Let $\mathcal{P}=\{P_1,\cdots,P_n\}$ be a collection of point-sets in $\mathbb{R}^d$. Given a point $x\in\mathbb{R}^d$, we define the following (convex) function
$F_z(x) = \sum_{i\in[n]}\max_{p\in P_i}\Vert x-p\Vert^z$, where $z\in\{1,2\}$.
It is easy to see that $\vert F_1(x) - F_1(y)\vert \leq n\Vert x-y\Vert$. This follows from the triangle inequality. My question is that if there is a similar relation for $F_2(\cdot)$, i.e. whether $\exists$ some paramter $K>0$, which is independent of $x$ and $y$ s.t $\vert F_2(x) - F_2(y)\vert \leq K\Vert x-y\Vert^2$. Note that triangle inequality does not hold in this case, although approximate triangle inequality does exists, i.e $\Vert x-y\Vert^2 \leq 2(\Vert x-z\Vert^2 + \Vert z-y\Vert^2)$.
The following two inequalities are well known
$\vert \Vert p-c\Vert^2 - \Vert q-c\Vert^2\vert \leq \epsilon\Vert p-c\Vert^2 + (1+1/\epsilon)\Vert p-q\Vert^2$, for any $\epsilon >0$
$2\langle w-v,u-v\rangle = \Vert w-v\Vert^2 - \Vert w-u\Vert^2 + \Vert u-v\Vert^2 \geq \Vert w-v\Vert^2 - \Vert w-u\Vert^2 $.
Using point 1, it is not hard to derive the following bound
$\vert F_2(x) - F_2(y)\vert \leq (1+1/\epsilon)n\Vert x-y\Vert^2 + \epsilon.F_2(x)$
Using point 2 along with Cauchy–Schwarz inequality, one can derive the following bound
$F_2(x) - F_2(y) \leq 2\Vert x-y\Vert\sqrt{nF_2(x)}\leq \Vert x-y\Vert^2 + nF_2(x)$.