Remark 2.3. The regularity $\rho \in\mathcal{C}([0,T],\mathcal{P}_4(\mathbb{R}^d\times \mathbb{R}^d))$ obtained in Theorem 2.2 above is an immediate consequence of the regularity of the initial condition $\rho _0 \in\mathcal{P}_4(\mathbb{R}^d\times \mathbb{R}^d)$ . It allows to extend the test function space $\mathcal{C}^{\infty }_{c}(\mathbb{R}^d\times \mathbb{R}^d)$ in Definition 2.1 to the larger space

(2.6) \begin{equation} \begin{aligned}\mathcal{C}^2_{*}(\mathbb{R}^d\times \mathbb{R}^d) \,:\!=\,\big \{\phi \in\mathcal{C}^2(\mathbb{R}^d\times \mathbb{R}^d):\, &\left |{\partial _{x_k}\phi (x,y)}\right | \leq C_{\phi} (1\!+\!\|{x}\|_2\!+\!\|{y}\|_2) \text{ for some } C_{\phi} \!\gt \!0\\ &\,\text{and } \sup _{(x,y)\in \mathbb{R}^d\times \mathbb{R}^d}\max _{k=1,\dots,d}|\partial ^2_{x_kx_k} \phi (x,y)| \lt \infty \big \}, \end{aligned} \end{equation}

as can be seen from the proof of Theorem 2.2, which we sketch in what follows.

Proof sketch of Theorem 2.2. The proof is based on the Leray-Schauder fixed point theorem and follows the steps taken in [Reference Carrillo, Choi, Totzeck and Tse12, Theorems 3.1, 3.2].

Step 1: For a given function $u\in \mathcal{C}([0,T],\mathbb{R}^d)$ and an initial measure $\rho _0\in\mathcal{P}_4(\mathbb{R}^d)$ , according to standard SDE theory [Reference Arnold3, Chapters 6], we can uniquely solve the auxiliary SDE

\begin{align*} &\begin{aligned} d\widetilde{X}_{t} = \begin{aligned}[t] &\!-\lambda _1\big (\widetilde{X}_{t}-u_t\big )\, dt -\lambda _2\big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big )\, dt -\lambda _3\nabla{\mathcal{E}}(\widetilde{X}_{t}) \,dt\\ &\!+\sigma _1 D\big (\widetilde{X}_{t}-u_t\big )\, d B_{t}^{1} +\sigma _2 D\big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big )\, d B_{t}^{2} +\sigma _3 D\big (\nabla{\mathcal{E}}(\widetilde{X}_{t})\big )\, d B_{t}^{3} \end{aligned} \end{aligned}\\ &\mspace{1mu}d\widetilde{Y}_{t} = \kappa \big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big )\, S^{\beta,\theta }\big (\widetilde{X}_{t}, \widetilde{Y}_{t}\big )\, dt \end{align*}

with $(\widetilde{X}_0,\widetilde{Y}_0)\sim \rho _0$ . This is due to the fact that the coefficients of the drift and diffusion terms are locally Lipschitz continuous and have at most linear growth, which, in turn, is a consequence of the assumptions on $\mathcal{E}$ as well as the smoothness of $S^{\beta,\theta }$ as defined in (1.4). This induces $\widetilde{\rho} _t=\textrm{Law}\big ((\widetilde{X}_t,\widetilde{Y}_t)\big )$ . Moreover, the assumed regularity of the initial distribution $\rho _0 \in\mathcal{P}_4(\mathbb{R}^d\times \mathbb{R}^d)$ allows to obtain a fourth-order moment estimate of the form $\mathbb{E}\big [\|{\widetilde{X}_t}\|_2^4+\|{\widetilde{Y}_t}\|_2^4\big ] \leqslant \big (1+2\mathbb{E}\big [\|{\widetilde{X}_0}\|_2^4+\|{\widetilde{Y}_0}\|_2^4\big ]\big )e^{ct}$ , see, e.g. [Reference Arnold3, Chapter 7]. So, in particular, $\widetilde{\rho} \in\mathcal{C}([0,T],\mathcal{P}_4(\mathbb{R}^d\times \mathbb{R}^d))$ .

Step 2: For some test function $\phi \in\mathcal{C}^2_{*}(\mathbb{R}^d\times \mathbb{R}^d)$ as defined in (2.6), by Itô’s formula, we derive

\begin{equation*} \begin{split} &d\phi (\widetilde{X}_t,\widetilde{Y}_t) = \nabla _x\phi (\widetilde{X}_t,\widetilde{Y}_t) \cdot \left ({-}\lambda _1\big (\widetilde{X}_{t}-u_t\big ) -\lambda _2\big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big ) -\lambda _3\nabla{\mathcal{E}}(\widetilde{X}_{t})\right ) dt \\ &\quad \quad \, + \nabla _y\phi (\widetilde{X}_t,\widetilde{Y}_t) \cdot \left (\kappa \big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big )\, S^{\beta,\theta }\big (\widetilde{X}_{t}, \widetilde{Y}_{t}\big )\right ) dt \\ &\quad \quad \, + \frac{1}{2} \sum _{k=1}^d \partial ^2_{x_kx_k}\phi (\widetilde{X}_t,\widetilde{Y}_t)\left (\sigma _1^2 D\big (\widetilde{X}_{t}-u_t\big )_{kk}^2 \!+\! \sigma _2^2 D\big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big )_{kk}^2 \!+\! \sigma _3^2 D\big (\nabla{\mathcal{E}}(\widetilde{X}_{t})\big )_{kk}^2\right )dt \\ &\quad \quad \, + \nabla _x\phi (\widetilde{X}_t,\widetilde{Y}_t) \cdot \left (\sigma _1 D\big (\widetilde{X}_{t}-u_t\big )\, d B_{t}^{1} + \sigma _2 D\big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big )\, d B_{t}^{2} + \sigma _3 D\big (\nabla{\mathcal{E}}(\widetilde{X}_{t})\big )\, d B_{t}^{3}\right ) \end{split} \end{equation*}

After taking the expectation, applying Fubini’s theorem and observing that the stochastic integrals of the form $\mathbb{E}\int _0^t\nabla _x\phi (\widetilde{X}_t,\widetilde{Y}_t)\cdot D(\,\cdot \,)\,dB_t$ vanish as a consequence of [Reference Øksendal50, Theorem 3.2.1(iii)] due to the established regularity $\widetilde{\rho} \in\mathcal{C}([0,T],\mathcal{P}_4(\mathbb{R}^d\times \mathbb{R}^d))$ and $\phi \in\mathcal{C}^2_{*}(\mathbb{R}^d\times \mathbb{R}^d)$ , we obtain

\begin{equation*} \begin{split} &\frac{d}{dt}\mathbb{E}\phi (\widetilde{X}_t,\widetilde{Y}_t) = -\mathbb{E}\nabla _x\phi (\widetilde{X}_t,\widetilde{Y}_t) \cdot \left (\lambda _1\big (\widetilde{X}_{t}-u_t\big ) +\lambda _2\big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big ) +\lambda _3\nabla{\mathcal{E}}(\widetilde{X}_{t})\right ) \\ &\quad \ + \mathbb{E} \nabla _y\phi (\widetilde{X}_t,\widetilde{Y}_t) \cdot \left (\kappa \big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big ) S^{\beta,\theta }\big (\widetilde{X}_{t}, \widetilde{Y}_{t}\big )\right ) \\ &\quad \ + \frac{1}{2} \sum _{k=1}^d \partial ^2_{x_kx_k}\phi (\widetilde{X}_t,\widetilde{Y}_t)\left (\sigma _1^2 D\big (\widetilde{X}_{t}-u_t\big )_{kk}^2 \!+\! \sigma _2^2 D\big (\widetilde{X}_{t}-\widetilde{Y}_{t}\big )_{kk}^2 \!+\! \sigma _3^2 D\big (\nabla{\mathcal{E}}(\widetilde{X}_{t})\big )_{kk}^2\right ) \end{split} \end{equation*}

according to the fundamental theorem of calculus. This shows that $\widetilde{\rho} \in\mathcal{C}([0,T],\mathcal{P}_4(\mathbb{R}^d\times \mathbb{R}^d))$ satisfies the Fokker-Planck equation

(2.7) \begin{equation} \begin{aligned} &\frac{d}{dt}\iint \phi (x,y) \,d\widetilde{\rho} _t(x,y) = - \iint \kappa S^{\beta,\theta }(x,y) \left \langle y-x,\nabla _y\phi (x,y) \right \rangle d\widetilde{\rho} _t(x,y)\\ &\quad \ - \iint \lambda _1\left \langle x - u_t, \nabla _x \phi (x,y) \right \rangle + \lambda _2\left \langle x-y,\nabla _x\phi (x,y) \right \rangle + \lambda _3\left \langle \nabla{\mathcal{E}}(x), \nabla _x \phi (x,y) \right \rangle d\widetilde{\rho} _t(x,y)\\ &\quad \ + \frac{1}{2} \iint \sum _{k=1}^d \left (\sigma _1^2D\!\left (x-u_t\right )_{kk}^2 + \sigma _2^2D\!\left (x-y\right )_{kk}^2 + \sigma _3^2D\!\left (\nabla{\mathcal{E}}(x)\right )_{kk}^2 \right ) \partial ^2_{x_kx_k} \phi (x,y) \,d\widetilde{\rho} _t(x,y) \end{aligned} \end{equation}

The remainder is identical to the cited reference and is summarised below for completeness.

Step 3: Setting $\mathcal{T} u\,:\!=\,y_{\alpha }({\widetilde{\rho} _Y})\in \mathcal{C}([0,T],\mathbb{R}^d)$ provides the self-mapping property of the map

\begin{align*}\mathcal{T}\;:\;\mathcal{C}([0,T],\mathbb{R}^d)\rightarrow\mathcal{C}([0,T],\mathbb{R}^d), \quad u\mapsto \mathcal{T}u=y_{\alpha }({\widetilde{\rho} _Y}), \end{align*}

which is compact as a consequence of a stability estimate for the consensus point [Reference Carrillo, Choi, Totzeck and Tse12, Lemma 3.2]. More precisely, as shown in the cited result, it holds $\|{y_{\alpha }({\widetilde{\rho} _{Y,t}})-y_{\alpha }({\widetilde{\rho} _{Y,s}})}\|_2 \lesssim W_2(\widetilde{\rho} _{Y,t},\widetilde{\rho} _{Y,s})$ for $\widetilde{\rho} _{Y,t},\widetilde{\rho} _{Y,s}\in\mathcal{P}_4(\mathbb{R}^d)$ . Together with the Hölder- $1/2$ continuity of the Wasserstein- $2$ distance $W_2(\widetilde{\rho} _{Y,t},\widetilde{\rho} _{Y,s})$ , this ensures the claimed compactness of $\mathcal{T}$ .

