[proofplan] The finite-cluster realisation input supplies the two smooth one-dimensional examples: choosing two different cluster separations gives two non-translate strict local minimisers, and the same construction is reused for the stationary energy-dissipation example. The singular attractive example is independent and follows by rescaling a compactly supported smooth probability density against the Riesz kernel $-|x|^{-s}$. The only extra point for the stationary curves is that strict local minimality modulo translations forces the descending metric slope to vanish at the selected minimisers, while translation invariance accounts for nearby points on the same orbit. [/proofplan] [step:Use the finite-cluster input to produce two non-translate local minimisers] Apply the finite-cluster realisation input with $d_0:=1$ and $d_1:=2$. It gives an even function $W:\mathbb R\to\mathbb R$ with $W\in C^\infty(\mathbb R;\mathbb R)$ and at most polynomial growth such that $\mathcal E_W$ is not displacement convex on $\mathcal P_2(\mathbb R)$ and such that the measures \begin{align*} \rho_0:=\frac{1}{2}\delta_0+\frac{1}{2}\delta_1 \end{align*} and \begin{align*} \rho_1:=\frac{1}{2}\delta_0+\frac{1}{2}\delta_2 \end{align*} are strict $W_2$-local minimisers of $\mathcal E_W$ modulo translations. Since $C^\infty(\mathbb R;\mathbb R)\subset C(\mathbb R;\mathbb R)$, the same potential has the regularity required in the first assertion. It remains to prove that $\rho_1\notin\mathcal O(\rho_0)$. Suppose, toward a contradiction, that there exists $a\in\mathbb R$ such that $\rho_1=(\tau_a)_\#\rho_0$. The support of $(\tau_a)_\#\rho_0$ is $\{a,a+1\}$, whose two distinct points have distance $1$. The support of $\rho_1$ is $\{0,2\}$, whose two distinct points have distance $2$. Equality of the measures would imply equality of their supports, contradicting preservation of the distance between the two support points under translation. Hence $\rho_1\notin\mathcal O(\rho_0)$, proving the first assertion. [/step] [step:Rescale a compactly supported density to force the attractive Riesz energy to diverge] Assume $0<s<n$. Let $C_c^\infty(\mathbb R^n;[0,\infty))$ denote the set of smooth compactly supported maps from $\mathbb R^n$ to $[0,\infty)$, and let $\operatorname{supp}\eta$ denote the closed support of a function $\eta$. Choose $\eta\in C_c^\infty(\mathbb R^n;[0,\infty))$ such that \begin{align*} \int_{\mathbb R^n}\eta(x)\,d\mathcal L^n(x)=1. \end{align*} For each $\varepsilon\in(0,1)$, define a density $r_\varepsilon:\mathbb R^n\to[0,\infty)$ by \begin{align*} r_\varepsilon(x):=\varepsilon^{-n}\eta(x/\varepsilon) \end{align*} and define the Borel probability measure $\rho_\varepsilon$ on $\mathbb R^n$ by \begin{align*} \rho_\varepsilon(A):=\int_A r_\varepsilon(x)\,d\mathcal L^n(x) \end{align*} for every Borel set $A\subset\mathbb R^n$. The change of variables $x=\varepsilon u$ gives \begin{align*} \int_{\mathbb R^n}r_\varepsilon(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}\eta(u)\,d\mathcal L^n(u)=1, \end{align*} so $\rho_\varepsilon$ is a probability measure. Since $\operatorname{supp}r_\varepsilon=\varepsilon\operatorname{supp}\eta$ is compact, $\rho_\varepsilon$ is compactly supported and belongs to $\mathcal P_2(\mathbb R^n)$. Define the number $I_\eta\in[0,\infty]$ by \begin{align*} I_\eta:=\int_{\mathbb R^n}\int_{\mathbb R^n}|u-v|^{-s}\eta(u)\eta(v)\,d\mathcal L^n(u)\,d\mathcal L^n(v). \end{align*} This number is finite and positive. To see