On Explicit $L^2$-Convergence Rate Estimate for Underdamped Langevin Dynamics

Abstract

We provide a refined explicit estimate of the exponential decay rate of underdamped Langevin dynamics in the $L^2$ distance, based on a framework developed in Albritton et al. (Variational methods for the kinetic Fokker–Planck equation, arXiv arXiv:1902.04037, 2019). To achieve this, we first prove a Poincaré-type inequality with a Gibbs measure in space and a Gaussian measure in momentum. Our estimate provides a more explicit and simpler expression of the decay rate; moreover, when the potential is convex with a Poincaré constant $m \ll 1$, our estimate shows the decay rate of $O(\sqrt{m})$ after optimizing the choice of the friction coefficient, which is much faster than m for the overdamped Langevin dynamics.

Appendices

Appendix A. The Decay Rate for Isotropic Quadratic Potential

For isotropic quadratic potential, an explicit expression for the spectral gap of ${\mathcal {L}}$ is available (thus also the decay rate in (7)). Note that while the result is stated for $d = 1$, it trivially extends to arbitrary dimension for isotropic quadratic potential as different coordinates are independent. The spectrum is also explicitly known for $V=0$ and $x\in {\mathbb {T}}^d$ on a torus, see [35].

Theorem 3

([46, (10.83)], [42, Theorem 3.1]) When $U(x) = \frac{m}{2}|x|^2$, $d = 1$, the spectrum of the operator $-{\mathcal {L}}$ is given by

$$\begin{aligned} \left\{ \lambda _{i,j} {:}{=} \frac{\gamma }{2} (i + j) + \frac{\sqrt{\gamma ^2 - 4 m}}{2}(i-j), \qquad i,j = 0, 1, 2, \cdots .\right\} . \end{aligned}$$

Let $\lambda _{\text {exact}}$ be the spectral gap for the real component of $\{\lambda _{i,j}\}_{i,j\geqslant 0}$. Notice that the spectral gap is always achieved when $i = 0$ and $j = 1$, thus

$$\begin{aligned} \lambda _{\text {exact}} = \mathfrak {Re}\left( \frac{\gamma }{2} - \frac{\sqrt{\gamma ^2-4m}}{2}\right) . \end{aligned}$$

(45)

Corollary A.1

For any dimension d, for isotropic potential $U(x) = \frac{m}{2}|x|^2$, (7) holds with the decay rate $\lambda _{\text {exact}}$.

Appendix B. The DMS Hypocoercive Estimation

In this section, we will revisit the decay rate by DMS estimation [18, 19], adapted and summarized for underdamped Langevin equation in [47, Sec. 2]. In the first part of this section, we will review the main result based on [47]; in addition, we will provide a new estimate of the operator norm of $\left\Vert {\mathcal {A}}{\mathcal {L}}_{\text {ham}}(1-\Pi _v)\right\Vert _{L^2(\rho _{\infty }) \rightarrow L^2(\rho _{\infty })}$, which leads into a more explicit expression of the decay rate. In the second part, we will present the asymptotic analysis of the decay rate with respect to m and $\gamma$, under the assumption that $\nabla _x^2 U \geqslant -2\, \textrm{Id}$.

1.1 Revisiting the DMS Hypocoercive Estimation in $L^2(\rho _{\infty })$

Let us first define an operator

$$\begin{aligned} {\mathcal {A}}= \left( 1 + ({\mathcal {L}}_{\text {ham}}\Pi _v)^* ({\mathcal {L}}_{\text {ham}}\Pi _v)\right) ^{-1} ({\mathcal {L}}_{\text {ham}}\Pi _v)^* \end{aligned}$$

(46)

and a Lyapunov function $\textsf{E}$ for $\phi (x, v)$ by

$$\begin{aligned} \textsf{E}(\phi ) = \frac{1}{2} \left\Vert \phi \right\Vert ^2_{L^2(\rho _{\infty })} -\epsilon \left( {\mathcal {A}}\phi , \phi \right) _{L^2(\rho _{\infty })}, \end{aligned}$$

(47)

where $\epsilon \in (-1, 1)$ is some quantity depending on ${\mathcal {L}}$, to be specified below. The functional $\textsf{E}$ is equivalent to $L^2(\rho _{\infty })$ norm in the following sense (see e.g., [47, Eq. (17)]),

$$\begin{aligned} \frac{1- |\epsilon |}{2} \left\Vert \phi \right\Vert _{L^2(\rho _{\infty })}^2 \leqslant \textsf{E}(\phi ) \leqslant \frac{1+|\epsilon |}{2} \left\Vert \phi \right\Vert _{L^2(\rho _{\infty })}^2. \end{aligned}$$

