Structural Stability of the RG Flow in the Gross–Neveu Model

Abstract

We study flow of renormalization group (RG) transformations for the massless Gross–Neveu model in a non-perturbative formulation. The model is defined on a two-dimensional Euclidean space with a finite volume. The quadratic approximation to the flow stays bounded after suitable renormalization. We show that for weak coupling this property also is true for the complete flow. As an application we prove an ultraviolet stability bound for the model. Our treatment is an application of a method of Bauerschmidt, Brydges, and Slade. The method was developed for an infrared problem and is now applied to an ultraviolet problem.

Appendices

A Fourier Series Conventions

We consider functions on the d-dimensional torus ${\mathbb {T}}_j \equiv {\mathbb {R}}^d/L^j {\mathbb {Z}}^d$. If $p \in {\mathbb {T}}_j^* \equiv 2\pi L^{-j} {\mathbb {Z}}^d$ then $e^{ipx}$ is invariant under $x_{\mu } \rightarrow x_{\mu } + L^j$ and so defines a function on the torus. We have $\int e^{-ipx} e^{iqx} \textrm{d}x = L^{dj} \delta _{p,q}$ so $e_p(x) = L^{-\frac{1}{2} dj} e^{ipx}$ form an orthonormal basis for $L^2({\mathbb {T}}_j)$ and there is the expansion

$$\begin{aligned} f(x) = \sum _{p \in {\mathbb {T}}_j^*} (e_p,f ) e_p(x) = \sum _{p \in {\mathbb {T}}_j^*} L^{-dj} \Big [ \int e^{-ipx} f(x) \textrm{d}x \Big ] e^{ipx} \end{aligned}$$

(464)

We write this as

$$\begin{aligned} f(x) = \sum _{p \in {\mathbb {T}}_j^*} L^{-dj} {\tilde{f}}( p) e^{ipx} \equiv {\sum \limits _{p \in {\mathbb {T}}_j^*}}^{\prime } {\tilde{f}}(p) e^{ipx} \end{aligned}$$

(465)

where

$$\begin{aligned} {\tilde{f}}(p) = \int e^{-ipx} f(x) \textrm{d}x \end{aligned}$$

(466)

We also have $\int {\overline{f}}(x) g(x) \textrm{d}x = \sum _p (f,e_p)(e_p, g)$ which we write as

$$\begin{aligned} \int \overline{ {\bar{f}}(x)} g(x) \textrm{d}x = \sum _{p \in {\mathbb {T}}_j^*} L^{-dj} \overline{ {\tilde{f}}(p) } {\tilde{g}}(p) \equiv {\sum \limits _{p \in {\mathbb {T}}_{j}^{*}}}^{\prime } \overline{ {\tilde{f}}(p) } {\tilde{g}}(p) \end{aligned}$$

(467)

B Characteristic Functions of Grassmann Integrals

Let ${{\mathcal {G}}}_h$ be the Grassmann algebra on some ${\mathbb {T}}_j$ as defined in Sect. 2.3. We consider integrals $\int [ \cdots ] \textrm{d}\mu$ on ${{\mathcal {G}}}_h$ defined to be continuous linear functionals on ${{\mathcal {G}}}_h$. For F of the form (46) we have for some constant c

$$\begin{aligned} \left| \int F \textrm{d} \mu \right| \le c\Vert F \Vert _h \end{aligned}$$

(468)

If $J \in {{\mathcal {C}}}'( {\mathbb {T}}_j \times {\mathbb {T}}_j)$ and $<\psi , J \bar{\psi }> = \int \psi (x) J(x,y) \bar{\psi }(y) \textrm{d}x \textrm{d}y$ then

$$\begin{aligned} \Vert <\psi , J \bar{\psi }> \Vert _h = h^2 \Vert J \Vert _{{{\mathcal {C}}}'} \end{aligned}$$

(469)

and

$$\begin{aligned} \begin{aligned}&\left| \int e^ {<\psi , J \bar{\psi }> } \textrm{d} \mu (\psi )\right| \le c \Vert e^ {<\psi , J \bar{\psi }> } \Vert _{h} \le c \exp \Big ( \Vert <\psi , J \bar{\psi }> \Vert _{h }\Big ) \le ce^{h^2 \Vert J\Vert _{{{\mathcal {C}}}'} } \end{aligned} \end{aligned}$$

(470)

This $\int e^ { <\psi , J \bar{\psi }> } \textrm{d} \mu (\psi )$ is defined to be the characteristic function of the integral. This generalizes the finite-dimensional case, see [16].

