Smoothness constant of $f$

Question

Let $f(x, y)=g(x)+g(y)+\frac{\lambda}{2}\|x-y\|^2$ where $x \in R^d$ and $y \in R^d$. Suppose $g$ is $\beta$-Lipschitz smooth. What will be the smoothness constant of $f$ in terms of $\beta$, $\lambda$ and $d$? I am okay with a loose upper bound.

Attempt: $$\ \nabla^2f(x, y)= \begin{bmatrix}\nabla_x \nabla_xf(x, y) & \nabla_x \nabla_yf(x, y)\\ \nabla_x \nabla_yf(x, y) & \nabla_y \nabla_yf(x, y)\end{bmatrix}= \begin{bmatrix}\nabla^2g(x)+\lambda I_{d\times d} & -\lambda I_{d\times d}\\ -\lambda I_{d\times d} & \nabla^2g(y) + \lambda I_{d\times d}\end{bmatrix} $$

$$\ \implies \text{Smoothness Constant}=\|\nabla^2f(x, y)\|\leq \ \biggr\|\begin{bmatrix}\nabla^2g(x) & 0\\ 0 & \nabla^2g(y)\end{bmatrix}\biggr\| \ + \lambda\biggr\|\begin{bmatrix}I_{d\times d} & -I_{d\times d}\\ -I_{d\times d} & I_{d\times d}\end{bmatrix}\biggr\| $$ The matrix norm used above is Spectral Norm. We have $\|\nabla^2g(x)\| \leq \beta$. How to use this to bound $\| \nabla^2f(x, y) \|$?

Answer 1

We can directly bound the matrix norm by subadditive property giving us $\|\nabla^2f(x, y)\| \leq \beta +2\lambda$

Answer 2

$|f(x,y)-f(x',y')| \leq |g(x)-g(x')|+|g(y)-g(y')|+\frac{\lambda}{2}|\|x-y\|^2-\|x'-y'\|^2| \leq \beta(\|x-x'\|+\|y-y'\|)+\frac{\lambda}{2}|\|x-y\|^2-\|x'-y'\|^2|$

Thus we only have to estimate $|\|x-y\|^2-\|x'-y'\|^2|$.

Writing it as $|(\|x-y\|-\|x'-y'\|)|(\|x-y\|+\|x'-y'\|)$ we can estimate the first factor with $\|x-x'\|+\|y-y'\|$ and the second one with $\|x\|+\|x'\|+\|y\|+\|y'\|$ which can also be estimate by $K(d)(\|x-x'\|+\|y-y'\|)$ where $K(d)$ is an opportune positive costant that exists thanks to the fact that in $\mathbb{R}^d$ all the norms are equivalent. Thus it is easy to find the Lipschitz-costant $H(d,\beta,\lambda)$ such that: $$|f(x,y)-f(x',y')| \leq H(d,\beta,\lambda)(\|x-x'\|+\|y-y'\|)$$ and then we have concluded.