(48)

Theorem 4

(See [47, Theorem 1]) Assume that the Poincaré inequality (10) holds and there exists $\textsf{R}_{\text {ham}}< \infty$ such that

$$\begin{aligned} \left\Vert {\mathcal {A}}{\mathcal {L}}_{\text {ham}}(1-\Pi _v)\right\Vert _{L^2(\rho _{\infty }) \rightarrow L^2(\rho _{\infty })} \leqslant \textsf{R}_{\text {ham}}. \end{aligned}$$

(49)

Suppose $\epsilon \in (-1, 1)$ is chosen such that $\lambda _{\text {DMS}} = \lambda _{\text {DMS}} (\gamma , m, \textsf{R}_{\text {ham}}, \epsilon ) > 0$, where

$$\begin{aligned} \lambda _{\text {DMS}} {:}{=} \frac{\gamma - \frac{\epsilon }{1 + m} - \sqrt{\epsilon ^2 (\textsf{R}_{\text {ham}}+ \frac{\gamma }{2})^2 + \left( \gamma - \frac{2 m + 1}{m+1}\epsilon \right) ^2}}{2(1+|\epsilon |)}. \end{aligned}$$

(50)

Then for any solution f(t, x, v) of (4) with $\int f_0\ \,\textrm{d}\rho _{\infty }= 0$, we have

$$\begin{aligned} \left\Vert f(t, \cdot , \cdot )\right\Vert _{L^2(\rho _{\infty })} \leqslant \sqrt{\frac{1+|\epsilon |}{1-|\epsilon |}} \left\Vert f_0\right\Vert _{L^2(\rho _{\infty })} e^{- \lambda _{\text {DMS}}\ t}. \end{aligned}$$

Notice that when $\epsilon = 0$, the rate $\lambda _{\text {DMS}} = 0$, which reduces to the conclusion that $\left\Vert f(t, \cdot , \cdot )\right\Vert _{L^2(\rho _{\infty })}$ is non-increasing in time t. The existence of $\textsf{R}_{\text {ham}}$ has been studied under fairly general assumptions on the potential U(x) in [19, Sec. 2]. In the Proposition B.1 below, we provide a simpler estimation of $\textsf{R}_{\text {ham}}$ only under the assumption of lower bound on Hessian; see the Appendix B.3 for its proof. The first part of the proof is the same as [19, Lemma 4]; the simplicity in our approach comes from the application of Bochner’s formula. It is interesting to observe that $\textsf{R}_{\text {ham}}$ does not depend on m when U is an isotropic quadratic potential.

Proposition B.1

Assume there exists $K \in {\mathbb {R}}$ such that $\nabla _x^2 U \geqslant -K\, \textrm{Id}$ for all $x \in {\mathbb {R}}^d$, then we can choose

$$\begin{aligned} \textsf{R}_{\text {ham}}= \sqrt{\max \{K, 2\}}. \end{aligned}$$

(51)

such that (49) is satisfied.

For the isotropic case $U(x) = \frac{m}{2}|x|^2$, we have

$$\begin{aligned} \left\Vert {\mathcal {A}}{\mathcal {L}}_{\text {ham}}(1-\Pi _v)\right\Vert _{L^2(\rho _{\infty }) \rightarrow L^2(\rho _{\infty })} = \sqrt{2}. \end{aligned}$$

Thus the optimal choice of $\textsf{R}_{\text {ham}}$ is $\sqrt{2}$ and (51) is tight in this case.

As an immediate consequence, if it holds that $\nabla _x^2 U \geqslant -2\, \textrm{Id}$, we can take $\textsf{R}_{\text {ham}}= \sqrt{2}$, which is tight for the isotropic case.

