Gamma Calculus Beyond Villani and Explicit Convergence Estimates for Langevin Dynamics with Singular Potentials

Abstract

We apply Gamma calculus to the hypoelliptic and non-symmetric setting of Langevin dynamics under general conditions on the potential. This extension allows us to provide explicit estimates on the convergence rate (which is exponential) to equilibrium for the dynamics in a weighted $H^1(\mu )$ sense, $\mu$ denoting the unique invariant probability measure of the system. The general result holds for singular potentials, such as the well-known Lennard–Jones interaction and confining well, and it is applied in such a case to estimate the rate of convergence when the number of particles N in the system is large.

Appendix

      Here we provide details behind some of the more technical estimates in the paper.

1.1 Quantitative Inequalities for $\mu$

      Recall that $\mu$ denotes the product Gibbs measure (2.13), and let $\mu _1$ and $\mu _2$ denote the marginal measures given by

$$\begin{aligned} \mu _1(A) \!=\! \int _A \int _{\mathscr {O}} \frac{1}{{\mathscr {N}}} e^{-\frac{1}{T} H(x,v)} \,\mathrm{d}x \, \mathrm{d}v \,\,\, \,\, \text { and } \,\,\,\,\, \mu _2(B) \!=\! \int _B \int _{({\mathbf {R}}^k)^N} \frac{1}{{\mathscr {N}}} e^{-\frac{1}{T} H(x,v)} \,\mathrm{d}v \, \mathrm{d}x \end{aligned}$$

defined for Borel subsets $A\subseteq ({\mathbf {R}}^k)^N$ and $B\subseteq {\mathscr {O}}$.

Proposition 5.9

Suppose that U satisfies Assumptions 2.6 and 2.7 and recall the constant $R_2>0$ defined in the statement of Theorem 5.1. Then we have

$$\begin{aligned} \int _{\mathscr {O}} |\nabla U|^2 \, \mathrm{d}\mu _2 \!\leqslant \! \frac{ \kappa '' T \sqrt{d}}{1-\frac{1}{16 \sqrt{d}}}\qquad \text { and } \qquad \mu _2(\{ x \!\in \! {\mathscr {O}} \, : \, U(x) \!\geqslant \! R_2 \}) \!\leqslant \! \frac{1}{2(10e^4+1)}. \end{aligned}$$

Moreover, if $K\subseteq {\mathscr {X}}$ denotes the set in Theorem 5.1, then

$$\begin{aligned} \mu (K^c) \leqslant \frac{1}{10e^4 +1}. \end{aligned}$$

(5.10)

Proof

Suppose that $U:({\mathbf {R}}^k)^N\rightarrow [0, \infty ]$ satisfies Assumptions 2.6 and 2.7. Consider the following gradient system on ${\mathscr {O}}$

$$\begin{aligned} \mathrm{d}X(t) = -\nabla U(X(t)) \, \mathrm{d}t + \sqrt{2 T} \, dW(t) \end{aligned}$$

(5.11)

where W(t) is a standard, Nk-dimensional Brownian motion on $(\Omega , {\mathscr {F}}, {\mathbf {P}})$. Under the assumptions on the potential U, it is not hard to show that, like Equation 2.1, Equation (5.11) has unique pathwise solutions on the state space ${\mathscr {O}}$ for all finite times $t\geqslant 0$. Moreover, $\mu _2$ is the unique invariant probability measure for the Markov process X(t). This can be seen by using U itself as a Lyapunov functional employing the hypotheses of the statement. For $n\in {\mathbf {N}}$, let $\xi _n$ be the first exit time of the process (5.11) from the set $\{ U \leqslant n \}$ and observe that

$$\begin{aligned} {\mathbf {E}}_x U(X(t\wedge \xi _n))&= U(x) + {\mathbf {E}}_x \int _0^{t\wedge \xi _n} -|\nabla U(X(s))|^2 + T \Delta U(X(s)) \, ds . \end{aligned}$$

Next, to bound the quantity above, note that $|\Delta U | \leqslant \sqrt{d} \sup _{|y|\leqq 1} |\nabla ^2 U y|$. Hence applying Assumption 2.7 gives

$$\begin{aligned} {\mathbf {E}}_x U(X(t\wedge \xi _n))&\leqslant U(x) + {\mathbf {E}}_x \int _0^{t\wedge \xi _n} -\bigg (1- \frac{1}{16 \sqrt{d}}\bigg ) |\nabla U(X(s))|^2 + \sqrt{d} \kappa '' T \, ds. \end{aligned}$$