Step 4: Then, for $u=\vartheta\mathcal{T} u$ with $\vartheta \in [0,1]$ , there exists $\rho \in\mathcal{C}([0,T],\mathcal{P}_4(\mathbb{R}^d\times \mathbb{R}^d))$ satisfying (2.7) with marginal $\rho _Y$ such that $u=\vartheta y_{\alpha }({\rho _Y})$ . For such $u$ , a uniform bound can be obtained either thanks to the boundedness or the growth condition of $\mathcal{E}$ required in the statement. An application of the Leray-Schauder fixed point theorem concludes the proof by providing a solution to (1.5).

Proof. We note that the functions $\phi _{\mathcal{X}}(x,y) = 1/2\left \|{x-x^*}\right \|_2^2$ and $\phi _{\mathcal{Y}}(x,y) = 1/2\left \|{y-x}\right \|_2^2$ are in $\mathcal{C}^2_{*}(\mathbb{R}^d\times \mathbb{R}^d)$ and recall that $\rho$ satisfies the weak solution identity (2.1) for such test functions. Hence, by applying (2.1) with $\phi _{\mathcal{X}}$ and $\phi _{\mathcal{Y}}$ as above, we obtain for the evolution of $\mathcal{X}(\rho _t)$

(3.1) \begin{equation} \begin{aligned} &\frac{d}{dt}\mathcal{X}(\rho _t) = - \iint \lambda _1\!\left \langle x \!-\! y_{\alpha }({\rho _{Y,t}}), x\!-\!x^* \right \rangle + \lambda _2 \!\left \langle x\!-\!y, x\!-\!x^* \right \rangle + \lambda _3 \!\left \langle \nabla{\mathcal{E}}(x), x\!-\!x^* \right \rangle d\rho _t(x,y)\\ &\qquad \qquad \;\; + \frac{1}{2} \iint \sigma _1^2\left \|{x\!-\!y_{\alpha }({\rho _{Y,t}})}\right \|_2^2 + \sigma _2^2\left \|{x\!-\!y}\right \|_2^2 + \sigma _3^2\left \|{\nabla{\mathcal{E}}(x)}\right \|_2^2 d\rho _t(x,y) \end{aligned} \end{equation}

and for the evolution of $\mathcal{Y}(\rho _t)$

(3.2) \begin{equation} \begin{aligned} &\frac{d}{dt}\mathcal{Y}(\rho _t) = - \iint \kappa S^{\beta,\theta }(x,y) \left \|{x-y}\right \|_2^2 d\rho _t(x,y)\\ &\qquad \qquad \;\, - \iint \lambda _1\left \langle x - y_{\alpha }({\rho _{Y,t}}), x-y \right \rangle + \lambda _2 \left \|{x-y}\right \|_2^2 + \lambda _3\left \langle \nabla{\mathcal{E}}(x), x-y \right \rangle d\rho _t(x,y)\\ &\qquad \qquad \;\, + \frac{1}{2} \iint \sigma _1^2\left \|{x-y_{\alpha }({\rho _{Y,t}})}\right \|_2^2 + \sigma _2^2\left \|{x-y}\right \|_2^2 + \sigma _3^2\left \|{\nabla{\mathcal{E}}(x)}\right \|_2^2 d\rho _t(x,y). \end{aligned} \end{equation}

Here we used $\nabla _x\phi _{\mathcal{X}}(x,y) = x-x^*$ , $\nabla _y\phi _{\mathcal{X}}(x,y)=0$ , $\partial ^2_{x_kx_k} \phi _{\mathcal{X}}(x,y) = 1$ , $\nabla _x\phi _{\mathcal{Y}}(x,y) = x-y$ , $\nabla _y\phi _{\mathcal{Y}}(x,y)=y-x$ and $\partial ^2_{x_kx_k} \phi _{\mathcal{Y}}(x,y) = 1$ . Let us now collect auxiliary estimates in (3.3a)–(3.3g), which turn out to be useful in establishing upper bounds for (3.1) and (3.2). Using standard tools such as Cauchy-Schwarz and Young’s inequality we have

(3.3a) \begin{align} -\left \langle x-y, x-x^* \right \rangle &\leq \left \|{x-y}\right \|_2\left \|{x-x^*}\right \|_2 \leq \frac{1}{2}\!\left (\left \|{x-y}\right \|_2^2+\left \|{x-x^*}\right \|_2^2\right ), \end{align}

(3.3b) \begin{align} -\left \langle x-y_{\alpha }({\rho _{Y,t}}), x-x^* \right \rangle &= -\left \|{x-x^*}\right \|_2^2 - \left \langle x^*-y_{\alpha }({\rho _{Y,t}}), x-x^* \right \rangle \nonumber \\ &\leq -\left \|{x-x^*}\right \|_2^2 + \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2 \left \|{x-x^*}\right \|_2, \end{align}

(3.3c) \begin{align} -\left \langle x-y_{\alpha }({\rho _{Y,t}}), x-y \right \rangle &= -\left \langle x-x^*, x-y \right \rangle - \left \langle x^*-y_{\alpha }({\rho _{Y,t}}), x-y \right \rangle \nonumber \\ &\leq \frac{1}{2}\!\left (\left \|{x-y}\right \|_2^2+\left \|{x-x^*}\right \|_2^2\right ) + \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2\left \|{x-y}\right \|_2, \end{align}

(3.3d) \begin{align} \left \|{x-y_{\alpha }({\rho _{Y,t}})}\right \|_2^2 &= \left \|{x-x^*}\right \|_2^2 - 2\left \langle y_{\alpha }({\rho _{Y,t}})-x^*,x-x^*\right \rangle + \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2^2 \nonumber \\ &\leq \left \|{x-x^*}\right \|_2^2 + 2\left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2\left \|{x-x^*}\right \|_2 + \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2^2, \end{align}

where in (3.3b)–(3.3d) we expanded the left-hand side of the scalar product and the norm by subtracting and adding $x^*$ . Furthermore, by means of A3 we obtain

(3.3e) \begin{align} -\left \langle \nabla{\mathcal{E}}(x), x-x^* \right \rangle &\leq \left \|{\nabla{\mathcal{E}}(x)}\right \|_2\left \|{x-x^*}\right \|_2 \leq C_{\nabla{\mathcal{E}}}\left \|{x-x^*}\right \|_2^2, \end{align}

(3.3f) \begin{align} -\left \langle \nabla{\mathcal{E}}(x), x-y \right \rangle &\leq \left \|{\nabla{\mathcal{E}}(x)}\right \|_2\left \|{x-y}\right \|_2 \leq C_{\nabla{\mathcal{E}}}\left \|{x-x^*}\right \|_2\left \|{x-y}\right \|_2 \nonumber \\ &\leq \frac{C_{\nabla{\mathcal{E}}}}{2}\left (\left \|{x-y}\right \|_2^2+\left \|{x-x^*}\right \|_2^2\right ), \end{align}

(3.3g) \begin{align} \left \|{\nabla{\mathcal{E}}(x)}\right \|_2^2 &\leq C_{\nabla{\mathcal{E}}}^2\left \|{x-x^*}\right \|_2^2. \end{align}

Integrating the bounds (3.3a), (3.3b), (3.3d), (3.3e) and (3.3g) into equation (3.1) results in the upper bound

\begin{equation*} \begin {aligned} &\frac {d}{dt} \mathcal{X}(\rho _t) \leq - \left ( 2\lambda _1 - \lambda _2 - 2\lambda _3 C_{\nabla {\mathcal {E}}} - \sigma _1^2 - \sigma _3^2C_{\nabla {\mathcal {E}}}^2 \right ) \mathcal{X}(\rho _t) + \left (\lambda _2+\sigma _2^2\right ) \mathcal{Y}(\rho _t) \\ &\qquad \qquad \; + \sqrt {2}\left (\lambda _1+\sigma _1^2\right ) \sqrt {\mathcal{X}(\rho _t)} \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2 + \frac {\sigma _1^2}{2}\left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2^2, \end {aligned} \end{equation*}

where we furthermore used that by Jensen’s inequality

(3.4) \begin{equation} \iint \left \|{x-x^*}\right \|_2 d\rho _t(x,y) \leq \sqrt{\iint \left \|{x-x^*}\right \|_2^2 d\rho _t(x,y)} =\sqrt{2\mathcal{X}(\rho _t)}. \end{equation}

For equation (3.2), we first note that, by definition, $S^{\beta,\theta }\geq \theta$ uniformly. This combined with the bounds (3.3c), (3.3d), (3.3f) and (3.3g) allows to derive

\begin{equation*} \begin {aligned} &\frac {d}{dt} \mathcal{Y}(\rho _t) \leq - \left (2\kappa \theta + 2\lambda _2 - \lambda _1 - \lambda _3 C_{\nabla {\mathcal {E}}} - \sigma _2^2\right ) \!\mathcal{Y}(\rho _t) + \left (\lambda _1 + \lambda _3 C_{\nabla {\mathcal {E}}} + \sigma _1^2 + \sigma _3^2C_{\nabla {\mathcal {E}}}^2 \right ) \!\mathcal{X}(\rho _t)\\ &\qquad \qquad \, + \sqrt {2}\left (\lambda _1 \sqrt {\mathcal{Y}(\rho _t)} + \sigma _1^2 \sqrt {\mathcal{X}(\rho _t)}\right ) \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2 + \frac {\sigma _1^2}{2}\left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2^2, \end {aligned} \end{equation*}

where we used (3.4) together with an analogous bound for $\iint \left \|{x-y}\right \|_2 d\rho _t(x,y)$ .

