Show that $g:\mathbb{R}^n\rightarrow\mathbb{R}$ defined by $g(x)=f(Ax+b)$ satisfies $g\in C^{1,1}_\tilde{L}(\mathbb{R}^n)$ where $\tilde{L}=||A||^2L$.

Question

Let $f\in C_L^{1,1}(\mathbb{R}^m)$, and let $A\in\mathbb{R}^{m\times n}$, $b\in\mathbb{R}^m$. Show that the function $g:\mathbb{R}^n\rightarrow\mathbb{R}$ defined by $g(x)=f(Ax+b)$ satisfies $g\in C^{1,1}_\tilde{L}(\mathbb{R}^n)$, where $\tilde{L}=||A||^2L$.

My Attempt

Given that $f\in C^{1,1}_L$, we know that $|f(x)-f(y)|\leq L|x-y|$. If we want to show that $g\in C_\tilde{L}^{1,1}$ then we have to show that $|f(Ax+b)-f(Ay+b)|\leq \tilde{L}|x-y|=||A||^2L|x-y|$ so we want to show that $\frac{|f(Ax+b)-f(Ay+b)|}{||A||^2}\leq |f(x)-f(y)|$.

I'm not sure if I'm headed in the right direction. Can I get some pointers on how to approach this problem?

Answer 1

Since $f$ is continuously differentiable, the multivariate chain rule yields $\nabla g(x) = A^{T} \nabla f(Ax + b)$. From there,

$$ \| \nabla g(x) - \nabla g(y) \| = \| A^{T} (\nabla f(Ax + b) - \nabla f(Ay + b)) \| \leq \| A \| \| \nabla f(Ax + b) - \nabla f(Ay + b) \|, $$ using the identity $\|A v \| \leq \|A\| \| v \|$. Finally, use the Lipschitz continuity of $\nabla f$ to get the claimed result.