1.2 Asymptotic Analysis of the Decay Rate

In this subsection, we shall assume that $\nabla _x^2 U \geqslant -2\, \textrm{Id}$, thus we can choose $\textsf{R}_{\text {ham}}= \sqrt{2}$, according to the Proposition B.1. To remove the dependence on the parameter $\epsilon$ and to find the optimal decay rate, let us introduce

$$\begin{aligned} \begin{aligned} \Lambda _{\text {DMS}}(\gamma , m)&{:}{=} \sup _{\epsilon \in (-1, 1)}\ \lambda _{\text {DMS}}(\gamma , m, \sqrt{2}, \epsilon )\\&= \sup _{\epsilon \in (-1, 1)} \frac{\gamma - \frac{\epsilon }{1 + m} - \sqrt{\epsilon ^2 (\sqrt{2} + \frac{\gamma }{2})^2 + \left( \gamma - \frac{2 m + 1}{m+1}\epsilon \right) ^2}}{2(1+|\epsilon |)}, \end{aligned} \end{aligned}$$

(52)

provided that the supremum is not achieved at the boundary i.e., $\epsilon = 1^{-}$ or $\epsilon = (-1)^{+}$. Observe that

• When $\epsilon = 0$, $\lambda _{\text {DMS}}(\gamma , m, \sqrt{2}, 0) = 0$;

• When $\epsilon = (-1)^{+}$, $\lambda _{\text {DMS}}(\gamma , m, \sqrt{2}, (-1)^{+}) < 0$.

Therefore, the supremum can only be achieved at $\epsilon = 1^{-}$, or the critical points of the expression on the right hand side of (52). In general, it is hard to obtain a simple explicit expression of $\Lambda _{\text {DMS}}(\gamma , m)$. Therefore, we shall consider the following asymptotic regions:

Proposition B.2

• (i) For fixed $m = O(1)$, we have

  $$\begin{aligned} \Lambda _{\text {DMS}}(\gamma , m) = \left\{ \begin{aligned} \left( \frac{-(1+m) \sqrt{3m^2+4m+1} + 3m^2+3m+1}{6m^2+8m+3}\right) \gamma +O(\gamma ^2),&\qquad \text{ when } \ \gamma \rightarrow 0;\\ \frac{4m^2}{(1+m)^2 }\gamma ^{-1} + O(\gamma ^{-2}),&\qquad \text{ when } \ \gamma \rightarrow \infty .\\ \end{aligned} \right. \end{aligned}$$

  (53)

• (ii) Consider coupled asymptotic regime $\gamma = b\sqrt{m}$ (or equivalently $m =\left( \gamma /b\right) ^2$) for some $b = O(1)$, we have

  $$\begin{aligned} \Lambda _{\text {DMS}}(\gamma , m) = \left\{ \begin{aligned}&\frac{\gamma ^5}{2b^4} + O(\gamma ^6),&\qquad \text { when } \gamma \rightarrow 0;\\ \frac{4}{\gamma } + O(\gamma ^{-2}),&\qquad \text { when } \gamma \rightarrow \infty .\\ \end{aligned}\right. \end{aligned}$$

  (54)

The proof can be found in Appendix B.3. The scaling in the first case is already known in e.g., [17, 26, 47]; in the above proposition, we simply explicitly calculate the leading order term. The second case is relevant when we choose $\gamma$ to optimize the convergence rate according to m and for the regime $m\rightarrow 0$.

1.3 Proofs of the Propositions in Appendix

Proof of Proposition B.1

We first consider the case that Hessian is bounded from below. It is equivalent to consider the operator norm of

$$\begin{aligned} -(1-\Pi _v) {\mathcal {L}}_{\text {ham}}{\mathcal {A}}^* = -(1-\Pi _v) {\mathcal {L}}_{\text {ham}}^2 \Pi _v \left( 1 + ({\mathcal {L}}_{\text {ham}}\Pi _v)^* ({\mathcal {L}}_{\text {ham}}\Pi _v)\right) ^{-1}. \end{aligned}$$

Notice that this operator is supported on $\text {Ran}(\Pi _v)$ from the observation that ${\mathcal {A}}= \Pi _v {\mathcal {A}}$, it is then equivalent to find the smallest $\textsf{R}_{\text {ham}}$ such that for any $\phi (x,v)$ with $\Pi _v \phi = \phi$ (i.e., $\phi (x,v) \equiv \phi (x)$ is a function of x only), we have

$$\begin{aligned} \left\Vert -(1-\Pi _v) {\mathcal {L}}_{\text {ham}}{\mathcal {A}}^* \phi \right\Vert _{L^2(\rho _{\infty })}\leqslant \textsf{R}_{\text {ham}}\left\Vert \phi \right\Vert _{L^2(\rho _{\infty })} = \textsf{R}_{\text {ham}}\left\Vert \phi \right\Vert _{L^2(\mu )}. \end{aligned}$$