finiteness, choose $R>0$ such that $\operatorname{supp}\eta\subset B(0,R)$ and set $M:=\|\eta\|_\infty$. The integrand vanishes unless $(u,v)\in B(0,R)\times B(0,R)$. For every $u\in B(0,R)$, the change of variables $z=u-v$ bounds the inner singular integral by \begin{align*} \int_{B(0,R)}|u-v|^{-s}\,d\mathcal L^n(v)\le \int_{B(0,2R)}|z|^{-s}\,d\mathcal L^n(z)<\infty, \end{align*} because $s<n$ makes $z\mapsto |z|^{-s}$ locally integrable near $0$ in $\mathbb R^n$. Hence $I_\eta<\infty$. Positivity follows because $\eta\ge0$, the total mass of $\eta$ is $1$, and the kernel $|u-v|^{-s}$ is positive off the diagonal, while $\mathcal L^n\otimes\mathcal L^n$ gives the diagonal zero measure. For $W(x)=-|x|^{-s}$ on $\mathbb R^n_0$ and $W(0)=-\infty$, the product measure $\rho_\varepsilon\otimes\rho_\varepsilon$ gives the diagonal zero mass because $\rho_\varepsilon\ll\mathcal L^n$. Since $s<n$ and $r_\varepsilon$ is bounded with compact support, the same local-integrability estimate across the diagonal shows that the positive kernel integral is finite. Therefore \begin{align*} \mathcal E_W[\rho_\varepsilon]= -\frac{1}{2}\int_{\mathbb R^n}\int_{\mathbb R^n}|x-y|^{-s}r_\varepsilon(x)r_\varepsilon(y)\,d\mathcal L^n(x)\,d\mathcal L^n(y). \end{align*} Use the change of variables $x=\varepsilon u$ and $y=\varepsilon v$. Under this substitution, the product measure transforms as $d\mathcal L^n(x)d\mathcal L^n(y)=\varepsilon^{2n}d\mathcal L^n(u)d\mathcal L^n(v)$, and $|x-y|^{-s}=\varepsilon^{-s}|u-v|^{-s}$. Hence \begin{align*} \mathcal E_W[\rho_\varepsilon]= -\frac{1}{2}\varepsilon^{-s}I_\eta. \end{align*} Choose $\varepsilon_k:=1/k$ for $k\in\mathbb N$ and set $\rho_k:=\rho_{\varepsilon_k}$. Since $I_\eta>0$ and $\varepsilon_k^{-s}\to\infty$, we obtain \begin{align*} \mathcal E_W[\rho_k]\to-\infty. \end{align*} It remains to verify weak convergence. Let $\varphi:\mathbb R^n\to\mathbb R$ be bounded and continuous. The change of variables $x=\varepsilon u$ gives \begin{align*} \int_{\mathbb R^n}\varphi(x)\,d\rho_\varepsilon(x)=\int_{\mathbb R^n}\varphi(\varepsilon u)\eta(u)\,d\mathcal L^n(u). \end{align*} The integrand converges pointwise to $\varphi(0)\eta(u)$ as $\varepsilon\downarrow0$ and is dominated by $\|\varphi\|_\infty\eta(u)$, which is integrable with respect to $\mathcal L^n$. The dominated convergence theorem therefore yields \begin{align*} \int_{\mathbb R^n}\varphi(x)\,d\rho_\varepsilon(x)\to\varphi(0)=\int_{\mathbb R^n}\varphi(x)\,d\delta_0(x). \end{align*} Thus $\rho_k\rightharpoonup\delta_0$ weakly as Borel probability measures. This proves the second assertion. [guided] We construct the concentrating measures by taking one fixed smooth probability density and squeezing it into smaller balls around the origin. Let $C_c^\infty(\mathbb R^n;[0,\infty))$ be the set of smooth compactly supported maps from $\mathbb R^n$ to $[0,\infty)$, and let $\operatorname{supp}\eta$ denote the closed support of $\eta$. Choose $\eta\in C_c^\infty(\mathbb R^n;[0,\infty))$ with total mass \begin{align*} \int_{\mathbb R^n}\eta(x)\,d\mathcal L^n(x)=1. \end{align*} For $\varepsilon\in(0,1)$, define $r_\varepsilon:\mathbb R^n\to[0,\infty)$ by \begin{align*} r_\varepsilon(x):=\varepsilon^{-n}\eta(x/\varepsilon). \end{align*} The factor $\varepsilon^{-n}$ is chosen exactly so that the mass stays equal to $1$ after the change of scale. Indeed, with $x=\varepsilon u$ and $d\mathcal L^n(x)=\varepsilon^n d\mathcal L^n(u)$, \begin{align*} \int_{\mathbb