Proof. By following the lines of the proof of Lemma 3.2 and noticing that in analogy to the estimates (3.3), it hold

(3.6a) \begin{align} -\left \langle x-y, x-x^* \right \rangle &\geq -\left \|{x-y}\right \|_2\left \|{x-x^*}\right \|_2 \geq -\frac{1}{2}\!\left (\left \|{x-y}\right \|_2^2+\left \|{x-x^*}\right \|_2^2\right ), \end{align}

(3.6b) \begin{align} -\left \langle x-y_{\alpha }({\rho _{Y,t}}), x-x^* \right \rangle &= -\left \|{x-x^*}\right \|_2^2 - \left \langle x^*-y_{\alpha }({\rho _{Y,t}}), x-x^* \right \rangle \nonumber \\ &\geq -\left \|{x-x^*}\right \|_2^2 - \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2 \left \|{x-x^*}\right \|_2, \end{align}

(3.6c) \begin{align} -\left \langle x-y_{\alpha }({\rho _{Y,t}}), x-y \right \rangle &= -\left \langle x-x^*, x-y \right \rangle - \left \langle x^*-y_{\alpha }({\rho _{Y,t}}), x-y \right \rangle \nonumber \\ &\geq -\frac{1}{2}\!\left (\left \|{x-y}\right \|_2^2+\left \|{x-x^*}\right \|_2^2\right ) - \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2\left \|{x-y}\right \|_2, \end{align}

(3.6d) \begin{align} \left \|{x-y_{\alpha }({\rho _{Y,t}})}\right \|_2^2 &= \left \|{x-x^*}\right \|_2^2 - 2\left \langle y_{\alpha }({\rho _{Y,t}})-x^*,x-x^*\right \rangle + \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2^2 \nonumber \\ &\geq \left \|{x-x^*}\right \|_2^2 - 2\left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2\left \|{x-x^*}\right \|_2, \end{align}

as well as

(3.6e) \begin{align} -\left \langle \nabla{\mathcal{E}}(x), x-x^* \right \rangle &\geq -\left \|{\nabla{\mathcal{E}}(x)}\right \|_2\left \|{x-x^*}\right \|_2 \geq -C_{\nabla{\mathcal{E}}}\left \|{x-x^*}\right \|_2^2, \end{align}

(3.6f) \begin{align} -\left \langle \nabla{\mathcal{E}}(x), x-y \right \rangle &\geq -\left \|{\nabla{\mathcal{E}}(x)}\right \|_2\left \|{x-y}\right \|_2 \geq -C_{\nabla{\mathcal{E}}}\left \|{x-x^*}\right \|_2\left \|{x-y}\right \|_2 \nonumber \\ &\geq -\frac{C_{\nabla{\mathcal{E}}}}{2}\left (\left \|{x-y}\right \|_2^2+\left \|{x-x^*}\right \|_2^2\right ), \end{align}

(3.6g) \begin{align} \left \|{\nabla{\mathcal{E}}(x)}\right \|_2^2 &\geq -C_{\nabla{\mathcal{E}}}^2\left \|{x-x^*}\right \|_2^2. \end{align}

we obtain the statement by integrating the bounds into equations (3.1) and (3.2).

Proof. The proof is a mere reformulation of the one of [Reference Fornasier, Klock, Riedl, Jiménez Laredo, Hidalgo and Babaagba29, Proposition 1], which is presented in what follows for the sake of completeness.

For any $a \gt 0$ , Markov’s inequality gives $\|{\omega _{\alpha }}\|_{L_1(\varrho )} \geq a \varrho (\{y\, :\, \exp ({-}\alpha{\mathcal{E}}(y)) \geq a\})$ . By choosing $a = \exp ({-}\alpha{\mathcal{E}}_{r})$ and noting that

\begin{align*} \varrho \left (\left \{y \in \mathbb{R}^d: \exp ({-}\alpha{\mathcal{E}}(y)) \geq \exp ({-}\alpha{\mathcal{E}}_{r})\right \}\right ) &= \varrho \left (\left \{y \in \mathbb{R}^d:{\mathcal{E}}(y) \leq{\mathcal{E}}_{r} \right \}\right ) \geq \varrho (B^\infty _{r}(x^*)), \end{align*}

we get $\left \|{\omega _{\alpha }}\right \|_{L_1(\varrho )} \geq \exp ({-}\alpha{\mathcal{E}}_{r})\varrho (B^\infty _{r}(x^*))$ . Now let $\widetilde r \geq r \gt 0$ . With the definition of the consensus point $y_{\alpha }({\varrho }) = \int y \omega _{\alpha }(y)/\|{\omega _{\alpha }}\|_{L_1(\varrho )}\,d\varrho (y)$ and Jensen’s inequality, we can decompose

\begin{align*} \begin{split} \left \|{y_{\alpha }({\varrho }) - x^*}\right \|_{\infty} &\leq \int _{B^\infty _{\widetilde r}(x^*)} \left \|{y-x^*}\right \|_{\infty} \frac{\omega _{\alpha }(y)}{\left \|{\omega _{\alpha }}\right \|_{L_1(\varrho )}}\,d\varrho (y) \\ &\quad \, + \int _{\left (B^\infty _{\widetilde r}(x^*)\right )^c} \left \|{y-x^*}\right \|_{\infty} \frac{\omega _{\alpha }(y)}{\left \|{\omega _{\alpha }}\right \|_{L_1(\varrho )}}\,d\varrho (y). \end{split} \end{align*}

After noticing that the first term is bounded by $\widetilde r$ since $ \left \|{y-x^*}\right \|_{\infty} \leq \widetilde r$ for all $y \in B^\infty _{\widetilde r}(x^*)$ , we can continue the former with

(3.9) \begin{equation} \begin{split} \left \|{y_{\alpha }({\varrho }) - x^*}\right \|_{\infty} &\leq \widetilde r + \frac{1}{ \exp ({-}\alpha{\mathcal{E}}_{r})\varrho (B^\infty _{r}(x^*))}\!\int _{(B^\infty _{\widetilde r}(x^*))^c} \left \|{y-x^*}\right \|_{\infty} \omega _{\alpha }(y)\,d\varrho (y)\\ &\leq \widetilde r + \frac{\exp \left ({-}\alpha \inf _{y \in (B^\infty _{\widetilde r}(x^*))^c}{\mathcal{E}}(y)\right )}{\exp \left ({-}\alpha{\mathcal{E}}_{r})\varrho (B^\infty _{r}(x^*)\right )}\!\int _{(B^\infty _{\widetilde r}(x^*))^c}\left \|{y-x^*}\right \|_{\infty} d\varrho (y)\\ &= \widetilde r + \frac{\exp \left ({-}\alpha \left (\inf _{y \in (B^\infty _{\widetilde r}(x^*))^c}{\mathcal{E}}(y) -{\mathcal{E}}_{r}\right )\right )}{\varrho (B^\infty _{r}(x^*))}\!\int \left \|{y-x^*}\right \|_{\infty} d\varrho (y), \end{split} \end{equation}

where for the second term we used $\left \|{\omega _{\alpha }}\right \|_{L_1(\varrho )} \geq \exp ({-}\alpha{\mathcal{E}}_{r})\varrho (B^\infty _{r}(x^*))$ from above. Let us now choose $\widetilde r = (q+{\mathcal{E}}_r)^{\nu }/\eta$ , which satisfies $\widetilde r \geq r$ , since A2 with $\underline{\mathcal{E}} = 0$ and $r \leq R_0$ implies

\begin{align*} \widetilde r = \frac{(q+{\mathcal{E}}_r)^{\nu }}{\eta } \geq \frac{{\mathcal{E}}_r^{\nu }}{\eta } = \frac{\left (\sup _{y \in B^\infty _{r}(x^*)}{\mathcal{E}}(y)\right )^{\nu }}{\eta } \geq \sup _{y \in B_{r}^\infty (x^*)}\left \|{y-x^*}\right \|_{\infty} = r. \end{align*}

Furthermore, due to the assumption $q+{\mathcal{E}}_r \leq{\mathcal{E}}_{\infty }$ in the statement we have $\widetilde r \leq{\mathcal{E}}_{\infty }^{\nu }/\eta$ , which together with the two cases of A2 with $\underline{\mathcal{E}} = 0$ allows to bound the infimum in (3.9) as follows

\begin{align*} \inf _{y \in (B^\infty _{\widetilde r}(x^*))^c}{\mathcal{E}}(y) -{\mathcal{E}}_r \geq \min \left \{{\mathcal{E}}_{\infty }, (\eta \widetilde r)^{\frac{1}{\nu }}\right \} -{\mathcal{E}}_r = (\eta \widetilde r)^{\frac{1}{\nu }} -{\mathcal{E}}_r = (q +{\mathcal{E}}_r) -{\mathcal{E}}_r = q. \end{align*}

Inserting this and the definition of $\,\widetilde r$ into (3.9), we get the result as a consequence of the norm equivalence $\left \|{\,\cdot \,}\right \|_{\infty} \leq \left \|{\,\cdot \,}\right \|_2\leq \sqrt{d}\left \|{\,\cdot \,}\right \|_{\infty}$ .

Proof. It is straightforward to check the properties of $\phi _r$ as it is a tensor product of classical well-studied mollifiers.

Remark 3.9. In order to ensure a finite decay rate $p\lt \infty$ in Proposition 3.8 , it is crucial to have non-vanishing diffusions $\sigma _1\gt 0$ , $\sigma _2\gt 0$ if $\lambda _2\not =0$ and $\sigma _3\gt 0$ if $\lambda _3\not =0$ . As apparent from the formulation of the statement as well as the proof below, $\sigma _2$ or $\sigma _3$ may be $0$ if the corresponding drift parameter, $\lambda _2$ or $\lambda _3$ , respectively, vanishes.