In particular let $\int [ \cdots ] \textrm{d}\mu _G$ be a Gaussian integral with covariance G of the form

$$\begin{aligned} G(x-y) = \int G_1(x-z) G_2(z-y) \textrm{d}z \end{aligned}$$

(471)

with $G_1, G_2 \in {{\mathcal {C}}}^3({\mathbb {T}}_j)$ so that h(G) defined in (105) is finite. In particular we could take $G=G_k$ defined in (13) or $G=G_k^z$ defined in (146). Then as in Lemma 7 the integral $\int F \textrm{d} \mu _{G}$ is defined on ${{\mathcal {G}}}_{h(C)}$ and $| \int F \textrm{d} \mu _{G} | \le \Vert F\Vert _{h(G)}$. If $h(G) < h$ then $\Vert F\Vert _{h(G)} \le \Vert F\Vert _h$ and ${{\mathcal {G}}}_h \subset {{\mathcal {G}}}_{h(G)}$ and so the Gaussian integral is defined on ${{\mathcal {G}}}_h$.

Lemma 23

Let G, h satisfy the above conditions An integral $\int [ \cdots ] \textrm{d} \mu$ on the Grassmann algebra ${{\mathcal {G}}}_h$ is Gaussian with covariance G if and only if the characteristic function satisfies

$$\begin{aligned} \int e^{ <\psi , J \bar{\psi }> } \textrm{d} \mu = \det ( I + J^T G) \end{aligned}$$

(472)

for all smooth functions J.

Remarks

The determinant is the Fredholm determinant, see Simon [27]. This can also be written

$$\begin{aligned} \int e^{ - <\bar{\psi }, J \psi > } \textrm{d} \mu = \det ( I + J G) \end{aligned}$$

(473)

Proof

First we note that $J^TG$ is trace class so the determinant is well defined. In fact G alone is trace class since $Gf = G_1 * G_2* f$ and each of these convolutions is Hilbert–Schmidt. Indeed the kernel of $f \rightarrow G_i*f$ has Hilbert–Schmidt norm squared

$$\begin{aligned} \int |G_i(x-y)|^2 \textrm{d}x \textrm{d}y = \Vert G_i\Vert ^2_2 |{\mathbb {T}}_j| < \infty \end{aligned}$$

(474)

Now we compute

$$\begin{aligned} \begin{aligned} \int e^{ <\psi , J \bar{\psi }> } \textrm{d} \mu _{G}&= \sum _{n=0} ^{\infty } \frac{1}{n!} \int J( x_1, y_1 ) \cdots J( x_n, y_n ) \det \Big \{ G (x_i,y_i) \Big \} \textrm{d}x \textrm{d}y\\&= \sum _{n=0} ^{\infty } \frac{1}{n!} \int \det \Big \{ (J^T G ) (x_i,x_j) \Big \} \textrm{d}x \\&= \det ( I + J^T G ) \end{aligned} \end{aligned}$$

(475)

The last step is the Fredholm formula [27].

For the converse suppose that $\int e^{ <\psi , J \bar{\psi }>} d \mu = \det ( I + J^T G)$. We compare the terms $n^{th}$ order in J and find

$$\begin{aligned} \begin{aligned}&\int \Big [ \int \psi (x_1) \bar{\psi }(y_1) \cdots \psi (x_n) \bar{\psi }(y_n) \ \textrm{d} \mu \Big ] \prod _{i=1}^n J(x_i,y_i ) \textrm{d}x \textrm{d}y\\&\quad = \int \det \Big \{ G (x_i,y_j) \Big \} \prod _{i=1}^n J(x_i,y_i) \textrm{d}x \textrm{d}y \end{aligned} \end{aligned}$$

(476)

Both sides of this equation are homogeneous functions of J of degree n, that is, $T(aJ) = a^n T(J)$. It follows by the polarization formula that they determine multilinear functions of J. Thus in this equation we can replace $\prod _{i=1}^n J(x_i,y_j )$ by $\prod _{i=1}^n J_i(x_i,y_j )$. Furthermore we can specialize and take $J_i(x_i,y_i) = f_i(x_i) g_i(y_i)$ with $f_i,g_i \in {{\mathcal {C}}}^{\infty } ({\mathbb {T}})$. Thus we have an identity between multilinear functions on ${{\mathcal {C}}}^{\infty } ({\mathbb {T}})$. By the kernel theorem the associated kernels are equal as distributions and hence as functions. Thus

$$\begin{aligned} \int \psi (x_1) \bar{\psi }(y_1) \cdots \psi (x_n) \bar{\psi }(y_n) \ \textrm{d} \mu =\det \Big \{ G (x_i,y_j) \Big \} \end{aligned}$$

(477)

which is our result. $\square$

Lemma 24

Let $G_k$ on ${\mathbb {T}}_{N+M-k}$ be as in the text.

