Lipschitz costant of linear maps optimal?

Question

Consider the linear transformation $f: x \mapsto Ax$ where $f: \mathbb R^n \to \mathbb R^m$ and $A$ is the corresponding matrix. It is easy to see that any finite dimensional linear map is Lipschitz continuous by using the inequality $$||Ax|| \leq ||A|| \cdot ||x||$$ for any vector norm $||.||$ (and the corresponding induced matrix norm). So we find

$$||f(x)-f(y)|| = ||Ax - Ay|| = ||A(x-y)|| \leq ||A|| \cdot ||x-y||$$ and we can use $||A||$ as the Lipschitz constant.

Now my question is: Is $L=||A||$ always optimal as a Lipschitz constant, or are there examples where $||Ax|| \leq L' ||x|| \forall x$ where $L' < ||A||$?

Answer 1

I just noticed that it is indeed optimal due to the definition:

$$||A|| := \max_{||y||=1} ||Ay||$$

So if we take $x:= \operatorname{argmax}_{||y||=1} ||Ay||$ then we see that indeed

$$||Ax|| = ||A||\cdot ||x|| = ||A||$$

Answer 2

This Lipschitz constant is optimal. Indeed, choose $x$ such that $\|Ax\|=\|A\|\|x\|$ (which is possible in finite dimensions), and observe that $$ \|Ax - A0\| = \|Ax\| = \|A\|\|x\| = \|A\|\|(x-0\|. $$