(55)

Given such a function $\phi$ with $\Pi _v \phi = \phi$, define

$$\begin{aligned}\varphi {:}{=} \left( 1 + ({\mathcal {L}}_{\text {ham}}\Pi _v)^* ({\mathcal {L}}_{\text {ham}}\Pi _v)\right) ^{-1} \phi .\end{aligned}$$

It is easy to check that $\Pi _v \varphi = \varphi$. By simplifying the above equation with (5) and (9),

$$\begin{aligned} \phi (x) = \varphi (x) - \Delta _x \varphi (x) + \nabla _x U(x) \cdot \nabla _x \varphi = \varphi (x) + \nabla _x^* \nabla _x \varphi (x). \end{aligned}$$

(56)

Furthermore, by some straightforward calculation, we have

$$\begin{aligned} -(1-\Pi _v) {\mathcal {L}}_{\text {ham}}{\mathcal {A}}^* \phi = - (1-\Pi _v) {\mathcal {L}}_{\text {ham}}^2 \Pi _v \varphi = - \sum _{i,j} (v_i v_j - \delta _{i,j}) \partial _{x_i, x_j} \varphi . \end{aligned}$$

Thus

$$\begin{aligned} \left\Vert -(1-\Pi _v) {\mathcal {L}}_{\text {ham}}{\mathcal {A}}^* \phi \right\Vert _{L^2(\rho _{\infty })}^2&= \int \left( \sum _{i,j} (v_i v_j - \delta _{i,j}) \partial _{x_i, x_j} \varphi \right) ^2\,\textrm{d}\rho _{\infty }\\&= 2 \sum _{i,j} \int \left( \partial _{x_i, x_j} {\varphi }\right) ^2\,\textrm{d}\mu . \end{aligned}$$

Then, by Bochner’s formula,

$$\begin{aligned} \Vert -(1-\Pi _v) {\mathcal {L}}_{\text {ham}}&{\mathcal {A}}^* (\phi ) \Vert _{L^2(\rho _{\infty })}^2 \\ =&\ 2 \int \nabla _x \varphi \cdot \nabla _x \nabla _x^* \nabla _x {\varphi } - \nabla _x {\varphi } \cdot \nabla _x^2 U \nabla _x {\varphi } - \nabla _x^* \nabla _x \left( \frac{|\nabla _x {\varphi }|^2}{2}\right) \,\textrm{d}\mu \\ =&\ 2 \int |\nabla _x^*\nabla _x {\varphi }|^2 - \nabla _x {\varphi } \cdot \nabla _x^2 U \nabla _x {\varphi }\,\textrm{d}\mu \\ \leqslant&\ 2 \left( \int |\nabla _x^*\nabla _x {\varphi }|^2\,\textrm{d}\mu + K \int |\nabla _x {\varphi }|^2\,\textrm{d}\mu \right) \\ \leqslant&\ \max \left\{ K, 2\right\} \left( \int |\nabla _x^* \nabla _x {\varphi }|^2\,\textrm{d}\mu + 2 \int |\nabla _x {\varphi }|^2\,\textrm{d}\mu \right) . \end{aligned}$$

From (56), we have

$$\begin{aligned} \left\Vert {\phi }\right\Vert _{L^2(\mu )}^2&= \int {\varphi }^2 + 2 {\varphi }\ \nabla _x^* \nabla _x {\varphi } + \left|\nabla _x^*\nabla _x {\varphi }\right|^2\ \textrm{d}\mu \\&\geqslant 2 \int |\nabla _x{\varphi }|^2\,\textrm{d}\mu + \int |\nabla _x^* \nabla _x {\varphi }|^2\,\textrm{d}\mu . \end{aligned}$$

By combining the last two equations,

$$\begin{aligned} \left\Vert -(1-\Pi _v) {\mathcal {L}}_{\text {ham}}{\mathcal {A}}^* (\phi )\right\Vert _{L^2(\rho _{\infty })}^2&\leqslant \max \{K, 2\} \left\Vert {\phi }\right\Vert _{L^2(\mu )}^2, \end{aligned}$$

which yields (51).