Proof of Proposition 3.8 . By the definition of the marginal $\rho _Y$ and the properties of the mollifier $\phi _r$ defined in Lemma 3.7, we have

\begin{align*} \rho _{Y,t}(B^\infty _{r}(x^*)) = \rho _t\big (\mathbb{R}^d \times B^\infty _{r}(x^*)\big ) \geq \rho _t(\Omega _r) \geq \iint \phi _{r}(x,y)\,d\rho _t(x,y). \end{align*}

Our strategy is to derive a lower bound for the right-hand side of this inequality. Using the weak solution property of $\rho$ as in Definition 2.1 and the fact that $\phi _{r}\in\mathcal{C}^{\infty }_c(\mathbb{R}^d\times \mathbb{R}^d)$ , we obtain

(3.13) \begin{equation} \begin{split} &\frac{d}{dt}\iint \phi _{r}(x,y)\,d\rho _t(x,y) =\sum _{k=1}^d \iint T^s_{k}(x,y) \,d\rho _t(x,y)\\ &\quad +\! \sum _{k=1}^d \iint \big (T^c_{1k}(x,y) \!+\! T^c_{2k}(x,y) \!+\! T^\ell _{1k}(x,y) \!+\! T^\ell _{2k}(x,y) \!+\! T^g_{1k}(x,y) \!+\! T^g_{2k}(x,y)\big ) \,d\rho _t(x,y), \end{split} \end{equation}

where $T^s_{k}(x,y) \,:\!=\, -\kappa S^{\beta,\theta }(x,y) \left (y-x\right )_k \partial _{y_k}\phi _r(x,y)$ and

\begin{equation*} \begin{aligned} &\, T^c_{1k}(x,y) \!\,:\!=\, \!-\lambda _1\left (x\!-\!y_{\alpha }({\rho _{Y,t}})\right )_k \partial _{x_k}\phi _r(x,y), &\!\!\!\, T^c_{2k}(x,y) &\!\,:\!=\, \!\frac{\sigma _1^2}{2} \left (x\!-\!y_{\alpha }({\rho _{Y,t}})\right )_k^2 \partial ^2_{x_kx_k} \phi _{r}(x,y), \\ &\, T^\ell _{1k}(x,y) \!\,:\!=\, \!-\lambda _2\left (x\!-\!y\right )_k \partial _{x_k}\phi _r(x,y), &\!\!\!\, T^\ell _{2k}(x,y) &\!\,:\!=\, \!\frac{\sigma _2^2}{2} \left (x\!-\!y\right )_k^2 \partial ^2_{x_kx_k} \phi _{r}(x,y), \\ &\, T^g_{1k}(x,y) \!\,:\!=\, -\lambda _3\partial _{x_k}{\mathcal{E}}(x) \partial _{x_k}\phi _r(x,y), &\!\!\!\, T^g_{2k}(x,y) &\!\,:\!=\, \!\frac{\sigma _3^2}{2} \left (\partial _{x_k}{\mathcal{E}}(x)\right )^2 \partial ^2_{x_kx_k} \phi _{r}(x,y) \end{aligned} \end{equation*}

for $k\in \{1,\dots,d\}$ . Since the mollifier $\phi _r$ and its derivatives vanish outside of $\Omega _r$ , we restrict our attention to $\Omega _r$ and aim for showing for all $k\in \{1,\dots,d\}$ that

• • $T^s_{k}(x,y) \geq 0$ ,

• • $T^c_{1k}(x,y) + T^c_{2k}(x,y) \geq -p^c\phi _r(x,y)$ ,

• • $T^\ell _{1k}(x,y) + T^\ell _{2k}(x,y) \geq -p^\ell \phi _r(x,y)$ ,

• • $T^g_{1k}(x,y) + T^g_{2k}(x,y) \geq -p^g\phi _r(x,y)$

pointwise for all $(x,y)\in \Omega _r$ with suitable constants $0 \leq p^\ell, p^c, p^g \lt \infty$ .

Term $T^s_{k}$ : Using the expression for $\partial _{y_k} \phi _{r}$ from Lemma 3.7 and the fact that $S^{\beta,\theta }\geq \theta/2\geq 0$ , it is easy to see that

(3.14) \begin{equation} T^s_{k}(x,y) = \frac{r^2\kappa }{2} S^{\beta,\theta }(x,y) \frac{\left (y-x\right )^2_k}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2}\phi _{r}(x,y) \geq 0. \end{equation}

Terms $T^c_{1k}+T^c_{2k}$ , $T^\ell _{1k}+T^\ell _{2k}$ and $T^g_{1k}+T^g_{2k}$ : We first note that the third inequality from above holds with $p^\ell =0$ if $\lambda _2=\sigma _2=0$ and the fourth with $p^g=0$ if $\lambda _3=\sigma _3=0$ .

Therefore, in what follows we assume that $\lambda _2,\sigma _2,\lambda _3,\sigma _3\gt 0$ . In order to lower bound the three terms from above, we arrange the summands by using the abbreviations introduced in (3.10) as follows. For $T^c_{1k}+T^c_{2k}$ , we have

(3.15a) \begin{align} &T^c_{1k}(x,y)+T^c_{2k}(x,y)\nonumber \\ &\qquad \,= -\lambda _1\left (x-y_{\alpha }({\rho _{Y,t}})\right )_k \delta ^{*}_{x_k} \phi _{r}(x,y) + \frac{\sigma _1^2}{2} \left (x-y_{\alpha }({\rho _{Y,t}})\right )_k^2 \delta ^{2,*}_{x_kx_k} \phi _{r}(x,y) \end{align}

(3.15b) \begin{align} &\qquad \quad \,\,-\lambda _1\left (x-y_{\alpha }({\rho _{Y,t}})\right )_k \delta ^Y_{x_k} \phi _{r}(x,y) + \frac{\sigma _1^2}{2} \left (x-y_{\alpha }({\rho _{Y,t}})\right )_k^2 \delta ^{2,Y}_{x_kx_k} \phi _{r}(x,y), \end{align}

for $T^\ell _{1k}+T^\ell _{2k}$ we have

(3.16a) \begin{align} &T^\ell _{1k}(x,y)+T^\ell _{2k}(x,y)\nonumber \\ &\qquad \,=-\lambda _2\left (x-y\right )_k \delta ^{*}_{x_k} \phi _{r}(x,y) + \frac{\sigma _2^2}{2} \left (x-y\right )_k^2 \delta ^{2,*}_{x_kx_k} \phi _{r}(x,y) \end{align}

(3.16b) \begin{align} &\qquad \quad \,\,-\lambda _2\left (x-y\right )_k \delta ^Y_{x_k} \phi _{r}(x,y) + \frac{\sigma _2^2}{2} \left (x-y\right )_k^2 \delta ^{2,Y}_{x_kx_k} \phi _{r}(x,y) \end{align}

and for $T^g_{1k}+T^g_{2k}$ we have

(3.17a) \begin{align} &T^g_{1k}(x,y)+T^g_{2k}(x,y)\nonumber \\ &\qquad \,=-\lambda _3\partial _{x_k}{\mathcal{E}}(x) \delta ^{*}_{x_k} \phi _{r}(x,y) + \frac{\sigma _3^2}{2} (\partial _{x_k}{\mathcal{E}}(x))^2 \delta ^{2,*}_{x_kx_k} \phi _{r}(x,y) \end{align}

(3.17b) \begin{align} &\qquad \quad \,\,-\lambda _3\partial _{x_k}{\mathcal{E}}(x) \delta ^Y_{x_k} \phi _{r}(x,y) + \frac{\sigma _3^2}{2} (\partial _{x_k}{\mathcal{E}}(x))^2 \delta ^{2,Y}_{x_kx_k} \phi _{r}(x,y). \end{align}

We now treat each of the two-part sums in (3.15a), (3.15b), (3.16a), (3.16b), (3.17a) and (3.17b) separately by employing a technique similar to the one used in the proof of [Reference Fornasier, Klock, Riedl, Jiménez Laredo, Hidalgo and Babaagba29, Proposition 2], which was developed originally to prove [Reference Fornasier, Klock and Riedl28, Proposition 20].

Terms (3.15a), (3.16a) and (3.17a): Owed to their similar structure (in particular with respect to the denominator of the derivatives $\delta ^{*}_{x_k} \phi _{r}$ and $\delta ^{2,*}_{x_kx_k} \phi _{r}$ ), we can treat the three sums (3.15a), (3.16a) and (3.17a) simultaneously. Therefore, we consider the general formulation

(3.18) \begin{equation} -\lambda \Upsilon _k(x,y) \delta ^{*}_{x_k} \phi _{r}(x,y) + \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) \delta ^{2,*}_{x_kx_k} \phi _{r}(x,y) \,=\!:\, T^*_{1k}(x,y)+T^*_{2k}(x,y), \end{equation}

which matches (3.15a) when $\Upsilon _k(x,y) = (x-y_{\alpha }({\rho _{Y,t}}))_k$ , $\lambda =\lambda _1$ and $\sigma =\sigma _1$ , (3.16a) when $\Upsilon _k(x,y) = (x-y)_k$ , $\lambda =\lambda _2$ and $\sigma =\sigma _2$ , and (3.17a) when $\Upsilon _k(x,y) = \partial _{x_k}{\mathcal{E}}(x)$ , $\lambda =\lambda _3$ and $\sigma =\sigma _3$ .

To achieve the desired lower bound over $\Omega _r$ , we introduce the subsets

\begin{align*} K^*_{1k} &\,:\!=\, \left \{(x,y) \in \mathbb{R}^d\times \mathbb{R}^d\, :\, \left |{\left (x-x^*\right )_k}\right | \gt \frac{\sqrt{c}}{2}r\right \} \end{align*}

and

\begin{align*} \begin{aligned} &K^*_{2k} \,:\!=\, \Bigg \{(x,y) \in \mathbb{R}^d\times \mathbb{R}^d\, :\, -\lambda \Upsilon _k(x,y) \left (x-x^*\right )_k \left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )_k^2\right )^2 \\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \gt \tilde{c}\left (\frac{r}{2}\right )^2\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-x^*\right )_k^2\Bigg \}, \end{aligned} \end{align*}

where $\tilde{c} \,:\!=\, 2c-1\in (0,1)$ . For fixed $k$ , we now decompose $\Omega _r$ according to

\begin{align*} \Omega _r = \big ((K_{1k}^{*})^c \cap \Omega _r\big ) \cup \big (K^*_{1k} \cap (K_{2k}^{*})^c \cap \Omega _r\big ) \cup \big (K^*_{1k} \cap K^*_{2k} \cap \Omega _r\big ). \end{align*}

In the following, we treat each of these three subsets separately.