• 1. There is $\varepsilon '_k$ with $|\varepsilon '_k| \le {{\mathcal {O}}}(1)|z_k|$ such that

  

  (478)

• 2. There is an identity between Gaussian integrals

  

  (479)

  where

  

  (480)

Proof

Start with the estimate

(481)

Then the integral in (478) is finite and by (470)

(482)

The first identity in (478) is (473) which still holds for the more singular . Note in this case is still the kernel of a trace class operator. For the second identity in (478) we note that

(483)

The operator has eigenvalues $\{ z_k e^{-p^2} \}$ with multiplicity 2n for each $p \in {\mathbb {T}}^*_{M+N-k}$. Hence

(484)

But for $z_k$ small $|\log ( 1 + z_k e^{-p^2} )| \le {{\mathcal {O}}}(1)|z_k| e^{-p^2}$ which gives $|\varepsilon '_k| \le {{\mathcal {O}}}(1)|z_k|$. So now part 1 is established.

For part 2 both sides define integrals on ${{\mathcal {G}}}_h$ and it suffices to check that they have the same characteristic function in the form (473). On the one hand as in Lemma 23 we have the characteristic function of the right side of (479)

$$\begin{aligned} \int e^{-< \bar{\psi }, J \psi > } \textrm{d} \mu _{G^z_k } = \det ( I + J G^z_k ) \end{aligned}$$

(485)

On the other hand we have similarly

(486)

The characteristic function of the left side of (479) is this divided by or multiplied by which is

(487)

But implies . Thus the last determinant is again $\det ( I + J G^z_k )$. The characteristic functions are equal. $\square$

C Gamma Matrices

Let $\gamma _0, \gamma _1$ be $2 \times 2$ traceless matrices satisfying $\{ \gamma _{\mu }, \gamma _{\nu } \} = 2 \delta _{\mu \nu }$. If R is a reflection through a line $x_{\mu } =c$ or a rotation by a multiple of $\pi /2$ then there is a matrix S in the group Pin(2) such that $S^{-1} \gamma _{\mu } S = \sum _{\nu } R_{\mu \nu } \gamma _{\nu }$. (Actually there are two such.) We also have

$$\begin{aligned} \gamma _5 = i \gamma _0 \gamma _1 = \frac{i}{2} \sum _{\mu \nu }\epsilon _{\mu \nu } \gamma _{\mu } \gamma _{\nu } \end{aligned}$$

(488)

which satisfies $S^{-1} \gamma _5 S = ( \det R ) \gamma _5$. For the spinor group Pin(n) and generalizations see for example [10] or [11].

We relabel $( I, \gamma _0, \gamma _ 1, \gamma _5)$ as $( \Gamma _1, \Gamma _2, \Gamma _3, \Gamma _4)$. Then the $\Gamma _i$ are traceless, self-adjoint matrices satisfying $\Gamma _i^2 = I$ and $\text { tr }( \Gamma _i \Gamma _j ) =2 \delta _{ij}$. They are linearly independent since if $\sum _i c_i \Gamma _i = 0$ then $c_j = \sum _i c_i \text { tr }( \Gamma _i \Gamma _j) =0$. They form a basis for the space $M_2$ of $2 \times 2$ complex matrices. Any matrix A can be written

$$\begin{aligned} A=\sum _i c_i \Gamma _i \quad \quad c_i = \frac{1}{2} \text { tr }( A \Gamma _i) \end{aligned}$$

(489)

Lemma 25

• 1. If $\alpha$ in $M_2$ satisfies $S^{-1} \alpha S = \alpha$ for all S, then there is a constant c such that $\alpha = c I$.

• 2. If $\alpha _{\mu } \in M_2$ satisfies $S^{-1} \alpha _{\mu } S=\sum _{\nu } R_{\mu \nu } \alpha _{\nu }$ for all S, then there is a constant c such that $\alpha _{\mu } = c \gamma _{\mu }$.

