|x-y\|$$ for all vectors $$x$$ and $$y$$.

I wonder if there are any generalizations to two different linear transformations. Does there exist some constant $$c$$ maybe as a function of $$\|A\|$$ and $$\|B\|$$such that: $$\|Ax - By\| \leq c\|x-y\|$$ for all vectors $$x$$ and $$y$$.

If everything is scalar, I can see that $$|Ax - By| \leq \max(A, B)|x-y|$$. But for multivariate cases, are there similar results? Thank you!

\begin{align}\|Ax - By\| &= \|(A-B)x + B(x-y)\| \\ &\leq \|(A-B)x\| + \|B(x-y)\| \\& \leq \|A-B\|\|x\| + \|B\|\|x-y\| \end{align} Essentially this shows that the magnitude of the difference depends on both the magnitude of the difference between $$A$$ and $$B$$ and on the magnitude of the difference between $$x$$ and $$y$$. In the counterexample by Mike the difference between $$x$$ and $$y$$ is small but the difference between $$A$$ and $$B$$ is large.
HINT: Take $$B=-A=I$$ and $$y=x$$ with $$||y||=||x||$$ nonzero. Then note on the one hand $$||y-x||=0$$ and on the other hand $$||Ay-Bx|| = ||2Ay|| = ||2y|| \not = 0 \not = c||y-x||; \ \forall c \in \mathbb{R}^+$$.
And even your scalar case is wrong: Take $$A=1$$ and $$B=-1$$ and $$y$$ be an arbitrarily large real number, and $$x=y$$.