Subset $(K_{1k}^{*})^c \cap \Omega _r$ : We have $\left |{\left (x-x^*\right )_k}\right | \leq \frac{\sqrt{c}}{2}r$ for each $(x,y) \in (K_{1k}^*)^c$ , which can be used to independently derive lower bounds for both summands in (3.18). For the first, we insert the expression for $\delta ^{*}_{x_k} \phi _{r}(x,y)$ to get

(3.19) \begin{equation} \begin{aligned} T^*_{1k}(x,y) &= \frac{r^2}{2}\lambda \Upsilon _k(x,y) \frac{\left (x-x^*\right )_k}{\left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )^2_k\right )^2}\phi _{r}(x,y) \\ &\geq -\frac{r^2}{2} \!\lambda \frac{\left |{\Upsilon _k(x,y)}\right |\!\left |{\left (x-x^*\right )_k}\right |}{\left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )^2_k\right )^2}\phi _r(x,y) \geq - \frac{2\lambda C_{\Upsilon} \sqrt{c}}{(1-c)^2\frac{r}{2}}\phi _r(x,y) \\ &\,=\!:\, -p^{*,\Upsilon }_{1}\phi _r(x,y), \end{aligned} \end{equation}

where, in the last inequality, we used that $(x,y)\in \Omega _r$ , the definition of $B$ and Assumption A3 to get the bound

(3.20) \begin{equation} \begin{split} \left |{\Upsilon _k(x,y)}\right | &= \begin{cases} \left |{(x-y_{\alpha }({\rho _{Y,t}}))_k}\right | \leq \frac{r}{2}+B, &\text{if }\Upsilon _k(x,y) = (x-y_{\alpha }({\rho _{Y,t}}))_k, \\ \left |{(x-y)_k}\right | \leq \frac{r}{2}, &\text{if }\Upsilon _k(x,y) = (x-y)_k, \\ \left |{\partial _{x_k}{\mathcal{E}}(x)}\right | \leq \left \|{\nabla{\mathcal{E}}(x)}\right \|_2 \leq C_{\nabla{\mathcal{E}}}\left \|{x-x^*}\right \|_2\\ \qquad \leq C_{\nabla{\mathcal{E}}}d\left \|{x-x^*}\right \|_{\infty} \leq C_{\nabla{\mathcal{E}}}d\frac{r}{2} &\text{if }\Upsilon _k(x,y) = \partial _{x_k}{\mathcal{E}}(x). \end{cases}\\ &\leq \max \left \{\frac{r}{2}+B, C_{\nabla{\mathcal{E}}}d\frac{r}{2}\right \} \,=\!:\, C_{\Upsilon} (r,B,d,C_{\nabla{\mathcal{E}}}). \end{split} \end{equation}

For the second summand, we insert the expression for $\delta ^{2,*}_{x_kx_k} \phi _{r}(x,y)$ to obtain

(3.21) \begin{equation} \begin{aligned} T^*_{2k}(x,y) &=\frac{\sigma ^2}{2} \Upsilon _k^2(x,y) \delta ^{2,*}_{x_kx_k} \phi _{r}(x,y)\\ &=\sigma ^2 \left (\frac{r}{2}\right )^2 \Upsilon _k^2(x,y) \frac{2\left (2\left (x\!-\!x^*\right )^2_k\!-\!\left (\frac{r}{2}\right )^2\right )\left (x\!-\!x^*\right )_k^2 \!-\! \left (\left (\frac{r}{2}\right )^2\!-\!\left (x\!-\!x^*\right )^2_k\right )^2}{\left (\left (\frac{r}{2}\right )^2-\left (x\!-\!x^*\right )^2_k\right )^4} \phi _{r}(x,y)\\ &\geq - \frac{\sigma ^2C_{\Upsilon} ^2}{(1-c)^4\left (\frac{r}{2}\right )^2}\phi _{r}(x,y) \,=\!:\, -p^{*,\Upsilon }_{2}\phi _r(x,y), \end{aligned} \end{equation}

where the last inequality uses $\Upsilon _k^2(x,y) \leq C_{\Upsilon} ^2$ .

Subset $K^*_{1k} \cap (K_{2k}^{*})^c \cap \Omega _r$ : As $(x,y) \in K^*_{1k}$ , we have $\left |{\left (x-x^*\right )_k}\right | \gt \frac{\sqrt{c}}{2}r$ . We observe that the sum in (3.18) is nonnegative for all $(x,y)$ in this subset whenever

(3.22) \begin{equation} \begin{aligned} &\left ({-}\lambda \Upsilon _k(x,y)\left (x-x^*\right )_k + \frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\right )\left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )^2_k\right )^2 \\ &\qquad \qquad \qquad \qquad \qquad \, \leq \sigma ^2\Upsilon _k^2(x,y) \left (2\left (x-x^*\right )^2_k-\left (\frac{r}{2}\right )^2\right )\left (x-x^*\right )_k^2. \end{aligned} \end{equation}

The first term on the left-hand side in (3.22) can be bounded from above by exploiting that $v\in (K_{2k}^{*})^c$ and by using the relation $\tilde{c} = 2c-1$ . More precisely, we have

\begin{align*} &-\lambda \Upsilon _k(x,y)\left (x-x^*\right )_k \left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )^2_k\right )^2 \leq \tilde{c}\left (\frac{r}{2}\right )^2\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-x^*\right )_k^2\\ &\quad \;\; = (2c\!-\!1)\left (\frac{r}{2}\right )^2\!\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-x^*\right )_k^2 \leq \! \left (2\left (x-x^*\right )_k^2-\left (\frac{r}{2}\right )^2\right )\!\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-x^*\right )_k^2, \end{align*}

where the last inequality follows since $v\in K^*_{1k}$ . For the second term on the left-hand side in (3.22), we can use $(1-c)^2 \leq (2c-1)c$ as per assumption, to get

\begin{align*} &\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )^2_k\right )^2 \leq \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) (1-c)^2\left (\frac{r}{2}\right )^4 \\ &\qquad \leq \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) (2c-1)\left (\frac{r}{2}\right )^2c\left (\frac{r}{2}\right )^2 \leq \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) \left (2\left (x-x^*\right )_k^2-\left (\frac{r}{2}\right )^2\right )\left (x-x^*\right )_k^2. \end{align*}

Hence, (3.22) holds and we have that (3.18) is uniformly nonnegative on this subset.

Subset $K^*_{1k} \cap K^*_{2k} \cap \Omega _r$ : As $(x,y) \in K_{1k}^*$ , we have $\left |{\left (x-x^*\right )_k}\right | \gt \frac{\sqrt{c}}{2}r$ . To start with we note that the first summand of (3.18) vanishes whenever $\sigma ^2\Upsilon _k^2(x,y) = 0$ , provided $\sigma \gt 0$ , so nothing needs to be done if $\Upsilon _k(x,y)=0$ . Otherwise, if $\sigma ^2\Upsilon _k^2(x,y) \gt 0$ , we exploit $(x,y)\in K_{2k}^*$ to get

\begin{equation*} \begin{split} \frac{\Upsilon _k(x,y)\left (x-x^*\right )_k}{\left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )_k^2\right )^2} &\geq \frac{-\left |{\Upsilon _k(x,y)}\right |\left |{\left (x-x^*\right )_k}\right |}{\left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )_k^2\right )^2}\\ &\gt \frac{2\lambda \Upsilon _k(x,y) \left (x-x^*\right )_k}{\tilde{c}\left (\frac{r}{2}\right )^2\sigma ^2\left |{\Upsilon _k(x,y)}\right |\left |{\left (x-x^*\right )_k}\right |} \geq -\frac{8\lambda }{\tilde{c}r^2\sigma ^2}. \end{split} \end{equation*}

Using this, the first summand of (3.18) can be bounded from below by

(3.23) \begin{equation} \begin{split} T^*_{1k}(x,y) =\lambda \frac{r^2}{2}\frac{\Upsilon _k(x,y)\left (x-x^*\right )_k}{\left (\left (\frac{r}{2}\right )^2-\left (x-x^*\right )^2_k\right )^2}\phi _{r}(x,y) &\geq -\frac{4\lambda ^2}{\tilde{c}\sigma ^2}\phi _{r}(x,y) \,=\!:\, -p^{*,\Upsilon }_{3}\phi _r(x,y). \end{split} \end{equation}

For the second summand, the nonnegativity of $\sigma ^2\Upsilon _k^2(x,y)$ implies the nonnegativity, whenever

\begin{equation*} 2\left (2\left (x-x^*\right )^2_k-\left (\frac {r}{2}\right )^2\right )\left (x-x^*\right )_k^2 \geq \left (\left (\frac {r}{2}\right )^2-\left (x-x^*\right )^2_k\right )^2. \end{equation*}

This holds for $v\in K_{1k}^*$ , if $2(2c-1)c \geq (1-c)^2$ as implied by the assumption.