Since $U\geqslant 0$ and $\kappa ' \in (0,1)$, this then implies the estimate

$$\begin{aligned} {\mathbf {E}}_x \int _0^{t\wedge \xi _n} |\nabla U(x(s))|^2 \, ds \leqslant \frac{U(x)}{1-\frac{1}{16 \sqrt{d}}} +\frac{ t \kappa '' T \sqrt{d}}{1-\frac{1}{16 \sqrt{d}}} . \end{aligned}$$

Using Fatou’s lemma and the fact that $\xi _n\rightarrow \infty$ almost surely, we find that for all $t>0$

$$\begin{aligned} \frac{1}{t} {\mathbf {E}}_x \int _0^t |\nabla U(x(s))|^2 \, ds \leqslant \frac{U(x)}{t(1-\frac{1}{16 \sqrt{d}})} + \frac{ \kappa '' T \sqrt{d}}{1-\frac{1}{16 \sqrt{d}}} \end{aligned}$$

Thus, by another simple approximation argument using convergence of the Césaro means to $\mu _2$,

$$\begin{aligned} \int _{\mathscr {O}} |\nabla U|^2 \,d \mu _2 \leqslant \frac{ \kappa '' T \sqrt{d}}{1-\frac{1}{16 \sqrt{d}}}. \end{aligned}$$

This gives the first inequality.

      For the second, observe that if $f(R)= c_\infty R^{2-\frac{2}{\eta _\infty }} - d_\infty$, then

$$\begin{aligned} \{ U \geqslant R\} \subseteq \{ |\nabla U|^2 \geqslant f(R)\}. \end{aligned}$$

Hence, employing the first inequality

$$\begin{aligned} \int _{\{ U \geqslant R\}} \, \mathrm{d}\mu _2 \leqslant \int _{\{ |\nabla U|^2 \geqslant f(R)\}} \, \mathrm{d}\mu _2\leqslant \frac{1}{f(R)} \int |\nabla U|^2 \, \mathrm{d}\mu _2 \leqslant \frac{ \kappa '' T \sqrt{d}}{f(R)(1-\frac{1}{16 \sqrt{d}})}, \end{aligned}$$

from which we arrive at the second claimed bound by plugging in $R=R_2$.

      To obtain the final desired inequality, observe that

$$\begin{aligned} \mu (K^c) = \int _{K^c} \, \mathrm{d}\mu&\leqslant \int _{\{|v|^2 \geqslant (20e^4 +2)NkT\}} \, \mathrm{d}\mu _1 + \int _{\{ U \geqslant R_2\}} \, \mathrm{d}\mu _2 \\&\leqslant \int _{\{|v|^2 \geqslant (20e^4+2) NkT\}} \, \mathrm{d}\mu _1 + \frac{1}{2(10e^4+1)}. \end{aligned}$$

Now, to get a bound on the remaining quantity above, this time we consider the process V(t) on $({\mathbf {R}}^k)^N$ defined by

$$\begin{aligned} \mathrm{d}V(t) = - V(t) \, \mathrm{d}t + \sqrt{2T} \, \mathrm{d}W(t), \end{aligned}$$

where W(t) is a standard Brownian motion on $({\mathbf {R}}^k)^N$ on $(\Omega , {\mathscr {F}}, {\mathbf {P}})$. Note that this process is exactly in the same form as in (5.11) by setting $U(v)= |v|^2/2$. Hence, in exactly the same way as before, it follows that

$$\begin{aligned} \frac{1}{t} \int _0^t {\mathbf {E}}_v |V(s)|^2 \, ds \leqslant \frac{|v|^2}{2t} + NkT \end{aligned}$$

for all $t>0$. Consequently,

$$\begin{aligned} \int |v|^2 \, \mathrm{d}\mu _1(v) \leqslant NkT. \end{aligned}$$

Plugging this fact into the above gives

$$\begin{aligned} \mu (K^c)&\leqslant \int _{\{|v|^2 \geqslant (20e^4+2) NkT\}} \, \mathrm{d}\mu _1 + \frac{1}{20e^4+2} \\&\leqslant \frac{1}{(20e^4+2) NkT} \int |v|^2 \, \mathrm{d}\mu _1(v) + \frac{1}{20e^4+2}\\&\leqslant \frac{1}{10e^4+1}. \end{aligned}$$

