Let $\Omega \subset \mathbb R^d \times \cdots \times \mathbb R^d$ be the subspace containing all elements $(x_1,\ldots, x_n)$ s.t. $x_i\neq x_j$ for all $I\neq j$. Notice that $E$ is defined on $\Omega$ which is open and convex (although $E$ can be extended continuously on $\mathbb R^d \times \cdots \times \mathbb R^d$).

I have one more questions. In Step 6, you claim that we may choose arbitrarily small $c$, but could we ensure that, for any $c>0$, the iterated point $P-\nabla G(P)$ can always be lying on the spiral? My intuition is that, $\nabla G(P)$ denotes the step that is proportional to $c$. So if $c$ is too small, we might have a step which is not long enough to touch the spiral again. Thanks a lot!

@AryehKontorovich Yes. Consider for example the most common case, i.e. $\mu(dx)=\rho(x)dx$. What is the general method to identify $\mu_n$ which may involve numerical integration?

Thank you so much for such a detailed proof. It remains one thing that is not clear: Our condition reads $r' \cos(t' - t) = r - \partial_r G(P)$ , $r' \sin(t' - t) = -r \partial_t G(P)$. Why $P'=P-\nabla G(P)$ implies that? I remember that you confirm $\nabla$ corresponds to Cartesian coordinate. Why do you use the partial derivatives w.r.t. $r$ and $t$? Thanks again!

I'm still confused with the definition of $G$, could you please specify its construction with more details?

Thanks for the quick reply. As for 1), it seems that $G((1+e^{-t})e^{it})-\nabla G((1+e^{-t})e^{it})$ is not a point as $G((1+e^{-t})e^{it})$ is real value and $\nabla G((1+e^{-t})e^{it})$ takes values in $\mathbb R^2$. Is there an error here? Also, could you please explain the meaning of "lying on the same spiral (1+e^{-s})e^{is}"?

3) The function $G$ is constructed by distinguishing $r\le 1$, $1+e^{-t}\le r< 1+2e^{-t}$, $1+2e^{-t}\le r< 1+e^{-t+2\pi}$ and $r\ge 10$. Is it correct?

2) Here the polar coordinate is applied, is the gradient $\nabla G$ still corresponding to $(x,y)$ instead of $(r,t)$?

Thanks a lot for the prompt reply. I'm reading your constructed example, and have some questions: 1) What does "the point $G((1+e^{-t})e^{it})-\nabla G((1+e^{-t})e^{it})$ also lies on the same spiral $(1+e^{-s})e^{is}$" mean?

Thank you very kindly for the reply, and I find it a very elegant answer. Could you please explain a bit more where the set of accumulation points of $(x_n)$ is connected? Also, I'd like to confirm in your claim that, the set of critical points is discrete (NOT FINITE). Thanks again!

@PiotrHajlasz Yes. Actually the statement on $\nabla F$ implies implicitly that $F$ is differentiable.

@Steve Yes. $V_i$ are all convex polygons

@Steve Thanks for the reply. All $p_i$ are strictly positive, and the density $\rho$ could be assumed to be smooth enough or even have a bounded support.