Term (3.16b): Recall that this term has the structure

(3.24) \begin{equation} -\lambda _2\left (x-y\right )_k \delta ^Y_{x_k} \phi _{r}(x,y) + \frac{\sigma _2^2}{2} \left (x-y\right )_k^2 \delta ^{2,Y}_{x_kx_k} \phi _{r}(x,y)\,=\!:\,T^{Y,1}_{1k}(x,y)+T^{Y,1}_{2k}(x,y). \end{equation}

We first note that the first summand of (3.24) is always nonnegative since

(3.25) \begin{equation} T^{Y,1}_{1k}(x,y)= \lambda _2 \frac{r^2}{2}\frac{\left (x-y\right )_k^2}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2}\phi _{r}(x,y) \geq 0. \end{equation}

For the second summand of (3.24), a direct computation shows

\begin{equation*} \begin{split} T^{Y,1}_{2k}(x,y) = \sigma _2^2 \left (\frac{r}{2}\right )^2 \left (x-y\right )_k^2 \frac{3\left (x-y\right )^4_k-\left (\frac{r}{2}\right )^4}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^4} \phi _{r}(x,y), \end{split} \end{equation*}

which is nonnegative on the set

\begin{align*} K^Y_k \,:\!=\, \left \{(x,y) \in \mathbb{R}^d\times \mathbb{R}^d\, :\, \left |{\left (x-y\right )_k}\right | \gt \frac{\sqrt{c}}{2}r\right \} \end{align*}

for any $c\geq 1/\sqrt{3}$ , as ensured by $(1-c)^2 \leq (2c-1)c$ . On the complement $(K^Y_k)^c$ , we have $\left |{\left (x-y\right )_k}\right | \leq \frac{\sqrt{c}}{2}r$ , which can be used to bound

(3.26) \begin{equation} \begin{aligned} T^{Y,1}_{2k}(x,y) &=\sigma _2^2 \left (\frac{r}{2}\right )^2 \!\left (x\!-\!y\right )_k^2 \frac{3\left (x\!-\!y\right )^4_k\!-\!\left (\frac{r}{2}\right )^4}{\left (\left (\frac{r}{2}\right )^2\!-\!\left (x\!-\!y\right )^2_k\right )^4}\phi _{r}(x,y)\\ &\geq -\frac{\sigma _2^2 c}{\left (1\!-\!c\right )^4}\phi _{r}(x,y) \,=\!:\, -p^{Y,\Upsilon _{\ell} }\phi _r(x,y). \end{aligned} \end{equation}

Terms (3.15b) and (3.17b): The final two terms to be controlled have again a similar structure of the form

(3.27) \begin{equation} -\lambda \Upsilon _k(x,y) \delta ^Y_{x_k} \phi _{r}(x,y) + \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) \delta ^{2,Y}_{x_kx_k} \phi _{r}(x,y)\,=\!:\,T^{Y,2}_{1k}(x,y)+T^{Y,2}_{2k}(x,y), \end{equation}

where we recycle the notation introduced after (3.18), i.e., $\Upsilon _k(x,y) = (x-y_{\alpha }({\rho _{Y,t}}))_k$ , $\lambda =\lambda _1$ and $\sigma =\sigma _1$ in the case of (3.15b) and $\Upsilon _k(x,y) = \partial _{x_k}{\mathcal{E}}(x)$ , $\lambda =\lambda _3$ and $\sigma =\sigma _3$ in the case of (3.17b).

The procedure for deriving lower bounds is similar to the one at the beginning with the exception that the denominator of the derivatives $\delta ^Y_{x_k} \phi _{r}$ and $\delta ^{2,Y}_{x_kx_k} \phi _{r}$ requires to introduce an adapted decomposition of $\Omega _r$ . To be more specific, we define the subsets

\begin{align*} K^Y_{1k} &\,:\!=\, \left \{(x,y) \in \mathbb{R}^d\times \mathbb{R}^d\, :\, \left |{\left (x-y\right )_k}\right | \gt \frac{\sqrt{c}}{2}r\right \} \end{align*}

and

\begin{align*} \begin{aligned} &K^Y_{2k} \,:\!=\, \Bigg \{(x,y) \in \mathbb{R}^d\times \mathbb{R}^d\, :\, -\lambda \Upsilon _k(x,y) \left (x-y\right )_k \left (\left (\frac{r}{2}\right )^2-\left (x-y\right )_k^2\right )^2 \\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \gt \tilde{c}\left (\frac{r}{2}\right )^2\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-y\right )_k^2\Bigg \}, \end{aligned} \end{align*}

where $\tilde{c} \,:\!=\, 2c-1\in (0,1)$ . For fixed $k$ , we now decompose $\Omega _r$ according to

\begin{align*} \Omega _r = \big ((K_{1k}^Y)^c \cap \Omega _r\big ) \cup \big (K^Y_{1k} \cap (K_{2k}^Y)^c \cap \Omega _r\big ) \cup \big (K^Y_{1k} \cap K^Y_{2k} \cap \Omega _r\big ). \end{align*}

In the following, we treat again each of these three subsets separately.

Subset $(K_{1k}^Y)^c \cap \Omega _r$ : We have $\left |{\left (x-y\right )_k}\right | \leq \frac{\sqrt{c}}{2}r$ for each $(x,y) \in (K_{1k}^Y)^c$ , which can be used to independently derive lower bounds for both summands in (3.27). For the first summand, we insert the expression for $\delta ^Y_{x_k} \phi _{r}(x,y)$ to get

(3.28) \begin{equation} \begin{aligned} T^{Y,2}_{1k}(x,y) &= \frac{r^2}{2}\lambda \Upsilon _k(x,y) \frac{\left (x-y\right )_k}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2}\phi _{r}(x,y) \\ &\geq -\frac{r^2}{2} \!\lambda \frac{\left |{\Upsilon _k(x,y)}\right |\!\left |{\left (x-y\right )_k}\right |}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2}\phi _r(x,y) \geq - \frac{2\lambda C_{\Upsilon} \sqrt{c}}{(1-c)^2\frac{r}{2}}\phi _r(x,y) \\ &\,=\!:\, -p^{Y,\Upsilon }_{1}\phi _r(x,y), \end{aligned} \end{equation}

where we recall from above that $\Upsilon _k(x,y) \leq C_{\Upsilon}$ , which was used in the last inequality. For the second summand, we insert the expression for $\delta ^{2,Y}_{x_kx_k} \phi _{r}(x,y)$ to obtain

(3.29) \begin{equation} \begin{aligned} T^{Y,2}_{2k}(x,y) &=\sigma ^2 \left (\frac{r}{2}\right )^2 \Upsilon _k^2(x,y) \frac{2\left (2\left (x\!-\!y\right )^2_k\!-\!\left (\frac{r}{2}\right )^2\right )\left (x\!-\!y\right )_k^2 \!-\! \left (\left (\frac{r}{2}\right )^2\!-\!\left (x\!-\!y\right )^2_k\right )^2}{\left (\left (\frac{r}{2}\right )^2\!-\!\left (x\!-\!y\right )^2_k\right )^4} \phi _{r}(x,y)\\ &\geq - \frac{\sigma ^2C_{\Upsilon} ^2}{(1-c)^4\left (\frac{r}{2}\right )^2}\phi _{r}(x,y) \,=\!:\, -p^{Y,\Upsilon }_{2}\phi _r(x,y), \end{aligned} \end{equation}

where the last inequality uses $\Upsilon _k^2(x,y) \leq C_{\Upsilon} ^2$ .

Subset $K^Y_{1k} \cap (K_{2k}^Y)^c \cap \Omega _r$ : As $(x,y) \in K^Y_{1k}$ , we have $\left |{\left (x-y\right )_k}\right | \gt \frac{\sqrt{c}}{2}r$ . We observe that the sum in (3.27) is nonnegative for all $(x,y)$ in this subset whenever

(3.30) \begin{equation} \begin{aligned} &\left ({-}\lambda \Upsilon _k(x,y)\left (x-y\right )_k + \frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\right )\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2 \\ &\qquad \qquad \qquad \qquad \qquad \, \leq \sigma ^2\Upsilon _k^2(x,y) \left (2\left (x-y\right )^2_k-\left (\frac{r}{2}\right )^2\right )\left (x-y\right )_k^2. \end{aligned} \end{equation}

The first term on the left-hand side in (3.30) can be bounded from above exploiting that $v\in (K_{2k}^Y)^c$ and by using the relation $\tilde{c} = 2c-1$ . More precisely, we have

\begin{align*} &-\lambda \Upsilon _k(x,y)\left (x-y\right )_k \left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2 \leq \tilde{c}\left (\frac{r}{2}\right )^2\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-y\right )_k^2\\ &\qquad = (2c-1)\left (\frac{r}{2}\right )^2\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-y\right )_k^2 \leq \! \left (2\left (x-y\right )_k^2-\left (\frac{r}{2}\right )^2\right )\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (x-y\right )_k^2, \end{align*}

where the last inequality follows since $v\in K^Y_{1k}$ . For the second term on the left-hand side in (3.30), we can use $(1-c)^2 \leq (2c-1)c$ as per assumption, to get

\begin{align*} &\frac{\sigma ^2}{2} \Upsilon _k^2(x,y)\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2 \leq \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) (1-c)^2\left (\frac{r}{2}\right )^4 \\ &\qquad \leq \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) (2c-1)\left (\frac{r}{2}\right )^2c\left (\frac{r}{2}\right )^2 \leq \frac{\sigma ^2}{2} \Upsilon _k^2(x,y) \left (2\left (x-y\right )_k^2-\left (\frac{r}{2}\right )^2\right )\left (x-y\right )_k^2. \end{align*}

Hence, (3.30) holds and we have that (3.27) is uniformly nonnegative on this subset.

Subset $K^Y_{1k} \cap K^Y_{2k} \cap \Omega _r$ : As $(x,y) \in K_{1k}^Y$ , we have $\left |{\left (x-y\right )_k}\right | \gt \frac{\sqrt{c}}{2}r$ . To start with we note that the first summand of (3.27) vanishes whenever $\sigma ^2\Upsilon _k^2(x,y) = 0$ , provided $\sigma \gt 0$ , so nothing needs to be done if $\Upsilon _k(x,y)=0$ . Otherwise, if $\sigma ^2\Upsilon _k^2(x,y) \gt 0$ , we exploit $(x,y)\in K_{2k}^Y$ to get

\begin{equation*} \begin{split} \frac{\Upsilon _k(x,y)\left (x-y\right )_k}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )_k^2\right )^2} &\geq \frac{-\left |{\Upsilon _k(x,y)}\right |\left |{\left (x-y\right )_k}\right |}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )_k^2\right )^2}\\ &\gt \frac{2\lambda \Upsilon _k(x,y) \left (x-y\right )_k}{\tilde{c}\left (\frac{r}{2}\right )^2\sigma ^2\left |{\Upsilon _k(x,y)}\right |\left |{\left (x-y\right )_k}\right |} \geq -\frac{8\lambda }{\tilde{c}r^2\sigma ^2}. \end{split} \end{equation*}

Using this, the first summand of (3.27) can be bounded from below by

(3.31) \begin{equation} \begin{split} T^{Y,2}_{1k}(x,y) =\lambda \frac{r^2}{2}\frac{\Upsilon _k(x,y)\left (x-y\right )_k}{\left (\left (\frac{r}{2}\right )^2-\left (x-y\right )^2_k\right )^2}\phi _{r}(x,y) &\geq -\frac{4\lambda ^2}{\tilde{c}\sigma ^2}\phi _{r}(x,y) \,=\!:\, -p^{Y,\Upsilon }_{3}\phi _r(x,y). \end{split} \end{equation}

For the second summand, the nonnegativity of $\sigma ^2\Upsilon _k^2(x,y)$ implies the nonnegativity, whenever

\begin{equation*} 2\left (2\left (x-y\right )^2_k-\left (\frac {r}{2}\right )^2\right )\left (x-y\right )_k^2 \geq \left (\left (\frac {r}{2}\right )^2-\left (x-y\right )^2_k\right )^2. \end{equation*}

This holds for $v\in K_{1k}^Y$ , if $2(2c-1)c \geq (1-c)^2$ as implied by the assumption.

