Bold Feynman Diagrams and the Luttinger–Ward Formalism Via Gibbs Measures: Non-perturbative Analysis

Abstract

Many-body perturbation theory (MBPT) is widely used in quantum physics, chemistry, and materials science. At the heart of MBPT is the Feynman diagrammatic expansion, which is, simply speaking, an elegant way of organizing the combinatorially growing number of terms of a certain Taylor expansion. In particular, the construction of the ‘bold Feynman diagrammatic expansion’ involves the partial resummation to infinite order of possibly divergent series of diagrams. This procedure demands investigation from both the combinatorial (perturbative) and the analytical (non-perturbative) viewpoints. In this paper, we approach the analytical investigation of the bold diagrammatic expansion in the simplified setting of Gibbs measures (known as the Euclidean lattice field theory in the physics literature). Using non-perturbative methods, we rigorously construct the Luttinger–Ward formalism for the first time, and we prove that the bold diagrammatic series can be obtained directly via an asymptotic expansion of the Luttinger–Ward functional, circumventing the partial resummation technique. Moreover we prove that the Dyson equation can be derived as the Euler–Lagrange equation associated with a variational problem involving the Luttinger–Ward functional. We also establish a number of key facts about the Luttinger–Ward functional, such as its transformation rule, its form in the setting of the impurity problem, and its continuous extension to the boundary of the domain of physical Green’s functions.

Appendices

Definitions and Results from Convex Analysis

In this section we review some definitions and results from convex analysis. In this paper many results are stated for concave functions, that is, functions f such that $-f$ are convex. The standard results of convex analysis can always be applied by considering negations. We state results below for convex functions to maintain consistency with the literature. Many results are stated in somewhat more generality than is needed for the purposes of this paper (for example, we do not simply conflate proper and non-proper convex functions). This is done to make sure that the reader can refer to the cited references. The discussion follows developments from Rockafellar [25] and Rockafellar and Wets [26].

1.1 Convex Sets and Functions

We begin with the definition of convex sets and functions.

Definition A.1

A set $C\subset \mathbb {R}^{n}$ is convex if $(1-t)x+ty\in C$ for every $x,y\in C$ and all $t\in [0,1]$.

Definition A.2

An extended real-valued function f on a convex set C, that is, a function $f:C \rightarrow [-\infty ,\infty ] = \mathbb {R}\cup \{ -\infty ,+\infty \}$, is convex if

$$\begin{aligned} f\left( (1-t)x+ty\right) \leqq (1-t)f(x)+tf(y) \end{aligned}$$

for all $x,y\in C$ and all $t\in (0,1)$, where we interpret $\infty - \infty = +\infty$ if necessary. We say that f is strictly convex on the convex set C if this inequality holds strictly whenever $x \ne y$.

Definition A.3

The (effective) domain of a convex function f on S, denoted $\mathrm {dom}\, f$, is the set $\mathrm {dom}\,f=\{x\in S\,:\, f(x)<+\infty \}$.

The following is an immediate consequence of the preceding definitions:

Lemma A.4

Let f be convex on $S\subset \mathbb {R}^{n}$. Then $\mathrm {dom}\, f$ is convex.

We note that when $f\in C^{2}(C)$, our definition of convexity coincides with the definition from multivariate calculus:

Theorem A.5

Let $f\in C^{2}(C)$, where $C\subset \mathbb {R}^{n}$ is open and convex. Then f is convex on C if and only if the Hessian matrix $\nabla ^{2}f(x)$ is positive semi-definite for all $x\in C$.

Proof

See Theorem 4.5 of Rockafellar [25].

Notice that for f convex on a convex set $C\subset \mathbb {R}^{n}$, we can extend to $\tilde{f}$ defined on $\mathbb {R}^{n}$ by taking $\tilde{f}\vert _{\mathbb {R}^{n}\backslash C}\equiv +\infty$. It is immediate that $\tilde{f}$ is convex on $\mathbb {R}^{n}$. Thus one loses no generality by considering only functions that are convex on $\mathbb {R}^{n}$.

