Given $A(x)$, with $A:\mathbb{R^n}\rightarrow\mathbb{R^{m\times n}}$ and $\mathbf{f}:\mathbb{R^n}\rightarrow \mathbb{R^n}$, with $\mathbf{f}$ Lipschitz, is it possible to derive a Lipschitz bound of $A(x)\mathbf{f}(x): \mathbb{R}^n \rightarrow \mathbb{R}^m$?
I thought of proceeding as follows, $$ \begin{align} \|A(x)\mathbf{f}(x)-A(y)\mathbf{f}(y)\| &= \|A(x)\mathbf{f}(x) - A(x)\mathbf{f}(y) + A(x)\mathbf{f}(y) -A(y)\mathbf{f}(y)\|\\ &\leq \|A(x)\|L_f\|x- y\|+ \|(A(x)-A(y))\mathbf{f}(y)\|\\ & \dots\\ & \leq L_{A,\mathbf{f}}\|x-y\| \end{align} $$ but from the last expression (right before the dots) I do not see a viable way of proceeding, even by adding additional requirements.
Edit (thanks to @MoisheKohan for pointing out): The problem stated above is a reframing of my original problem, where $A(x)=J_{\mathbf{g}}^+(x)$ is the pseudo inverse of the Jacobian $J_{\mathbf{g}}$ of a function $\mathbf{g}:\mathbb{R}^n\rightarrow\mathbb{R}^m$. Assuming the matrix-valued function is Lipschitz would obviously solve the problem, but I don't know whether it is a reasonable assumption).
(Trying to reason in a general manner, I thought of reframing the last line as $a\|x-y\|+b$, with $a,b>0$, for which searching for a bound of the type $C(a,b)\|x-y\|$ would lead to a contradiction $b<0$ in the case $\|x-y\|\rightarrow 0$, but this does not take into account the fact that $b=b(x,y)$ as in the original problem. I am a bit stuck.)