$\square$

1.2 Quantitative Bounds for Singular Potentials

      Next we aim to prove Proposition 2.40 which gives the claimed estimates on the singular potential in Example 2.37. We first need the following lower bound on the gradient:

Lemma 5.12

Consider the potential U and the open set ${\mathscr {O}}$ defined in Example 2.37. Then we have the estimate

$$\begin{aligned} | \nabla U(x)| \geqslant \frac{A a}{2 N^{3/2}} \sum _{i=1}^N |x_i|^{a -1} + \frac{Bb }{2N^{\frac{7}{2}}}\sum _{i<j} \frac{1}{|x_i-x_j|^{b+1}} -\frac{Aa}{\sqrt{N}}-B b N^{b+ \frac{5}{2}} \end{aligned}$$

(5.13)

for all $x\in {\mathscr {O}}$.

      The argument is a reworking of the proof of Lemma 4.12 of [5]. This proof is also in [16], but here we give explicit constants.

Proof

The idea behind the proof is to use the basic fact that, for $x\in {\mathscr {O}}$, $|\nabla U (x)| \geqslant |\nabla U(x) \cdot y|$ for all $y\in ({\mathbf {R}}^k)^N$ with $|y|=1$. Then we aim to pick a convenient direction $y\in ({\mathbf {R}}^k)^N$ with $|y|=1$. Notationally, we set ${\mathscr {Z}}_N= \{ 1, 2, \ldots , N\}$.

      We first claim that

$$\begin{aligned} |\nabla U(x)| \geqslant \frac{Aa}{\sqrt{N}} |x_i|^{a-1}- \frac{Aa}{\sqrt{N}}- Bb N^{b+\frac{5}{2}} \end{aligned}$$

(5.14)

for all $i=1,2, \ldots , N$ and $x\in {\mathscr {O}}$. From this, summing both sides of the previous inequality from 1 to N it follows that

$$\begin{aligned} |\nabla U (x)| \geqslant \frac{Aa}{N^{3/2}}\sum _{i=1}^N |x_i|^{a-1} - \frac{Aa}{\sqrt{N}}- Bb N^{b+ \frac{5}{2}} \end{aligned}$$

(5.15)

on ${\mathscr {O}}$. Note that without loss of generality it suffices to show the bound (5.14) above for $i=1$ and for $|x_1| > 1$. For $x=(x_1, x_2, \ldots , x_N)\in {\mathscr {O}}$, consider an increasing sequence of sets $S_i(x)$, $i=1,2, \ldots , N$, defined inductively as follows:

$$\begin{aligned} S_1(x)&= \{ j \in {\mathscr {Z}}_N\, : \, |x_1- x_j|< N^{-1}\} \\ S_m(x)&= \{ j \in {\mathscr {Z}}_N \, : \, |x_j - x_k|< N^{-1} \,\, \exists \,\,k \in S_{m-1}(x)\}, \,\,\,\, m =2, \ldots , N. \end{aligned}$$

Observe that $S_1(x)\ne \emptyset$ since $1\in S_1(x)$. Also note that for any $i,j \in S_N(x)$, $|x_i - x_j| < 1$. Consequently, combining $|x_1-x_j|^2=|x_1|^2+ |x_j|^2 - 2 x_1\cdot x_j$ with the inequality $|x_1-x_j|<1$, it follows that for $j\in S_N(x)$, $2 x_1 \cdot x_j \geqslant |x_1|^2 + |x_j|^2 - 1 \geqslant 0$ where the last inequality follows since $|x_1|>1$. Moreover, if $i\in S_N(x)$ while $j\notin S_N(x)$ we have that $|x_i - x_j | \geqslant N^{-1}$. Let $\sigma (x)=(\sigma _1(x), \ldots , \sigma _N (x)) \in ({\mathbf {R}}^k)^N$ be such that $\sigma _i(x)= x_1/|x_1|$ if $i\in S_N(x)$ and $\sigma _i=0$ otherwise. We thus have the bound