Proof

For the first item we have

$$\begin{aligned} \begin{aligned} \text { tr }( \alpha \gamma _{\mu } )&= \text { tr }(S^{-1} \alpha S S^{-1} \gamma _{\mu } S ) = \sum _{\nu } R _{\mu \nu } \text { tr }( \alpha \gamma _{\nu } ) \\ \text { tr }( \alpha \gamma _5 )&= \text { tr }(S^{-1} \alpha SS^{- 1} \gamma _5 S) = \det R \text { tr }( \alpha \gamma _5) \end{aligned} \end{aligned}$$

(490)

In all cases we can pick an r such that the right side is minus the left side and hence the expression is zero. For example for $\text { tr }( \alpha \gamma _0)$ the reflection $x_0 \rightarrow -x_0,x_1 \rightarrow x_1$ has this property. Thus in the expansion in our basis only multiples of the identity survive.

For the second item we have similarly

$$\begin{aligned} \begin{aligned} \text { tr }( \alpha _{\mu } )&= \sum _{\nu } R_{\mu \nu } \text { tr }( \alpha _{\nu } ) \\ \text { tr }( \alpha _{\mu } \gamma _{\mu ' } )&= \text { tr }(S^{-1} \alpha _{\mu }SS^{-1} \gamma _{\mu ' } S) = \sum _{\mu , \nu } R_{\mu \nu } R_{\mu ' \nu ' } \text { tr }( \alpha _{\nu } \gamma _{\nu '} ) \\ \text { tr }( \alpha _{\mu } \gamma _5 )&= \text { tr }(S^{-1} \alpha _{\mu }S S^{-1} \gamma _5 S ) = \sum _{ \nu } r_{\mu \nu } \det R \text { tr }( \alpha _{\nu } \gamma _5 ) \end{aligned} \end{aligned}$$

(491)

In the first and third case we can pick an R such that the right side is minus the left side and hence zero. This is also true for the second case if $\mu \ne \mu '$. If $\mu = \mu '$ we can choose R so that $\text { tr }( \alpha _{0}\gamma _{0} ) = \text { tr }( \alpha _2 \gamma _1)$, namely take the rotation $x_0 \rightarrow x_1, x_1 \rightarrow -x_0$. Then $\text { tr }(\alpha _{ \mu } \gamma _{\nu }) = 2c \delta _{\mu \nu }$ for some constant c and

$$\begin{aligned} \alpha _{\mu } = \sum _{\nu } \frac{1}{2} \text { tr }( \alpha _{\mu } \gamma _{\nu } )\gamma _{\nu } = c \gamma _{\mu } \end{aligned}$$

(492)

Now we consider the sixteen linear operators on ${\mathbb {C}}^2 \otimes {\mathbb {C}}^2$ of the form $\Gamma _{ij} \equiv \Gamma _i \otimes \Gamma _j$. These are self-adjoint and $\Gamma _i^2 =I \otimes I$. We also have $\text { tr }( \Gamma _{ij}) = \text { tr }( \Gamma _i \otimes \Gamma _j) = ( \text { tr }\Gamma _i )(\text { tr }\Gamma _j) =0$ and

$$\begin{aligned} \text { tr }( \Gamma _{ij}\Gamma _{i'j'} ) = \text { tr }( \Gamma _i\Gamma _{i'} \otimes \Gamma _j \Gamma _{j'} ) = \text { tr }( \Gamma _i\Gamma _{i'} ) \text { tr }( \Gamma _j \Gamma _{j'} ) = 4\delta _{ij} \delta _{i'j'} \end{aligned}$$

(493)

As before the $\Gamma _{ij}$ are a basis for the space ${{\mathcal {L}}}({\mathbb {C}}^2 \otimes {\mathbb {C}}^2)$ of linear operators on ${\mathbb {C}}^2 \otimes {\mathbb {C}}^2$, and every element A has an expansion

$$\begin{aligned} A=\sum _{ij} c_{ij} \Gamma _{ij} \quad \quad c_{ij} = \frac{1}{4} \text { tr }( A \Gamma _{ij} ) \quad \quad \end{aligned}$$

(494)

$\square$

Lemma 26

If $\alpha \in {{\mathcal {L}}}({\mathbb {C}}^2 \otimes {\mathbb {C}}^2)$ satisfies $( S^{-1} \otimes S^{-1} )\alpha ( S \otimes S ) = \alpha$ for all S, then there are constants $c, c_p, c_v$ such that

$$\begin{aligned} \alpha = c ( I \otimes I ) + c_p( \gamma _5 \otimes \gamma _5) + c_v \sum _{\mu } ( \gamma _{\mu } \otimes \gamma _{\mu } ) \end{aligned}$$

(495)