The following definitions are helpful for ruling out pathologies:

Definition A.6

A convex function f is called proper if $\mathrm {dom}\, f\ne \emptyset$ and $f(x)>-\infty$ for all x.

We will only ever need to consider proper convex functions.

Definition A.7

If f is a proper convex function, then f is called closed if it is also lower semi-continuous. (If f is a non-proper convex function, then f is called closed if it is either $f\equiv +\infty$ or $f\equiv -\infty$.)

Remark A.8

For the fact that this can be taken as the definition, see Theorem 7.1 of [25].

The convexity of a function guarantees its continuity in a certain sense:

Theorem A.9

A convex function f on $\mathbb {R}^{n}$ is continuous relative to any relatively open convex set in $\mathrm {dom}\, f$. In particular, f is continuous on $\mathrm {int\, dom}\, f$. In fact, it holds that a proper convex function f is locally Lipschitz on $\mathrm {int\, dom}\, f$.

Proof

See Theorems 10.1 and 10.4 of Rockafellar [25].

1.2 First-order Properties of Convex Functions

There is an extension of the notion of differentiability that is fundamental to the analysis of convex functions.

Definition A.10

Let f be a convex function on $\mathbb {R}^{n}$. $y\in \mathbb {R}^{n}$ is called a subgradient of f at $x\in \mathrm {dom}\,f$ if $f(z)\geqq f(x)+\left\langle y,z-x\right\rangle$ for all $z\in \mathbb {R}^{n}$. The subdifferential of f at $x\in \mathrm {dom}\,f$, denoted $\partial f(x)$, is the set of all subgradients of f at x. By convention $\partial f (x) = \emptyset$ for $x \notin \mathrm {dom}\,f$.

Theorem A.11

Let f be a proper convex function. $\partial f(x)$ is a non-empty bounded set if and only if $x\in \mathrm {int\, dom}\, f$.

Proof

See Theorem 23.4 of Rockafellar [25].

It is perhaps no surprise that the derivative and the subdifferential of a convex function coincide wherever it is differentiable.

Theorem A.12

Let f be a convex function, and let $x\in \mathbb {R}^{n}$ such that f(x) is finite. If f is differentiable at x, then $\nabla f(x)$ is the unique subgradient of f at x, where $\nabla$ is the gradient defined with respect to the inner product used to define the subgradient. Conversely, if f has a unique subgradient at x, then f is differentiable at x.

Proof

See Theorem 25.1 of Rockafellar [25].

1.3 The Convex Conjugate

A fundamental notion of convex analysis is convex conjugation, which extends the older notion of Legendre transformation.

Definition A.13

Let f be a function $\mathbb {R}^{n}\rightarrow [-\infty ,+\infty ]$. Then the convex conjugate (or, Legendre-Fenchel transform) $f^{*}:\mathbb {R}^{n}\rightarrow [-\infty ,+\infty ]$ with respect to an inner product $\langle \, \cdot \,,\,\cdot \,\rangle$ on $\mathbb {R}^n$ is defined by

$$\begin{aligned} f^{*}(y)=\sup _{x}\left\{ \left\langle x,y\right\rangle -f(x)\right\} =-\inf _{x}\left\{ f(x)-\left\langle x,y\right\rangle \right\} . \end{aligned}$$

Theorem A.14

Let f be a convex function. Then $f^{*}$ is a closed convex function, proper if and only if f is proper. Furthermore, if f is closed, then $f^{**}=f.$

Proof

See Theorem 12.2 of Rockafellar [25].

It is an important fact that the subgradients of f and $f^{*}$ are, in a sense, inverse mappings.

Theorem A.15

If f is a closed proper convex function, then $x\in \partial f^{*}(y)$ if and only if $y\in \partial f(x)$.

Proof

See Corollary 23.5.1 of Rockafellar [25].