$$\begin{aligned}&\sqrt{N} |\nabla U(x)| \nonumber \\&\geqslant \sigma (x) \cdot \nabla U(x)\nonumber \\&= A a \sum _{n \in S_N(x)} \frac{x_1\cdot x_n}{|x_1|}|x_n|^{a-2} + Bb \sum _{n\in S_N(x)} \sum _{\begin{array}{c} i=1\\ i\ne n \end{array}}^N \frac{x_1 |x_1|^{-1}\cdot (x_n - x_i)}{|x_n-x_i|^{b+2}}\nonumber \\&\geqslant Aa |x_1|^{a-1} +Bb \sum _{n\in S_N(x)} \sum _{\begin{array}{c} i=1\\ i\ne n\\ i \in S_N(x) \end{array}}^N \frac{x_1 |x_1|^{-1}\cdot (x_n - x_i)}{|x_n-x_i|^{b+2}} \nonumber \\&\qquad + Bb\sum _{n\in S_N(x)} \sum _{\begin{array}{c} i=1\\ i\ne n\\ i \notin S_N(x) \end{array}}^N \frac{x_1 |x_1|^{-1}\cdot (x_n - x_i)}{|x_n-x_i|^{b+2}}. \end{aligned}$$

(5.16)

Next note that

$$\begin{aligned} \sum _{n\in S_N(x)} \sum _{\begin{array}{c} i=1\\ i\ne n\\ i \in S_N(x) \end{array}}^N \frac{x_1 |x_1|^{-1}\cdot (x_n - x_i)}{|x_n-x_i|^{b+2}}&= \sum _{\begin{array}{c} n,i \in S_N(x)\\ n\ne i \end{array}} \frac{x_1 |x_1|^{-1}\cdot (x_n - x_i)}{|x_n-x_i|^{b+2}} \\&= \sum _{\begin{array}{c} n,i \in S_N(x)\\ n\ne i \end{array}} \frac{x_1 |x_1|^{-1}\cdot (x_i - x_n)}{|x_n-x_i|^{b+2}}.\end{aligned}$$

Consequently,

$$\begin{aligned} \sum _{n\in S_N(x)} \sum _{\begin{array}{c} i=1\\ i\ne n\\ i \in S_N(x) \end{array}}^N \frac{x_1 |x_1|^{-1}\cdot (x_n - x_i)}{|x_n-x_i|^{b+2}} =0. \end{aligned}$$

Plugging this back into (5.16), we obtain the following inequality

$$\begin{aligned} \sqrt{N} |\nabla U(x)|&\geqslant Aa |x_1|^{a-1} + 0 - Bb N^{b + 3}. \end{aligned}$$

This finishes the proof of the bound (5.14) when $i=1$, as desired.

      We next show that for all $i,m \in {\mathscr {Z}}_N$ with $i\ne m$

$$\begin{aligned} |\nabla U(x)| \geqslant \frac{2B b}{\sqrt{N}}|x_i-x_m|^{-b-1} - \frac{A a}{\sqrt{N}}\sum _{n=1}^N |x_n|^{a-1} \end{aligned}$$

(5.17)

for all $x =(x_1, x_2, \ldots , x_N) \in {\mathscr {O}}$. Note then that this estimate together with the bound (5.15) implies the lemma. Without loss of generality, we will prove the bound (5.17) for $i=1, m=2$. For $x\in {\mathscr {O}}$, let $\sigma (x)= (x_2-x_1)/|x_2-x_1|$ and $\xi _k(x)= c_k(x) \sigma (x)$ where the constants $c_k(x) \in \{-1,1\}$ are chosen to satisfy $c_k(x) = 1$ if $x_k\cdot \sigma (x) < x_2 \cdot \sigma (x)$ and $c_k(x)=-1$ otherwise. With this choice of direction $\xi (x):=(\xi _1(x), \ldots , \xi _N(x))$ we find that on ${\mathscr {O}}$:

$$\begin{aligned} \sqrt{N} |\nabla U(x)|&\geqslant \xi (x) \cdot \nabla U(x)\nonumber \\&= \sum _{n =1}^N A a \xi _n(x) \cdot x_n |x_n|^{a-2} + B b\sum _{i<n} (\xi _i(x)-\xi _n(x)) \cdot \frac{x_n-x_i}{|x_n-x_i|^{b+2}}\nonumber \\&\geqslant - Aa \sum _{n=1}^N |x_n|^{a-1}+ Bb \sum _{i<n} (\xi _i(x)-\xi _n(x)) \cdot \frac{x_n-x_i}{|x_n-x_i|^{b+2}}\nonumber \\&= - Aa \sum _{n=1}^N |x_n|^{a-1} \!+\! Bb \sum _{i <n} (c_i(x)\!-\!c_n(x)) \cdot \frac{x_n \cdot \sigma (x)\!-\!x_i\cdot \sigma (x)}{|x_n-x_i|^{b+2}}. \end{aligned}$$

