Lipschitz Property of Matrix Mulitiplication

Question

Fix $x \in \mathbb{R}^{d_1}$ and define the map $$ \begin{aligned} F:\mathbb{R}^{d_1\times d_2 + d_2} & \rightarrow \mathbb{R}^{d_2} \\ (A,b)&\mapsto Ax+b. \end{aligned} $$

Is this map Lipschitz? It seems to me that it should be but I'm having trouble showing it....

Answer 1

We adopt in both ${\mathbb R}^{d_2\times d_1+d_2}$ and in ${\mathbb R}^{d-2}$ the euclidean metric $|\cdot|$ . It is well known that the operator norm $\|A\|:=\sup_{x\ne0}{|Ax|\over|x|}$ of a linear map $A$ with matrix $[a_{ik}]$ satisfies $$\|A\|\leq |A|:=\left(\sum_{i, k}|a_{ik}^2|\right)^{1/2}\ .$$ This means that $|Ax|\leq |A|\>|x|$ for all $x$.

Denote the given (fixed) vector in ${\mathbb R}^{d_1}$ by $c$ (instead of $x$). Let two points $(A,b)$, $(A',b')\in{\mathbb R}^{d_2\times d_1+d_2}$ be given. Then we have to look at $$F(A,b)-F(A',b')=(A-A')c+(b-b')\ .\tag{1}$$ We have $|A-A'|\leq\bigl|(A,b)-(A',b')\bigr|\,$, and similarly for $|b-b'|$. From $(1)$ we therefore obtain $$|F(A,b)-F(A',b')|\leq|A-A'|\>|c|+|b-b'|\leq\bigl|(A,b)-(A',b')\bigr|\ (|c|+1)\ .$$ This shows that $F$ is Lipschitz continuous with $|c|+1$ an acceptable Lipschitz constant.