Proof

As in the previous lemma we can argue that $\text { tr }( \alpha \Gamma _{ij} ) =0$ except for $\Gamma _{ij} = I \otimes I$, $\gamma _{\mu } \otimes \gamma _{\mu }$, $\gamma _5 \otimes \gamma _5$. For example for any R

$$\begin{aligned} \text { tr }( \alpha (\gamma _{\mu } \otimes \gamma _5 ) ) = \det R \sum _{\nu } R_{\mu \nu } \text { tr }( \alpha (\gamma _{\nu } \otimes \gamma _5 ) ) \end{aligned}$$

(496)

If $\mu =0$ choose R to be $x_0 \rightarrow x_0, x_1 \rightarrow -x_1$ and conclude $\text { tr }( \alpha (\gamma _0 \otimes \gamma _5 ) )=0$. We can also argue that $\text { tr }( \alpha \gamma _{\mu } \otimes \gamma _{\mu } )$ is independent of $\mu$. Then our basic expansion (494) gives the result. $\square$

D Bounds on Propagators

Lemma 27

$$\begin{aligned} \begin{aligned} | C_k(x) |&\le {{\mathcal {O}}}(1)e^{ - |x|/L} \\ |w_k( x) |&\le {{\mathcal {O}}}(1)|x| ^{- 1} e^{-|x| } \end{aligned} \end{aligned}$$

(497)

Proof

(cf. [7]) We have

(498)

We write

$$\begin{aligned} e^{- p^2 } - e^{- L^2p^2 } = - \int _1^{L^2} \frac{\textrm{d}}{\textrm{d} \lambda } e^{-\lambda p^2} \textrm{d} \lambda = p^2 \int _1^{L^2} e^{-\lambda p^2} \textrm{d} \lambda \end{aligned}$$

(499)

which gives

(500)

Our definition of the weighted sum $\sum '_p$ explicitly excluded the $p=0$ term. Here we can assume it is restored since it gives a constant and we are taking a derivative. For the bracketed expression we use the identity

$$\begin{aligned} {\sum \limits _{p \in {\mathbb {T}}^*_{N+M-k }}}^{\!\!\!\!\!\!\!\!\!\prime } e^{ipx} e^{-\lambda p^2} = \frac{1}{4 \pi \lambda } \sum _{y \in L^{N+M -k } {\mathbb {Z}}^2} e^{ - |x-y|^2 / 4 \lambda } \end{aligned}$$

(501)

Both sides are periodic with period $L^{N+M -k }$ and so can be regarded as a functions on the torus ${\mathbb {T}}_{M+N -k}$. This is a special case of the Poisson summation formula. If we let $\lambda = kt$ then can be understood as two different representations of the fundamental solutions for the heat operator $\partial /\partial t - k \Delta$, either by Fourier series on the left or by Fourier transform on ${\mathbb {R}}^2$ followed by periodizing on the right.

With this identity we have

(502)

We work on the bracketed expression in the last equation. Since $\alpha ^2 - 4 \alpha \ge -4$ we have

$$\begin{aligned} \frac{ |x-y|^2 }{ 4 \lambda } - \frac{ 2 |x-y| }{ \sqrt{ \lambda }} \ge - 4 \end{aligned}$$

(503)

Therefore

$$\begin{aligned} \int _1^{L^2} \frac{ \textrm{d} \lambda }{ \lambda ^2} \ e^{ - |x-y|^2 / 4 \lambda } \le {{\mathcal {O}}}(1)\int _1^{L^2} \frac{ \textrm{d} \lambda }{ \lambda ^2} \ e^{ -2 |x-y| / \sqrt{\lambda }} \le {{\mathcal {O}}}(1)e^{-2 |x-y|/L} \end{aligned}$$

(504)

and so for $- \frac{1}{2} L^{N+M-k} \le x_{\mu } \le \frac{1}{2} L^{N+M-k}$

$$\begin{aligned} | C_k(x) | \le {{\mathcal {O}}}(1)\sum _{y \in L^{N+M -k } {\mathbb {Z}}^2} | x-y | e^{-2 |x-y|/L} \le {{\mathcal {O}}}(1)e^{- |x|/L} \end{aligned}$$

(505)

Now we turn to the bound on $w_k(x)$ given by

(506)

Following the same steps as before we have instead of (502)

(507)

Now in the bracketed expression

$$\begin{aligned} \frac{ |x-y|^2 }{4 \lambda } \ge \frac{3}{2} |x-y| + \frac{|x-y|}{ 2\sqrt{ \lambda } } - 4 \end{aligned}$$