Roughly speaking, differentiability of a convex function corresponds to the strict convexity of its conjugate. Indeed:

Theorem A.16

If f is a closed proper convex function, then the following are equivalent:

• 1. $\mathrm {int}\,\mathrm {dom}\,f$ is nonempty, f is differentiable on $\mathrm {int}\,\mathrm {dom}\,f$, and $\partial f (x) = \emptyset$ for all $x\in \mathrm {dom}\,f \,\backslash \, \mathrm {int}\,\mathrm {dom}\,f$.

• 2. $f^*$ is strictly convex on all convex subsets of $\mathrm {dom}\,\partial f^* := \{ y \,:\, \partial f^* (y) \ne \emptyset \}$.

Proof

See Theorem 11.13 of [26].

Note that for proper convex f, if $\mathrm {dom}\,f^*$ is open, then $\mathrm {dom}\,\partial f^* = \mathrm {dom}\,f^*$ by Theorem A.11, and under the additional assumption that $\mathrm {dom}\,f$ is open, Theorem A.16 simplifies to the following:

Theorem A.17

Let f is a lower semi-continuous, proper convex function, and suppose that $\mathrm {dom}\,f$ and $\mathrm {dom}\,f^*$ are open. Then the following are equivalent:

• 1. f is differentiable on $\mathrm {dom}\,f$.

• 2. $f^*$ is strictly convex on $\mathrm {dom}\,f^*$.

1.4 Sequences of Convex Functions

Pointwise convergence of convex functions entails a kind of convergence of their subgradients.

Theorem A.18

Let f be a convex function on $\mathbb {R}^{n}$, and let C be an open convex set on which f is finite. Let $f_{1},f_{2},\ldots$ be a sequence of convex functions finite on C and converging pointwise to f on C. Let $x\in C$, and let $x_{1},x_{2},\ldots$ be a sequence of points in C converging to x. Then for any $\varepsilon >0$, there exists N such that

$$\begin{aligned} \partial f_{i}(x_{i})\subset \partial f(x)+B_{\varepsilon }(0) \end{aligned}$$

for all $i\geqq N$.

Proof

See Theorem 24.5 of Rockafellar [25].

Besides pointwise convergence, there is in fact another nature of convergence for convex functions. This is the notion of epi-convergence, which is defined (even for non-convex functions) as follows:

Definition A.19

Let $f_i, f$ be extended-real-valued functions on $\mathbb {R}^n$. Then we say that the sequence $\{f_i\}$ epi-converges to f, written as $f = \mathrm {e}\lim _{i\rightarrow \infty } f_i$ or $f_i \overset{\mathrm {e}}{\rightarrow } f$ as $i\rightarrow \infty$, if for all $x \in \mathbb {R}^n$, the following two conditions are satisfied:

$$\begin{aligned} \begin{aligned} \liminf _i f_i (x_i) \geqq f(x) \quad \text{ for } \text{ every } \text{ sequence } x_i \rightarrow x \\ \limsup _i f_i( x_i) \leqq f(x) \quad \text{ for } \text{ some } \text{ sequence } x_i \rightarrow x. \end{aligned} \end{aligned}$$

We say that the sequence $\{f_i\}$ hypo-converges to f, written as $f = \mathrm {h}\lim _{i\rightarrow \infty } f_i$ or $f_i \overset{\mathrm {h}}{\rightarrow } f$ as $i\rightarrow \infty$, if $\{-f_i\}$ epi-converges to $-f$.

The notion of epi-convergence is particularly natural in the theory of convex functions; accordingly hypo-convergence is more relevant to concave functions. Note also that epi-convergence is neither stronger nor weaker than pointwise convergence. However, there is a useful theorem that relates the pointwise convergence and epi-convergence of convex functions.

Theorem A.20

Let $f_{i}$ be a sequence of convex functions on $\mathbb {R}^{n}$, and let f be a lower semi-continuous convex function on $\mathbb {R}^{n}$ such that $\mathrm {dom}\, f$ has non-empty interior. Then $f=\mathrm {e}\lim _{i\rightarrow \infty }f_{i}$ if and only if the $f_{i}$ converge uniformly to f on every compact set C that does not contain a boundary point of $\mathrm {dom}\, f$.

Proof

See Theorem 7.17 of Rockafellar and Wets [26].

