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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!

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There is a the following generalization

\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.

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No, this cannot be bounded as per the question.

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 $x=y$.

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    $\begingroup$ Thank you! Yes, you are right. in the scalar example, we need A, B to have the same signs. $\endgroup$
    – wutai
    Jan 25 at 20:07

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