We now consider the isotropic case. Recall that the operator norm of ${\mathcal {A}}{\mathcal {L}}_{\text {ham}}(1-\Pi _v)$ is the smallest $\textsf{R}_{\text {ham}}$ such that (55) holds. Let us consider the elliptic PDE (56). By the choice $U(x) = \frac{m}{2}|x|^2$,

$$\begin{aligned} {\phi }(x) = \left( 1 + m (x - \frac{1}{m} \nabla _x) \cdot \nabla _x \right) {\varphi }(x). \end{aligned}$$

Then by rescaling the variable $x = \frac{y}{\sqrt{m}}$ and rescaling the functions $\bar{\phi }(y) {:}{=} {\phi }(x) = \phi \left( \frac{y}{\sqrt{m}}\right)$, $\bar{\varphi }(y) {:}{=} {\varphi }(x) = {\varphi }\left( \frac{y}{\sqrt{m}}\right)$, we have

$$\begin{aligned} \bar{\phi }(y) = \left( 1 + m (y - \nabla _y) \cdot \nabla _y\right) \bar{\varphi }(y). \end{aligned}$$

(57)

In addition, by rewriting (55), we need to find the smallest $\textsf{R}_{\text {ham}}$ such that

$$\begin{aligned} 2 m^2 \sum _{i,j} \int \left|\partial _{y_i, y_j} \bar{\varphi }(y)\right|^2\ e^{-\frac{|y|^2}{2}}\,\textrm{d}y\leqslant \textsf{R}_{\text {ham}}^2 \int \bar{\phi }(y)^2 e^{-\frac{|y|^2}{2}}\,\textrm{d}y. \end{aligned}$$

(58)

Next, let us expand the last equation by probabilists’ Hermite polynomials $H_k(z) {:}{=} (z - \frac{\textrm{d}}{\textrm{d}z})^k \cdot 1$ for integers $k\geqslant 0$. Recall two important properties

$$\begin{aligned} H'_k(z) = k H_{k-1}(z), \qquad \frac{1}{\sqrt{2\pi }}\int H_j(z) H_k(z) e^{-\frac{z^2}{2}}\,\textrm{d}z = k!\, \delta _{j,k}. \end{aligned}$$

Given $\varvec{n} = (n_1, n_2, \cdots , n_d)$, define

$$\begin{aligned} H_{\varvec{n}}(y) {:}{=} H_{n_1}(y_1) H_{n_2}(y_2) \cdots H_{n_d}(y_d). \end{aligned}$$

By the above properties, it is easy to show that if $\bar{\varphi } = H_{\varvec{n}}$, then $\bar{\phi } = N_{\varvec{n}} H_{\varvec{n}}$, where $N_{\varvec{n}} {:}{=} 1 + m \sum _{i} n_i$. Thus if $\bar{\varphi }(y) = \sum _{\varvec{n}} a_{\varvec{n}}H_{\varvec{n}}$, then we have $\bar{\phi } = \sum _{n} a_{\varvec{n}} N_{\varvec{n}} H_{\varvec{n}}$. By such an expansion, (58) can be rewritten as

$$\begin{aligned} 2m^2 \sum _{i, j} \sum _{\varvec{n}} a_{\varvec{n}}^2 (n_i n_j - \delta _{i,j} n_i) \prod _{k=1}^{d} n_k! \leqslant \textsf{R}_{\text {ham}}^2 \sum _{\varvec{n}} a_{\varvec{n}}^2 N_{\varvec{n}}^2 \prod _{k=1}^{d} n_k! \end{aligned}$$

Then finding the operator norm of ${\mathcal {A}}{\mathcal {L}}_{\text {ham}}(1-\Pi _v)$ is equivalent to finding the smallest $\textsf{R}_{\text {ham}}$ such that for any $\varvec{n}$, one has

$$\begin{aligned} \sum _{i,j} (n_i n_j - \delta _{i,j} n_i) \leqslant \frac{\textsf{R}_{\text {ham}}^2}{2 m^2 } N_{\varvec{n}}^2 \equiv \frac{\textsf{R}_{\text {ham}}^2}{2 m^2 } \left( 1 + m \sum _i n_i\right) ^2. \end{aligned}$$

When $n_1 \rightarrow \infty$ and $n_2, n_3, \cdots , n_d = 0$, we know that $\frac{\textsf{R}_{\text {ham}}^2}{2} \geqslant 1$. Also observe that

$$\begin{aligned} \sum _{i,j} (n_i n_j - \delta _{i,j} n_i) \leqslant \left( \sum _{i} n_i\right) ^2 = \frac{1}{m^2} \left( m \sum _i n_i\right) ^2\leqslant \frac{1}{m^2} \left( 1 + m \sum _i n_i\right) ^2. \end{aligned}$$

Therefore, $\frac{\textsf{R}_{\text {ham}}^2}{2} = 1$ is sufficient.