Under certain mild conditions, the epi-convergence of a sequence of convex functions is equivalent to the epi-convergence of the corresponding sequence of conjugate functions. Indeed, the following theorem is a natural motivation for considering epi-convergence as opposed to pointwise convergence.

Theorem A.21

Let $f_{i}$ and f be lower semi-continuous, proper convex functions on $\mathbb {R}^{n}$. Then the $f_{i}$ epi-converge to f if and only if the $f_{i}^{*}$ epi-converge to $f^{*}$.

Proof

See Theorem 11.34 of Rockafellar and Wets [26].

Finally, under certain circumstances one can upgrade mere pointwise convergence of convex functions to uniform convergence on compact subsets:

Theorem A.22

Let $f_{i}$ and f be finite convex functions on an open convex set $O \subset \mathbb {R}^n$, and suppose that $f_i \rightarrow f$ pointwise on O. Then $f_i$ converges uniformly to f on every compact subset of O.

Proof

See Corollary 7.18 of Rockafellar and Wets [26].

Classical Results on Weak Convergence of Probability Measures

For completeness we recall here several classical results on the weak convergence of measures. For reference, see, for example, Billingsley [6].

Let S be a metric space, and let $\mathcal {P}(S)$ denote the set of probability measures on S (equipped with the Borel $\sigma$-algebra). We say that a sequence $\mu _k \in \mathcal {P}(S)$ converges weakly to $\mu \in \mathcal {P}(S)$, denoted $\mu _k \Rightarrow \mu$, if $\int f \, \,\mathrm {d}\mu _k \rightarrow \int f\,\,\mathrm {d}\mu$ as $k\rightarrow \infty$ for all bounded, continuous functions $f : S \rightarrow \mathbb {R}$. A number of equivalent characterizations of weak convergence are given in the following result, often known as the Portmanteau theorem:

Theorem A.1

(Portmanteau) Let S be a metric space, and let $\mu _k, \mu \in \mathcal {P}(S)$. The following are all equivalent conditions for the weak convergence $\mu _k \Rightarrow \mu$:

• 1. $\lim _{k\rightarrow \infty } \int f \, \,\mathrm {d}\mu _k = \int f\,\,\mathrm {d}\mu$ for all bounded, continuous functions $f : S \rightarrow \mathbb {R}$.

• 2. $\liminf _{k\rightarrow \infty } \int f \, \,\mathrm {d}\mu _k \geqq \int f\,\,\mathrm {d}\mu$ for all lower semi-continuous functions $f : S \rightarrow \mathbb {R}$ bounded from below.

• 3. $\liminf _{k\rightarrow \infty } \mu _k(U) \geqq \mu ( U)$ for all open sets $U \subset S$.

Remark A.2

There are several other equivalent conditions often included in the statement of this result.

A condition for extracting a weakly convergent subsequence, as guaranteed by Prokhorov’s theorem below, is given by the following notion of tightness:

Definition A.3

Let S be a metric space equipped with the Borel $\sigma$-algebra. A set $\mathcal {C}$ of measures on S is called $tight$ if for any $\varepsilon > 0$, there exists a compact subset $K \subset S$ such that $\mu (K) > 1-\varepsilon$ for all $\mu \in \mathcal {C}$. A sequence of measures is called tight if the set of terms in the sequence is tight.

Theorem A.4

(Prokhorov) Let S be a metric space equipped with the Borel $\sigma$-algebra. Then any tight sequence in $\mathcal {P}(S)$ admits a weakly convergent subsequence.