(5.18)

To bound the remaining term on the righthand side above, note that if i, n are either such that $x_i \cdot \sigma (x) < x_2(x) \cdot \sigma (x)$ and $x_n \cdot \sigma (x) < x_2\cdot \sigma (x)$ or such that $x_i \cdot \sigma (x) \geqslant x_2 \cdot \sigma (x)$ and $x_n \cdot \sigma (x) \geqslant x_2 \cdot \sigma (x)$, then $c_i(x)=c_n(x)$. Hence the corresponding term in the sum in (5.18) is zero. On the other hand, if i, n are either such that $x_i\cdot \sigma < x_2 \cdot \sigma \leqslant x_n \cdot \sigma$ or such that $x_n \cdot \sigma < x_2\cdot \sigma \leqslant x_i \cdot \sigma$, then the corresponding term in the sum (5.18) is nonnegative via the $c_i(x)$, $c_n(x)$. In particular, by these observations we arrive at the estimate

$$\begin{aligned} \sqrt{N} |\nabla U(x)|&\geqslant - Aa \sum _{n=1}^N |x_n|^{a-1} + Bb (c_1(x)-c_2(x)) \frac{(x_2 -x_1) \cdot \sigma }{|x_2-x_1|^{b +2}}\nonumber \\&= - A a \sum _{n=1}^N |x_n|^{a-1} + 2 Bb \frac{1}{|x_2-x_1|^{b +1}} \end{aligned}$$

(5.19)

as $c_1(x)-c_2(x) =2$ and $(x_2-x_1)\cdot \sigma = |x_2-x_1|$ by construction. $\quad \square$

      We can now use the previous result to conclude Proposition 2.40.

Proof of Proposition 2.40

We begin by computing $\nabla U$ and $\nabla ^2U$ on ${\mathscr {O}}$. Observe that for $x\in {\mathscr {O}}$ and $i,n=1,2, \ldots , N$, $\ell ,m =1,2,\ldots , k$,

$$\begin{aligned} \partial _{x_i^\ell } U(x)&= A a x_i^\ell |x_i|^{a-2} - B b \sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^N \frac{x_i^\ell -x_j^\ell }{|x_i -x_j|^{b+2}} \end{aligned}$$

and

$$\begin{aligned} \partial _{x_{n}^m x_i^\ell }^2 U(x)&= Aa \delta _{(i, \ell ), (n, m)} |x_i|^{a-2} + A a(a-2) \delta _{i, n} |x_i|^{a-4} x_i^\ell x_{n}^m\\&+ B b\sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^N \bigg \{\frac{\delta _{(j, \ell ), (n, m)} - \delta _{(i, \ell ), (n, m)} }{|x_i -x_{j}|^{b+2}} + (b+2) \frac{(x_i^\ell -x_{j}^\ell )(\delta _{i, n}(x_i^m-x_{j}^m) + \delta _{n, j} (x_{j}^m-x_i^m))}{|x_i -x_{j}|^{b+4}}\bigg \} \end{aligned}$$

where $\delta _{i,j}=1$ if $i=j$ and 0 otherwise. Also, the $(a-2)\times \cdots$ term is defined to be 0 when $a=2$. Thus for any $y\in ({\mathbf {R}}^k)^N$ with $|y| \leqslant 1$ and any $x\in {\mathscr {O}}$ we arrive at the estimate

$$\begin{aligned} |\nabla ^2 U(x) y|&\leqslant Aa(a-1)k \sum _{i=1}^N |x_i|^{a-2} + 4Bb (b+3) k \sum _{i<n} \frac{1}{|x_i -x_{n}|^{b+2}}.\qquad \end{aligned}$$

(5.20)

Applying Lemma 5.12 with Young’s inequality gives that for all $x\in {\mathscr {O}}$

$$\begin{aligned} |\nabla U(x)|^2 \geqslant \frac{A^2a^2}{8 N^{3}} \sum _{i=1}^N |x_i|^{2a -2} + \frac{B^2 b^2 }{8 N^7}\sum _{i<j} |x_i-x_j|^{-2b-2} -\frac{2A^2a^2}{N}-2 B^2b^2 N^{2b+5}. \end{aligned}$$