(508)

This follows from $\lambda \le 1$ and $\alpha ^2 - 3 \sqrt{ \lambda } \alpha -\alpha \ge \alpha ^2 - 4 \alpha \ge - 4$. Therefore

$$\begin{aligned} \begin{aligned} \int _{L^{-2k} }^{1} \frac{ \textrm{d} \lambda }{ \lambda ^2} \ e^{ - |x-y|^2 / 4 \lambda }&\le {{\mathcal {O}}}(1)e^{ - \frac{3}{2} |x-y| } \int _{L^{-2k} }^{1} \frac{ \textrm{d} \lambda }{ \lambda ^2} \ e^{ - |x-y|/2\sqrt{ \lambda } } \\&\le {{\mathcal {O}}}(1)e^{ - \frac{3}{2} |x-y| } \int _0^{\infty } \frac{ \textrm{d} \lambda }{ \lambda ^2} \ e^{ - |x-y|/2\sqrt{ \lambda } } \\&= {{\mathcal {O}}}(1)e^{ - \frac{3}{2} |x-y| } \frac{1}{ |x-y|^2 } \int _0^{\infty } \xi e^{-\xi } \textrm{d} \xi \\&= {{\mathcal {O}}}(1)e^{ - \frac{3}{2} |x-y| } \frac{1}{ |x-y|^2 } \end{aligned} \end{aligned}$$

(509)

This yields

$$\begin{aligned} |w_k(x)| \le {{\mathcal {O}}}(1)\sum _{y \in L^{N+M -k } {\mathbb {Z}}^2} |x-y|^{-1} e^{- \frac{3}{2} |x-y| } \le {{\mathcal {O}}}(1)|x|^{-1} e^{-|x|} \end{aligned}$$

(510)

$\square$

Lemma 28

For $|z_k|$ sufficiently small

$$\begin{aligned} | C^z_k(x) - C_k(x) | \le {{\mathcal {O}}}(1) |z_k| e^{ - |x|/L } \end{aligned}$$

(511)

Proof

We have with the sum over $p \in {\mathbb {T}}^*_{N+M-k}$

(512)

The $C_k(x)$ is the case $n=m=0$ so for the difference we can restriction the sum to $n+m \ge 1$. We use the identity

$$\begin{aligned} e^{-(n+L^2m+1) p^2} - e^{-(n+L^2m+L^2) p^2} =p^2 \int _{n+L^2m +1} ^{n + L^2 m + L^2} e^{-\lambda p^2} \textrm{d} \lambda \end{aligned}$$

(513)

and the Poisson summation formula (501) to obtain

(514)

where the sums are over $y \in L^{N+M -k} {\mathbb {Z}}^2$. We enlarge the upper limit of the $\lambda$ integral to $L^2(n+m+1)$ and then

$$\begin{aligned} | C^z_k (x) - C_k(x) | \le {{\mathcal {O}}}(1)\sum _{n,m \ge 1} |z_k|^{n+m} \sum _{y} |x-y| \ \ e^{ - |x-y|^2 / 4L^2 (n+m+1) }\nonumber \\ \end{aligned}$$

(515)

Now let $\ell = m+n \ge 1$. If $\ell =1$ we use the inequality (503) with $\lambda = 2L^2$ to get $e^{ - |x-y|^2 / 8\,L^2 } \le {{\mathcal {O}}}(1)e^{ -\sqrt{2} |x-y|/ L }$. This gives the required bound ${{\mathcal {O}}}(1) |z_k| e^{ - |x|/L }$. For $\ell \ge 2$ we make the estimate

$$\begin{aligned} |z_k|^{\frac{1}{2} \ell } e^{ - |x-y|^2 / 4L^2 (\ell +1) } = \exp \Big ( - \frac{\ell }{2} | \log |z_k|| - |x-y|^2 / 4L^2 (\ell +1) \Big ) \le e^{-2 |x-y| } \end{aligned}$$

(516)

The last inequality requires some explanation. If $\frac{1}{4} \ell | \log |z_k| | \ge |x-y|$ then the first term in exponential provides the decay. On the other hand if $|x-y| \ge \frac{1}{4} \ell | \log |z_k||$ then $|x-y|^2 / 4\,L^2\ell \ge |\log |z_k| | |x-y| / 24\,L^2 \ge 2 | x-y|$ and the second factor provides the decay. Inserting the bound in (515) gives the estimate ${{\mathcal {O}}}(1)|z_k| e^{-|x| }$ which suffices. $\square$