R^n}r_\varepsilon(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}\eta(u)\,d\mathcal L^n(u)=1. \end{align*} Define the Borel probability measure $\rho_\varepsilon$ by \begin{align*} \rho_\varepsilon(A):=\int_A r_\varepsilon(x)\,d\mathcal L^n(x) \end{align*} for every Borel set $A\subset\mathbb R^n$. Since $\eta$ has compact support, $r_\varepsilon$ has compact support as well. Compact support implies finite second moment, so $\rho_\varepsilon\in\mathcal P_2(\mathbb R^n)$. Now define \begin{align*} I_\eta:=\int_{\mathbb R^n}\int_{\mathbb R^n}|u-v|^{-s}\eta(u)\eta(v)\,d\mathcal L^n(u)\,d\mathcal L^n(v). \end{align*} We need two facts about this number: it is finite and it is positive. For finiteness, choose $R>0$ with $\operatorname{supp}\eta\subset B(0,R)$ and write $M:=\|\eta\|_\infty$. The integrand is supported in $B(0,R)\times B(0,R)$ and is bounded by $M^2|u-v|^{-s}$ there. For fixed $u\in B(0,R)$, the substitution $z=u-v$ gives \begin{align*} \int_{B(0,R)}|u-v|^{-s}\,d\mathcal L^n(v)\le \int_{B(0,2R)}|z|^{-s}\,d\mathcal L^n(z). \end{align*} The last integral is finite exactly because $s<n$: the singularity $|z|^{-s}$ is locally integrable in $\mathbb R^n$ when the exponent is smaller than the dimension. Positivity holds because $\eta$ has mass $1$, so it is positive on a set of positive Lebesgue measure, and $|u-v|^{-s}>0$ away from the diagonal. The diagonal has zero $\mathcal L^n\otimes\mathcal L^n$ measure, so it cannot remove this positivity. We next compute the energy. Since $\rho_\varepsilon$ is absolutely continuous with respect to $\mathcal L^n$, the product measure $\rho_\varepsilon\otimes\rho_\varepsilon$ gives the diagonal zero mass. Thus the assigned value $W(0)=-\infty$ does not create an atom of infinite contribution. The same local-integrability estimate used above shows that the integral across the diagonal is finite. Therefore \begin{align*} \mathcal E_W[\rho_\varepsilon]= -\frac{1}{2}\int_{\mathbb R^n}\int_{\mathbb R^n}|x-y|^{-s}r_\varepsilon(x)r_\varepsilon(y)\,d\mathcal L^n(x)\,d\mathcal L^n(y). \end{align*} Apply the change of variables $x=\varepsilon u$ and $y=\varepsilon v$. The two Lebesgue measures contribute a factor $\varepsilon^{2n}$, the two densities contribute a factor $\varepsilon^{-2n}$, and the kernel contributes a factor $\varepsilon^{-s}$. Hence \begin{align*} \mathcal E_W[\rho_\varepsilon]= -\frac{1}{2}\varepsilon^{-s}I_\eta. \end{align*} Taking $\varepsilon_k=1/k$ and $\rho_k=\rho_{\varepsilon_k}$ gives $\mathcal E_W[\rho_k]\to-\infty$, because $I_\eta>0$. Finally we prove weak convergence to $\delta_0$. Let $\varphi:\mathbb R^n\to\mathbb R$ be bounded and continuous. Again using $x=\varepsilon u$, \begin{align*} \int_{\mathbb R^n}\varphi(x)\,d\rho_\varepsilon(x)=\int_{\mathbb R^n}\varphi(\varepsilon u)\eta(u)\,d\mathcal L^n(u). \end{align*} For each fixed $u$, continuity of $\varphi$ gives $\varphi(\varepsilon u)\to\varphi(0)$. The integrable dominating function is $\|\varphi\|_\infty\eta(u)$. The dominated convergence theorem gives \begin{align*} \int_{\mathbb R^n}\varphi(x)\,d\rho_\varepsilon(x)\to\int_{\mathbb R^n}\varphi(0)\eta(u)\,d\mathcal L^n(u)=\varphi(0). \end{align*} Since $\varphi(0)=\int_{\mathbb R^n}\varphi(x)\,d\delta_0(x)$, this is precisely weak convergence $\rho_\varepsilon\rightharpoonup\delta_0$ as Borel probability measures. [/guided] [/step] [step:Show that strict local minimisers give stationary energy-dissipating curves] For the third assertion, use the same finite-cluster realisation input with $d_0:=1$ and $d_1:=2$. Let $W\in C^\infty(\mathbb R;\mathbb R)$ be the resulting even potential with at most polynomial growth, and define \begin{align*} \rho_0:=\frac{1}{2}\delta_0+\frac{1}{2}\delta_1 \end{align*} and \begin{align*} \sigma_0:=\frac{1}{2}\delta_0+\frac{1}{2}\delta_2. \end{align*} The realisation input gives that $\mathcal E_W$ is not displacement convex on $\mathcal P_2(\mathbb R)$ and that both $\rho_0$ and $\sigma_0$ are strict $W_2$-local minimisers modulo translations. The support argument from the first step gives $\sigma_0\notin\mathcal O(\rho_0)$. Let $\mu_0\in\{\rho_0,\sigma_0\}$, and define the constant curve $\mu:[0,\infty)\to\mathcal P_2(\mathbb R)$ by \begin{align*} \mu_t:=\mu_0 \end{align*} for every $t\in[0,\infty)$. A constant curve in a metric space is locally absolutely continuous, and its metric derivative satisfies \begin{align*} |\mu'|(r)=0 \end{align*} for $\mathcal L^1$-almost every $r\in[0,\infty)$. We prove that $|\partial\mathcal E_W|(\mu_0)=0$. Let $r_0>0$ be a strict-local-minimality radius for $\mu_0$ modulo translations. Translation invariance of the interaction energy gives \begin{align*} \mathcal E_W[(\tau_a)_\#\mu_0]=\mathcal E_W[\mu_0] \end{align*} for every $a\in\mathbb R$, because $(x+a)-(y+a)=x-y$ in the defining double integral. Choose $\eta\in\mathcal P_2(\mathbb R)$ with $\eta\ne\mu_0$, with $\mathcal E_W[\eta]$ well-defined, and with \begin{align*} 0<W_2(\eta,\mu_0)<r_0. \end{align*} Then \begin{align*} \operatorname{dist}_{W_2}(\eta,\mathcal O(\mu_0))\le W_2(\eta,\mu_0)<r_0. \end{align*} If $\operatorname{dist}_{W_2}(\eta,\mathcal O(\mu_0))>0$, strict local minimality gives $\mathcal E_W[\eta]>\mathcal E_W[\mu_0]$. If $\operatorname{dist}_{W_2}(\eta,\mathcal O(\mu_0))=0$, then $\eta\in\mathcal O(\mu_0)$: indeed, choose $a_j\in\mathbb R$ such that $W_2(\eta,(\tau_{a_j})_\#\mu_0)\to0$; since $W_2((\tau_a)_\#\mu_0,\mu_0)=|a|$, the sequence $(a_j)_{j=1}^{\infty}$ is bounded, and a convergent subsequence $a_{j_m}\to a$ yields $(\tau_{a_{j_m}})_\#\mu_0\to(\tau_a)_\#\mu_0$ in $W_2$, so $\eta=(\tau_a)_\#\mu_0$. In this second case, translation invariance gives $\mathcal E_W[\eta]=\mathcal E_W[\mu_0]$. Hence, for all sufficiently near admissible $\eta$, \begin{align*} (\mathcal E_W[\mu_0]-\mathcal E_W[\eta])^+=0. \end{align*} By the definition of the descending metric slope, $|\partial\mathcal E_W|(\mu_0)=0$. Since $\mu_t=\mu_0$ for all $t\ge0$, we have $|\partial\mathcal E_W|(\mu_t)=0$ for every $t\ge0$ and \begin{align*} \mathcal E_W[\mu_t]=\mathcal E_W[\mu_s] \end{align*} for every $0\le s\le t$. Therefore \begin{align*} \int_s^t |\mu'|^2(r)\,d\mathcal L^1(r)=0 \end{align*} and \begin{align*} \int_s^t |\partial\mathcal E_W|^2(\mu_r)\,d\mathcal L^1(r)=0. \end{align*} Substitution into the energy-dissipation inequality gives \begin{align*} \mathcal E_W[\mu_t]+\frac{1}{2}\int_s^t |\mu'|^2(r)\,d\mathcal L^1(r)+\frac{1}{2}\int_s^t |\partial\mathcal E_W|^2(\mu_r)\,d\mathcal L^1(r)=\mathcal E_W[\mu_s]. \end{align*} Thus the constant curves at $\rho_0$ and $\sigma_0$ are energy-dissipating Wasserstein gradient-flow solutions in the stated metric-slope sense. Finally, \begin{align*} \lim_{t\to\infty}\rho_t=\rho_0 \end{align*} and \begin{align*} \lim_{t\to\infty}\sigma_t=\sigma_0 \end{align*} in $W_2$, because both curves are constant. Since $\sigma_0\notin\mathcal O(\rho_0)$, in particular $\sigma_0\ne\rho_0$, so the two stationary limits are different. This proves the third assertion and completes the proof. [/step]