(5.21)

      In order to compare (5.21) with (5.20), next note that for any constants $C_i>0$ we have

$$\begin{aligned} \sum _{i=1}^N |x_i|^{a-2}&= \sum _{\{i\, :\, |x_i|^{a} \geqslant C_1\}} |x_i|^{a-2} + \sum _{\{i: |x_i|^{a} \leqslant C_1\}} |x_i|^{a-2} \leqslant C_1^{\frac{a-2}{a}} N+ \frac{1}{C_1}\sum _{i=1}^N |x_i|^{2a-2} \end{aligned}$$

and, similarly,

$$\begin{aligned} \sum _{i<j} \frac{1}{|x_i -x_j |^{b+2}}&=\sum _{\{i<j\, :\, |x_i-x_j|^{b} \leqslant 1/C_2\}} \frac{1}{|x_i -x_j |^{b+2}} + \sum _{\{i<j\, :\, |x_i-x_j|^{b} \geqslant 1/C_2\}} \frac{1}{|x_i -x_j |^{b+2}} \\&\leqslant C_2^{\frac{b+2}{b}}N^2 +\frac{1}{C_2} \sum _{i<n} \frac{1}{|x_i -x_n|^{2b+2}}. \end{aligned}$$

Thus picking $C_1=128(a-1)N^4k^2 T/(Aa)$, $C_2=512 N^8k^2 (b+3)T/(Bb)$ and combining (5.21) with (5.20) produces the following bound on ${\mathscr {O}}$

$$\begin{aligned} |\nabla ^2 U (x) \cdot y| \leqslant \frac{1}{16Td}|\nabla U(x)|^2 + \kappa '' \end{aligned}$$

(5.22)

for any $y \in ({\mathbf {R}}^k)^N$ with $|y| \leqslant 1$ where $\kappa ''$ is as in the statement of the result. Note that this validates the first part of Assumption 2.7.

In order to check the second part of Assumption 2.7, first note that for any $\eta _0> -1, \eta _0 \ne 0,$ we have the following estimate

$$\begin{aligned} U(x)^{2+\frac{2}{\eta _0}} \geqslant \frac{1}{2N} \sum _{i=1}^N A^{2+ \frac{2}{\eta _0}} |x_i|^{2a+\frac{2a}{\eta _0}} + \frac{1}{2N^2} \sum _{i<j} \frac{B^{2+\frac{2}{\eta _0}}}{|x_i -x_j|^{2b+\frac{2b}{\eta _0}}} \end{aligned}$$

(5.23)

for any $x\in {\mathscr {O}}$. Moreover, for any $\eta _\infty >1$ and any $x\in {\mathscr {O}}$

$$\begin{aligned} U(x)^{2-\frac{2}{\eta _\infty }} \leqslant (2N)^{2-\frac{2}{\eta _\infty }} \sum _{i=1}^N A^{2-\frac{2}{\eta _\infty }} |x_i|^{2a- \frac{2a}{\eta _\infty }} + (2N^2)^{2- \frac{2}{\eta _\infty }} \sum _{i<j} \frac{B^{2-\frac{2}{\eta _\infty }}}{|x_i -x_j|^{2b-\frac{2b}{\eta _\infty }}}. \end{aligned}$$

(5.24)

Also note that on ${\mathscr {O}}$

$$\begin{aligned} |\nabla U(x)|^2&\leqslant 2 A^2 a^2 \sum _{i=1}^N |x_i|^{2a-2} + 2 B^2 b^2 N \sum _{i<j} \frac{1}{|x_i-x_j|^{2b+2}} \end{aligned}$$

(5.25)

Combining the estimate (5.23) with (5.25) and picking $\eta _0=b$ produces the inequality

$$\begin{aligned} |\nabla U(x)|^2 \leqslant c_0 U(x)^{2+\frac{2}{b}} + d_0 \end{aligned}$$

(5.26)

where $c_0$ and $d_0$ are as in the statement of the result. Similarly, combining (5.24) with (5.21) and choosing $\eta _\infty =a$ gives

$$\begin{aligned} |\nabla U(x)|^2 \geqslant c_\infty U(x)^{2-\frac{2}{a}} - d_\infty \end{aligned}$$

(5.27)

where $c_\infty$ and $d_\infty$ are as in the statement of the result. $\quad \square$