Proof of Lemmas

1.1 Lemma 2.8

Proof

Suppose $\mu \ll \lambda$ is in $\mathcal {M}_2$ and write $\,\mathrm {d}\mu = \rho \,\,\mathrm {d}x$ where $\rho$ is the probability density. Since $\mu \ll \lambda$, $\mathrm {Cov}(\mu )$ must be positive definite. Let $\mu _G$ be the Gaussian measure with the same mean and covariance as $\mu$, and let $\rho _G$ be the corresponding probability density. Then one can compute that

$$\begin{aligned} \int \rho \log \rho _G \,\,\mathrm {d}x = -\frac{1}{2} \log \left( (2\pi e)^N \det \mathrm {Cov}(\mu ) \right) \end{aligned}$$

(and in particular this integral is absolutely convergent). Now

$$\begin{aligned} \rho \log \rho = \rho \log \rho _G + \rho \log \frac{\rho }{\rho _G}. \end{aligned}$$

The first term on the right-hand side of this equation is absolutely integrable, and the integral of the second term exists (in particular, the integral of the negative part of the second term is finite, and the value of the full integral is in fact $-H_{\mu _G}(\mu )$). Therefore the integral $\int \rho \log \rho \,\,\mathrm {d}x \in (-\infty ,\infty ]$ exists. Moreover,

$$\begin{aligned} H(\mu )= & {} -\int \rho \log \rho \,\,\mathrm {d}x = \frac{1}{2} \log \left( (2\pi e)^N \det \mathrm {Cov}(\mu ) \right) \\&+ H_{\mu _G}(\mu ) \leqq \frac{1}{2} \log \left( (2\pi e)^N \det \mathrm {Cov}(\mu ) \right) \end{aligned}$$

with equality if and only if $\mu _G = \mu$.

To prove the second inequality in the statement of the lemma, define $\overline{\mu }:=\int x\, \,\mathrm {d}\mu$ to be the mean of $\mu$. Then $\mathrm {Cov}(\mu ) = \mathcal {G}(\mu ) - \overline{\mu }\,\overline{\mu }^T$, so in particular $\det \mathrm {Cov}(\mu ) \leqq \det \mathcal {G}(\mu )$, with equality if and only if $\overline{\mu } = 0$.

1.2 Lemma 2.9

Proof

Without loss of generality we can assume that $\mu _j = \rho _j \,\,\mathrm {d}x$ for all j.

First, by the Portmanteau theorem for weak convergence of measures (Theorem A.1) we have, for any $z\in {\mathbb {R}^{N}}$, that

$$\begin{aligned} z^{T}\mathcal {G}(\mu ) z = \int (z^T x)^2 \,\,\mathrm {d}\mu\leqq & {} \liminf _{j\rightarrow \infty } \int (z^T x)^2 \,\,\mathrm {d}\mu ^{(j)} \\= & {} \liminf _{j\rightarrow \infty } \int z^T xx^T z \,\,\mathrm {d}\mu ^{(j)} = \liminf _{j\rightarrow \infty } z^T \mathcal {G}(\mu _j) z \leqq C \Vert z\Vert ^2. \end{aligned}$$

It follows that $\mu \in \mathcal {M}_2$ (and moreover $\mathcal {G}(\mu ) \preceq C\cdot I_n$).

Our goal is to put ourselves in a position to use the upper semi-continuity (note our sign convention) of the relative entropy with respect to the topology of weak convergence (see Fact 2.7). Let $\beta > 0$, and let $Z_\beta = \int e^{-\beta \Vert x\Vert ^2}\,\,\mathrm {d}x$. Let $\gamma _\beta$ be the Gaussian measure with density proportional to $e^{-\beta \Vert x\Vert ^2}$. Then

$$\begin{aligned} H(\mu _j)= & {} -\int \rho _j \log \rho _j \,\,\mathrm {d}x \\= & {} \log (Z_\beta ) - \int \rho _j(x) \log \frac{\rho _j(x)}{\frac{1}{Z_\beta } e^{-\beta \Vert x\Vert ^2 }} \,\,\mathrm {d}x + \beta \int \rho _j(x) \Vert x\Vert ^2 \,\,\mathrm {d}x \\= & {} \log (Z_\beta ) + H_{\gamma _\beta } (\mu _j) + \beta \mathrm {Tr}[\mathcal {G}(\mu _j)]. \end{aligned}$$

Then by the upper semi-continuity of the relative entropy with respect to the topology of weak convergence, we have

$$\begin{aligned} \limsup _{j\rightarrow \infty } H(\mu _j) \leqq \log (Z_\beta ) + H_{\gamma _\beta } (\mu ) + \beta C N = H(\mu ) + \beta \left( C N - \mathrm {Tr}[\mathcal {G}(\mu )] \right) . \end{aligned}$$

Since this inequality holds for any $\beta >0$, the lemma follows.