Concluding the proof: Combining the formerly established lower bounds (3.19), (3.21), (3.23), (3.25), (3.26), (3.28), (3.29) and (3.31), we obtain for the constants $p^c$ , $p^\ell$ and $p^g$ defined at the beginning of the proof

(3.32) \begin{equation} \begin{aligned} p^c &= p^{*,\Upsilon _c}_{1} \!+ p^{*,\Upsilon _c}_{2} \!+ p^{*,\Upsilon _c}_{3} \!+ p^{Y,\Upsilon _c}_{1} \!+ p^{Y,\Upsilon _c}_{2} \!+ p^{Y,\Upsilon _c}_{3} = 2\!\left (\frac{2\lambda _1 C_{\Upsilon} \sqrt{c}}{(1-c)^2\frac{r}{2}} \!+\! \frac{\sigma _1^2C_{\Upsilon} ^2}{(1-c)^4\!\left (\frac{r}{2}\right )^2} \!+\! \frac{4\lambda _1^2}{\tilde{c}\sigma _1^2}\right )\\ p^\ell &= p^{*,\Upsilon _{\ell} }_{1} \!+ p^{*,\Upsilon _{\ell} }_{2} \!+ p^{*,\Upsilon _{\ell} }_{3} \!+ p^{Y,\Upsilon _{\ell} } = \frac{2\lambda _2 C_{\Upsilon} \sqrt{c}}{(1-c)^2\frac{r}{2}} \!+\! \frac{\sigma _2^2C_{\Upsilon} ^2}{(1-c)^4\left (\frac{r}{2}\right )^2} \!+\! \frac{4\lambda _2^2}{\tilde{c}\sigma _2^2} \!+\! \frac{\sigma _2^2c}{(1-c)^4}\\ p^g &= p^{*,\Upsilon _g}_{1} \!+ p^{*,\Upsilon _g}_{2} \!+ p^{*,\Upsilon _g}_{3} \!+ p^{Y,\Upsilon _g}_{1} \!+ p^{Y,\Upsilon _g}_{2} \!+ p^{Y,\Upsilon _g}_{3} = 2\!\left (\frac{2\lambda _3 C_{\Upsilon} \sqrt{c}}{(1-c)^2\frac{r}{2}} \!+\! \frac{\sigma _3^2C_{\Upsilon} ^2}{(1-c)^4\!\left (\frac{r}{2}\right )^2} \!+\! \frac{4\lambda _3^2}{\tilde{c}\sigma _3^2}\right )\!. \end{aligned} \end{equation}

Together with (3.14) and by using the evolution of $\phi _r$ as in (3.13), we eventually obtain

\begin{equation*} \begin{split} &\frac{d}{dt}\iint \!\phi _{r}\,d\rho _t \geq -d\left (p^c + p^\ell + p^g\right ) \iint \!\phi _{r}\,d\rho _t\\ &\;\quad \,\geq -d\sum _{i=1}^3\omega _i\!\left (\!\left (1\!+\!\mathbb{1}_{i\not =2}\right )\!\left (\frac{2\lambda _i C_{\Upsilon} \sqrt{c}}{(1\!-\!c)^2\frac{r}{2}} \!+\! \frac{\sigma _i^2C_{\Upsilon} ^2}{(1\!-\!c)^4\!\left (\frac{r}{2}\right )^2} \!+\! \frac{4\lambda _i^2}{\tilde{c}\sigma _i^2}\right ) \!+\! \mathbb{1}_{i=2}\frac{\sigma _2^2c}{(1\!-\!c)^4} \!\right ) \!\!\iint \!\phi _{r}\,d\rho _t\\ &\;\quad \,= -q \iint \!\phi _{r}\,d\rho _t, \end{split} \end{equation*}

where $q$ is defined implicitly and where $\omega _i=\mathbb{1}_{\lambda _i\gt 0}$ for $i\in \{1,2,3\}$ . Notice that $\omega _1=1$ since $\lambda _1\gt 0$ by assumption. An application of Grönwall’s inequality concludes the proof.

Proof of Theorem 2.5. If $\mathcal{V}(\rho _0)=0$ , there is nothing to be shown since in this case $\rho _0=\delta _{(x^*,x^*)}$ . Thus, let $\mathcal{V}(\rho _0)\gt 0$ in what follows.

Let us first choose the parameter $\alpha$ such that

(3.33) \begin{align} \alpha \gt \alpha _0 \,:\!=\, \frac{1}{q_{\varepsilon} }\Bigg (\log \left (\frac{2^{d+2}\sqrt{d}}{c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )}\right ) + \max \left \{\frac{1}{2},\frac{p}{(1-\vartheta )\chi _1}\right \}\log \left (\frac{\mathcal{V}(\rho _0)}{\varepsilon }\right )- \log \rho _0\big (\Omega _{r_{\varepsilon}/2}\big )\!\Bigg ), \end{align}

where we introduce the definitions

(3.34) \begin{align} c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right ) \,:\!=\, \min \left \{ \frac{\vartheta }{2}\frac{\chi _1}{2\sqrt{2}\left (\lambda _1+\sigma _1^2\right )}, \sqrt{\frac{\vartheta }{2}\frac{\chi _1}{\sigma _1^2}} \right \} \end{align}

as well as

(3.35) \begin{align} q_{\varepsilon} \,:\!=\, \frac{1}{2}\min \bigg \{\left (\eta \frac{c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\varepsilon }}{2\sqrt{d}}\right )^{1/\nu }\!,{\mathcal{E}}_{\infty }\bigg \} \quad \text{and}\quad r_{\varepsilon} \,:\!=\, \!\max _{s \in [0,R_0]}\left \{\max _{v \in B_s^\infty (x^*)}{\mathcal{E}}(v) \leq q_{\varepsilon} \right \}\!. \end{align}

Moreover, $p$ is as given in (3.12) in Proposition 3.8 with $B=c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\mathcal{V}(\rho _0)}$ in $C_{\Upsilon}$ and with $r=r_{\varepsilon}$ . By construction, $q_{\varepsilon} \gt 0$ and $r_{\varepsilon} \leq R_0$ . Furthermore, recalling the notation ${\mathcal{E}}_{r}=\sup _{v \in B_{r}^\infty (x^*)}{\mathcal{E}}(v)$ from Proposition 3.6, we have $q_{\varepsilon} +{\mathcal{E}}_{r_{\varepsilon} } \leq 2q_{\varepsilon} \leq{\mathcal{E}}_{\infty }$ according to the definition of $r_{\varepsilon}$ . Since $q_{\varepsilon} \gt 0$ , the continuity of $\mathcal{E}$ ensures that there exists $s_{q_{\varepsilon} }\gt 0$ such that ${\mathcal{E}}(v)\leq q_{\varepsilon}$ for all $v\in B_{s_{q_{\varepsilon} }}^\infty (x^*)$ , yielding also $r_{\varepsilon} \gt 0$ .

Let us now define the time horizon $T_{\alpha} \geq 0$ by

(3.36) \begin{align} T_{\alpha} \,:\!=\, \sup \big \{t\geq 0 :\mathcal{V}(\rho _{t'}) \gt \varepsilon \text{ and } \left \|{y_{\alpha }({\rho _{Y,t'}})-x^*}\right \|_2 \lt C(t') \text{ for all } t' \in [0,t]\big \} \end{align}

with $C(t)\,:\!=\,c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\mathcal{V}(\rho _t)}$ . Notice for later use that $C(0)=B$ .

Our aim now is to show that $\mathcal{V}(\rho _{T_{\alpha} }) = \varepsilon$ with $T_{\alpha} \in \big [\frac{(1-\vartheta )\chi _1}{(1+\vartheta/2)\chi _2}T^*,T^*\big ]$ and that we have at least exponential decay of $\mathcal{V}(\rho _{t})$ until time $T_{\alpha}$ , i.e., until the accuracy $\varepsilon$ is reached.

First, however, we verify that $T_{\alpha} \gt 0$ , which is due to the continuity of $t\mapsto\mathcal{V}(\rho _{t})$ and $t\mapsto \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2$ since $\mathcal{V}(\rho _{0}) \gt \varepsilon$ and $\left \|{y_{\alpha }({\rho _{Y,0}})-x^*}\right \|_2 \lt C(0)$ at time $0$ . While the former is a consequence of the assumption, the latter follows from Proposition 3.6 with $q_{\varepsilon}$ and $r_{\varepsilon}$ as defined in (3.35), which allows to show that

\begin{align*} \left \|{y_{\alpha }({\rho _{Y,0}}) - x^*}\right \|_2 &\leq \frac{\sqrt{d}\left (q_{\varepsilon} \!+\!{\mathcal{E}}_{r_{\varepsilon} }\right )^\nu }{\eta } \!+\! \frac{\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _{Y,0}\big (B_{r_{\varepsilon} }^\infty (x^*)\big )}\int \left \|{y-x^*}\right \|_2d\rho _{Y,0}(y)\\ &\leq \frac{\sqrt{d}\left (q_{\varepsilon} \!+\!{\mathcal{E}}_{r_{\varepsilon} }\right )^\nu }{\eta } \!+\! \frac{\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _{Y,0}\big (B_{r_{\varepsilon} }^\infty (x^*)\big )}\iint \left \|{y-x}\right \|_2\!+\!\left \|{x-x^*}\right \|_2d\rho _{0}(x,y)\\ &\leq \frac{c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\varepsilon }}{2} \!+\! \frac{2\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _{Y,0}\big (B_{r_{\varepsilon} }^\infty (x^*)\big )}\sqrt{\mathcal{V}(\rho _0)}\\ &\leq c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\varepsilon } \lt c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\mathcal{V}(\rho _0)} = C(0). \end{align*}

The first inequality in the last line holds by the choice of $\alpha$ in (3.33) and by noticing that $\Omega _{r_{\varepsilon}/2}\subset \mathbb{R}^d\times B_{r_{\varepsilon} }^\infty (x^*)$ and thus $\rho _0(\Omega _{r_{\varepsilon}/2})\leq \rho _{Y,0}\big (B_{r_{\varepsilon} }^\infty (x^*)\big )$ .