In summary, $\left\Vert {\mathcal {A}}{\mathcal {L}}_{\text {ham}}(1-\Pi _v)\right\Vert _{L^2(\rho _{\infty }) \rightarrow L^2(\rho _{\infty })} = \sqrt{2}$ and the optimal choice of $\textsf{R}_{\text {ham}}$ is $\sqrt{2}$. $\quad \square$

Proof of Proposition B.2

We used Maple software to help verify the asymptotic expansion.

Part (i): $m = O(1)$.

• (when $\gamma \rightarrow 0$). Via asymptotic expansion, we have

  $$\begin{aligned} \lambda _{\text {DMS}}(\gamma , m, \sqrt{2}, 1^{-}) = - \frac{1+\sqrt{6 m^2 + 8m + 3}}{4(1+m)} + O(\gamma ) < 0. \end{aligned}$$

  Thus the supremum is not obtained at $\epsilon = 1^{-}$. Then let us consider critical points within the domain $(-1, 1)$, whose asymptotic expansions are

  $$\begin{aligned} \epsilon _{\pm } = \frac{(6m^2+5m+1 \pm \sqrt{3m^2+4m+1})(1+m)}{18m^3+30m^2+17m+3} \gamma + O(\gamma ^2) > 0. \end{aligned}$$

  After comparison, the larger decay rate is obtained at $\epsilon _{-}$ with the value in (53).

• (when $\gamma \rightarrow \infty$). Similarly, via asymptotic expansion, we have

  $$\begin{aligned} \lambda _{\text {DMS}}(\gamma , m, \sqrt{2}, 1^{-}) = -\frac{\frac{\sqrt{5}}{2}-1}{4}\gamma + O(1) < 0. \end{aligned}$$

  Thus we need to consider the critical points. It turns out, there is only one critical point within the domain $(-1, 1)$, which is $\varepsilon = \frac{8m}{1+m}\gamma ^{-1} + O(\gamma ^{-2})$ with the decay rate in (53).

Part (ii): $\gamma = b\sqrt{m}$ with $b=O(1)$.

• (when $\gamma \rightarrow 0$). Via asymptotic expansion, one could check that

  $$\begin{aligned} \lambda _{\text {DMS}}(\gamma , m=(\gamma /b)^2, \sqrt{2}, 1^{-}) = -\frac{1+\sqrt{3}}{4} + O(\gamma ) < 0. \end{aligned}$$

  Thus, we only need to consider the decay rate at critical points, which are given by

  $$\begin{aligned} \epsilon _1 = \frac{\gamma ^3}{b^2} + O(\gamma ^4), \qquad \epsilon _2 = \frac{2}{3} \gamma + O(\gamma ^2). \end{aligned}$$

  and the associated decay rates are

  $$\begin{aligned} \lambda _{\text {DMS}}(\gamma , m=(\gamma /b)^2, \sqrt{2}, \epsilon _1)&= \frac{\gamma ^5}{2b^4} + O(\gamma ^{6})> 0;\\ \lambda _{\text {DMS}}(\gamma , m=(\gamma /b)^2, \sqrt{2}, \epsilon _2)&= -\frac{1}{3} \gamma + O(\gamma ^2) < 0. \end{aligned}$$

  Therefore, the optimal decay rate is obtained at $\epsilon _1$, which gives (54).

• (when $\gamma \rightarrow \infty$). Via asymptotic expansion, one could obtain

  $$\begin{aligned} \lambda _{\text {DMS}}(\gamma , m=(\gamma /b)^2, \sqrt{2}, 1^{-}) = -\frac{\sqrt{5}-2}{8}\gamma + O(1) < 0. \end{aligned}$$

  Thus the supremum in (52) cannot be obtained at $\epsilon = 1^{-}$. Then, let us look at the critical points. It turns out there is only one within the interval $(-1, 1)$, which is $\epsilon _1 = \frac{8}{\gamma } + O(\gamma ^{-2})$. The optimal decay rate must be achieved at $\epsilon _1$, with the expression given in (54). $\quad \square$