Lipschitz continuity of parametric optimizer Consider the parametric optimal solution $x^{*}: \mathbb{R}^n \rightarrow \mathbb{R}^n$ defined as
$$ x^*( y ) := \arg\min_{x \in X } \ \ x^\top x + x^\top A y \\ \quad \qquad \text{subject to: } \ f(x) = 0$$
where $X \subset \mathbb{R}^n$ is compact and convex, $A = A^\top$, $f: \mathbb{R}^n \rightarrow Z \subset \mathbb{R}^n$ is continuously differentiable, hence Lipschitz continuous, with $Z$ compact.
Assume that, for all $y \in \mathbb{R}^n$, the optimal solution $x^*(y)$ is unique.
I am wondering if $x^*(\cdot)$ is Lipschitz continuous. And, if not, I am looking for a counterexample and additional assumptions under which the claim holds.
Comment: it can be shown that $x^*(\cdot)$ is continuous; if $f$ is affine, then the optimization problem is convex and, if the optimal solution is unique, then $x^*(\cdot)$ is Lipschitz continuous, because $x^*(y)$ would be the projection of $- \frac{1}{2}A y $ onto a convex set.
 A: For the case that $f(x)$ is linear (or more generally $f(x)=0$ is a convex set) indeed there exists a proof  here
http://www.opentradingsystem.com/quantNotes/Projection_on_convex_set_.html
So this  gives us a hint for a contr-example when $f(x)$ is not linear and the set $f(x)=0$ is not convex. 
So let's built a simple contr-example on $R^2$:
$f_1(x)=x_1^2+x_2^2-1, f_2(x)=x_1^2-1, A=I$. It defines only two feasible points: $(1,0)$ and $(-1,0)$. Now let $y=(0,0)$. Obviously for $y=(\epsilon,0)$ your solution is $(1,0)$ and for  $y=(-\epsilon,0)$ your solution is $(-1,0)$. This completes the contr-example. 
even simpler contr-example can be built for 1 dimensional $x$. Let $f(x)=x^2-1$ then the feasible $x$s are only $x=\pm 1$. For $A=1$ and $y\ne 0$ we have solution $x(y)=(y>0)1 +(y<0)(-1)$  and not unique solution $\pm 1$ for $y=0$.
EDITION
Unfortunately these examples are not good enough, because the condition of uniqueness of $x^*(y)$ is violated. I suspect that the uniqueness of $x^*(y)$ implies the convexity of the set $\{x|X\in X, f(x)=0\}$ but I don't see how to prove it yet.   Geometrically it seems obvious for me that if the set is not convex we can always can find a $y$ such that the projector $x(y)$ is not unique, so uniqueness of $x(y)$ for any $y$ guarantees convexity.   
