Why is the Barycenter operation in Hadamard spaces Lipschitz continuous? I am looking into exercise 9.2.22 of "A course in metric geometry" by Burago-Burago-Ivanov. 
For a  Hadamard space $H$ (a complete simply connected metric space of nonpositive curvature in the Alexandrov sense) 
and a fixed point $p \in H$ the function $f(x) = (d(p,x))^2$ is strongly convex, where $d$ denotes the metric of $H$. 
Therefore for a fixed collection of points $p_1, \ldots, p_n$ the function 
$$ f(x) = \sum_{i=1}^n (d(p_i,x))^2 $$ 
is also strictly convex and therefore has a unique point $y$ where $f$ attains its maximum. The Point $y$ is called the Barycenter of $p_1, \ldots ,p_n$. 
The Statement I want to prove is the following: 
The Barycenter operation is Lipschitz continuous, i.e if $q_1, \ldots q_n$ is another collection of points and $y'$ is their barycenter then one has a relation of the form 
$$ d(y,y') \leq L \cdot \left( d(p_1,q_1) + \ldots + d(p_n,q_n) \right) $$
I suspect that i need some statement of the form: If strongly convex functions $f,g$ satisfy $\Vert f-g\Vert_{\infty}< C $ then their Minima $y,y'$ satisfy $ d(y,y')< CL$. Does such a statement exist? 
 A: (1) In a Hardmard space $(X ,|\ |)$, define a convex set
$K$ wrt $|\ |$ Then then we have $$ f : X\rightarrow \mathbb{R},\
f(x)=|xK|\ ({\rm or}\ |xK|^2)
$$ Note that $f$ attains a minimum at $x_0$ and $x_0$ is unique So
we have
$$ \pi : X\rightarrow K,\ \pi (x)=x_0 $$
Then $\pi$ is distance nonexpanding
Proof : Note that $$\angle (xx_0,y_0x_0),\ \angle (yy_0,x_0y_0) >
\frac{\pi}{2}$$
If ${\rm geod}_{[ab]}(t)$ is unit speed geodesic from $a$ to $b$, then
we have first variation formula, since $X$ is Hadamard space, $$
|{\rm geod}_{[x_0x]} (t)- {\rm geod}_{[y_0y]} (\tau ) | = |x_0y_0|
-t\cos\ \angle (xx_0,y_0x_0) -\tau \cos\ \angle (yy_0,x_0y_0) +
o(t+\tau )
$$ Hence we complete the proof
(2) Now we will apply to our case : If we have $(H,d)$,
then define
$$ (X,|\ |)= \underbrace{(H,d)\times \cdots \times (H,d)}_{n\ times}
$$ where $$ |(p_i)(p_i')| := \sqrt{ \sum_i d(p_i,p_i')^2 } $$
Then we have a convex set $K:=\Delta H$ And let $f((p_i)):=|(p_i)K|=\inf_{x\in H}\ \sqrt{\sum_i d(p_i,x)^2}
$ 
By (1) since $\pi$ is distance nonexpanding so $$ \pi:X\rightarrow K,\ \pi((x_i))=(a_i),\ a_i=x_0,\ \pi((y_i))=(b_i),\ b_i=y_0$$ $$\sqrt{n}d(x_0,y_0)
\leq \sqrt{\sum_i d(x_i,y_i)^2} $$
This complete the proof