Next, we show that the functional $\mathcal{V}(\rho _t)$ is sandwiched between two exponentially decaying functions with rates $(1-\vartheta )\chi _1$ and $(1+\vartheta/2)\chi _2$ , respectively. More precisely, we prove that, up to time $T_{\alpha}$ , $\mathcal{V}(\rho _t)$ decays

1. (i) at least exponentially fast (with rate $(1-\vartheta )\chi _1)$ , and

2. (ii) at most exponentially fast (with rate $(1+\vartheta/2)\chi _2)$ .

To obtain (i), recall that Corollary 3.3 provides an upper bound on the time derivative of $\mathcal{V}(\rho _t)$ given by

(3.37) \begin{equation} \begin{split} \frac{d}{dt}\mathcal{V}(\rho _t) &\leq -\chi _1\mathcal{V}(\rho _t) + 2\sqrt{2}\left (\lambda _1 + \sigma _1^2\right ) \sqrt{\mathcal{V}(\rho _t)}\left \|{y_{\alpha }({\rho _{Y,t}}) - x^*}\right \|_2 + \sigma _1^2\left \|{y_{\alpha }({\rho _{Y,t}}) - x^*}\right \|_2^2 \end{split} \end{equation}

with $\chi _1$ as in (2.12a) being strictly positive by assumption. By combining (3.37) and the definition of $T_{\alpha}$ in (3.36), we have by construction

\begin{align*} \frac{d}{dt}\mathcal{V}(\rho _t) \leq -(1-\vartheta )\chi _1\mathcal{V}(\rho _t) \quad \text{ for all } t \in (0,T_{\alpha} ). \end{align*}

Analogously, for (ii), by Corollary 3.5, we obtain a lower bound on the time derivative of $\mathcal{V}(\rho _t)$ given by

(3.38) \begin{equation} \begin{aligned} \frac{d}{dt}\mathcal{V}(\rho _t) &\geq -\chi _2\mathcal{V}(\rho _t) - 2\sqrt{2}\left (\lambda _1 + \sigma _1^2\right ) \sqrt{\mathcal{V}(\rho _t)}\left \|{y_{\alpha }({\rho _{Y,t}}) - x^*}\right \|_2 \\ &\geq -(1+\vartheta/2)\chi _2\mathcal{V}(\rho _t) \quad \text{ for all } t \in (0,T_{\alpha} ), \end{aligned} \end{equation}

where the second inequality again exploits the definition of $T_{\alpha}$ . Grönwall’s inequality now implies for all $t \in [0,T_{\alpha} ]$ the upper and lower estimates

(3.39a) \begin{align}\mathcal{V}(\rho _t) &\leq\mathcal{V}(\rho _0) \exp \left ({-}(1-\vartheta )\chi _1 t\right ), \end{align}

(3.39b) \begin{align}\mathcal{V}(\rho _t) &\geq\mathcal{V}(\rho _0) \exp \left ({-}(1+\vartheta/2)\chi _2 t\right ), \end{align}

thereby proving (i) and (ii). The definition of $T_{\alpha}$ together with the one of $C(t)$ permits to control

(3.40) \begin{align} \max _{t \in [0,T_{\alpha} ]}\left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2 \leq \max _{t \in [0,T_{\alpha} ]} C(t)\leq C(0). \end{align}

To conclude, it remains to prove $\mathcal{V}(\rho _{T_{\alpha} }) = \varepsilon$ with $T_{\alpha} \in \big [\frac{(1-\vartheta )\chi _1}{(1+\vartheta/2)\chi _2}T^*,T^*\big ]$ . To this end, we consider the following three cases separately.

Case $T_{\alpha} \geq T^*$ : If $T_{\alpha} \geq T^*$ , the time-evolution bound of $\mathcal{V}(\rho _t)$ from (3.39a) combined with the definition of $T^*$ in (2.13) allows to immediately infer $\mathcal{V}(\rho _{T^*}) \leq \varepsilon$ . Therefore, with $\mathcal{V}(\rho _{t})$ being continuous, $\mathcal{V}(\rho _{T_{\alpha} }) = \varepsilon$ and $T_{\alpha} = T^*$ according to the definition of $T_{\alpha}$ in (3.36).

Case $T_{\alpha} \lt T^*$ and $\mathcal{V}(\rho _{T_{\alpha} }) \leq \varepsilon$ : By continuity of $\mathcal{V}(\rho _t)$ , it holds for $T_{\alpha}$ as defined in (3.36), $\mathcal{V}(\rho _{T_{\alpha} }) = \varepsilon$ . Thus, $\varepsilon =\mathcal{V}(\rho _{T_{\alpha} }) \geq\mathcal{V}(\rho _0) \exp \left ({-}(1+\vartheta/2)\chi _2 T_{\alpha} \right )$ as a consequence of the time-evolution bound (3.39b). The latter can be reordered as

\begin{equation*} \frac {(1-\vartheta )\chi _1}{(1+\vartheta/2)\chi _2}T^* = \frac {1}{(1+\vartheta/2) \chi _2}\log \left (\frac {\mathcal{V}(\rho _0)}{\varepsilon }\right ) \leq T_{\alpha} \lt T^*. \end{equation*}

Case $T_{\alpha} \lt T^*$ and $\mathcal{V}(\rho _{T_{\alpha} }) \gt \varepsilon$ : We will prove that this case can actually not occur by showing that $\left \|{y_{\alpha }({\rho _{Y,T_{\alpha} }})-x^*}\right \|_2 \lt C(T_{\alpha} )$ for the $\alpha$ chosen in (3.33). In fact, if both $\mathcal{V}(\rho _{T_{\alpha} })\gt \varepsilon$ and $\left \|{y_{\alpha }({\rho _{Y,T_{\alpha} }})-x^*}\right \|_2 \lt C(T_{\alpha} )$ held true simultaneously, this would contradict the definition of $T_{\alpha}$ in (3.36). To obtain this contradiction, we apply again Proposition 3.6 with $q_{\varepsilon}$ and $r_{\varepsilon}$ as before to get

(3.41) \begin{align} \begin{split} \left \|{y_{\alpha }({\rho _{Y,T_{\alpha} }})-x^*}\right \|_2 &\leq \frac{\sqrt{d}\left (q_{\varepsilon} \!+\!{\mathcal{E}}_{r_{\varepsilon} }\right )^\nu }{\eta } \!+\! \frac{\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _{Y,T_{\alpha} }\big (B_{r_{\varepsilon} }^\infty (x^*)\big )}\int \left \|{y-x^*}\right \|_2d\rho _{Y,T_{\alpha} }(y)\\ &\leq \frac{\sqrt{d}\left (q_{\varepsilon} \!+\!{\mathcal{E}}_{r_{\varepsilon} }\right )^\nu }{\eta } \!+\! \frac{\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _{Y,T_{\alpha} }\big (B_{r_{\varepsilon} }^\infty (x^*)\big )}\!\iint \!\left \|{y-x}\right \|_2\!+\!\left \|{x-x^*}\right \|_2d\rho _{T_{\alpha} }(x,y)\\ &\leq \frac{c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\varepsilon }}{2} \!+\! \frac{2\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _{Y,T_{\alpha} }\big (B_{r_{\varepsilon} }^\infty (x^*)\big )}\sqrt{\mathcal{V}(\rho _{T_{\alpha} })}\\ &\lt \frac{c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\mathcal{V}(\rho _{T_{\alpha} })}}{2} \!+\! \frac{2\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _{Y,T_{\alpha} }\big (B_{r_{\varepsilon} }^\infty (x^*)\big )}\sqrt{\mathcal{V}(\rho _{T_{\alpha} })}. \end{split} \end{align}

Since, thanks to (3.40), we have $\max _{t \in [0,T_{\alpha} ]}\|{y_{\alpha }({\rho _{Y,t}})-x^*}\|_2 \leq B$ for $B=C(0)$ , which in particular does not depend on $\alpha$ , Proposition 3.8 guarantees the existence of $p\gt 0$ independent of $\alpha$ (but dependent on $B$ and $r_{\varepsilon}$ ) with

\begin{align*} \rho _{Y,T_{\alpha} }(B_{r_{\varepsilon} }^\infty (x^*)) &\geq \left (\iint \phi _{r_{\varepsilon} }(x,y) \,d\rho _0(x,y)\right )\exp ({-}pT_{\alpha} ) \\ &\geq \frac{1}{2^d}\,\rho _0\big (\Omega _{r_{\varepsilon}/2}\big ) \exp ({-}pT^*) \gt 0. \end{align*}

Here we use that $(x^*,x^*)\in \operatorname{supp}(\rho _0)$ to bound the initial mass $\rho _0$ and the fact that $\phi _{r}$ from Lemma 3.7 is bounded from below on $\Omega _{r/2}$ by $1/2^d$ . With this, we can continue the chain of inequalities in (3.41)3 to obtain

\begin{align*} \begin{split} \left \|{y_{\alpha }({\rho _{Y,T_{\alpha} }})-x^*}\right \|_2 &\lt \frac{c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\mathcal{V}(\rho _{T_{\alpha} })}}{2} + \frac{2^{d+1}\sqrt{d}\exp \left ({-}\alpha q_{\varepsilon} \right )}{\rho _0\big (\Omega _{r_{\varepsilon}/2}\big ) \exp ({-}pT^*)}\sqrt{\mathcal{V}(\rho _{T_{\alpha} })}\\ &\leq c\left (\vartheta,\chi _1,\lambda _1,\sigma _1\right )\sqrt{\mathcal{V}(\rho _{T_{\alpha} })} = C(T_{\alpha} ), \end{split} \end{align*}

with the first inequality in the last line holding due to the choice of $\alpha$ in (3.33). This gives the desired contradiction, again thanks to the continuity of $t\mapsto\mathcal{V}(\rho _{t})$ and $t\mapsto \left \|{y_{\alpha }({\rho _{Y,t}})-x^*}\right \|_2$ .