I have seen this argument in a paper and I wanted to ask why this is correct.
Suppose that we have a vector $v(x) \in \mathbb{R}^n$ for $x \in [-1,1]$ and we know that $\|v(x)\|^2_2$ is $\alpha$-Lipschitz. We apply a contractive matrix $A \in \mathbb{R}^{n \times n}$ to $v(x)$. Why is the $\ell_2$ norm squared of this new vector $\|A v(x)\|_2^2$ also $\alpha$-Lipschitz?
What happens if we know that the vector $v(x)$ is also $\beta$-Lipschitz?
My attempt:
Suppose that $v(x)$ is $\beta$-Lipschitz. Let $f(x) = \|Av(x)\|_2^2$. We can write
\begin{align} |f(x) - f(y)| & = |\|Av(x)\|_2^2- \|Av(y)\|_2^2\\ & \leq |\|A(v(x)-v(y))\|^2_2|\\ & \leq \|A\|_2^2 \|v(x) - v(y)\|_2^2\\ & \leq \beta\|x-y\|_2^2 \end{align}
Therefore, $\|Av(x)\|_2^2$ is $\beta-$Lipschitz if $v(x)$ is $\beta-$Lipschitz. But, can we prove this only using $\|v(x)\|_2^2$ Lipschitz continuity?