1.3 Fact 2.11

Proof

We can assume that $\mu$ is absolutely continuous with respect to the Lebesgue measure, that is, has a density $\rho$ (otherwise $H(\mu ) = -\infty$ and the inequality is trivial). It follows that $\mu _i := \pi _i \# \mu$ are absolutely continuous with respect to the Lebesgue measure, that is, have densities $\rho _i$, for $i=1,2$. Let $x=(x_1,x_2)$ denote the splitting of $x\in \mathbb {R}^N$ according to the product structure ${\mathbb {R}^{N}}= \mathbb {R}^p \times \mathbb {R}^{N-p}$. Then using the fact that $\mu _1 \times \mu _2$ has density $\rho _1(x_1) \rho _2(x_2)$, one directly computes that

$$\begin{aligned} {\begin{matrix} &{} H(\mu _1) + H(\mu _2) + H_{\mu _1 \times \mu _2} (\mu ) \\ &{} \ \ = \int \rho _1(x_1) \log \rho _1(x_1) \,\mathrm {d}x_1 + \int \rho _2(x_2) \log \rho _2(x_2) \,\mathrm {d}x_2 + \int \rho (x) \log \frac{\rho (x)}{\rho _1(x_1) \rho _2(x_2)} \,\mathrm {d}x \\ &{} \ \ = \int \rho (x) \log \rho _1(x_1) \,\mathrm {d}x + \int \rho (x) \log \rho _2(x_2) \,\mathrm {d}x + \int \rho (x) \log \frac{\rho (x)}{\rho _1(x_1) \rho _2(x_2)} \,\mathrm {d}x \\ &{} \ \ = \int \rho (x) \log \rho (x) \,\mathrm {d}x\\ &{} \ \ = H(\mu ), \end{matrix}} \end{aligned}$$

but by Fact 2.7, the relative entropy term is non-negative.

1.4 Lemma 3.2

Proof

Upper semi-continuity follows directly from Fatou’s lemma. $\Omega$ is proper because its domain is nonempty and evidently $\Omega$ does not attain the value $+\infty$.

Now let $\theta \in [0,1]$ and $A_{1},A_{2}\in \mathrm {dom}\,\Omega$. Then

$$\begin{aligned} \Omega [\theta A_{1}+(1-\theta )A_{2}]= & {} -\log \int _{{\mathbb {R}^{N}}}\left( e^{-\frac{1}{2}x^{T}A_{1}x-U(x)}\right) ^{\theta }\left( e^{-\frac{1}{2}x^{T}A_{2}x-U(x)}\right) ^{1-\theta }\,\,\mathrm {d}x\\\geqq & {} -\log \left[ \left( \int _{{\mathbb {R}^{N}}}e^{-\frac{1}{2}x^{T}A_{1}x-U(x)}\,\,\mathrm {d}x\right) ^{\theta }\left( \int _{{\mathbb {R}^{N}}}e^{-\frac{1}{2}x^{T}A_{2}x-U(x)}\,\,\mathrm {d}x\right) ^{1-\theta }\right] \\= & {} \theta \Omega [A_{1}]+(1-\theta )\Omega [A_{2}], \end{aligned}$$

where we have used Hölder’s inequality in the second step. This establishes concavity. Strict concavity on $\mathrm {dom}\,\Omega$ follows from the following fact: Hölder’s inequality holds with equality in this scenario if and only if $e^{-\frac{1}{2}x^{T}A_{1}x-U(x)}=e^{-\frac{1}{2}x^{T}A_{2}x-U(x)}$ for all x, that is, if and only if $A_{1}=A_{2}$.

Lastly, observe that since $\mathrm {dom}\,\Omega$ is an open set, for any $A\in \mathrm {dom}\,\Omega$,

$$\begin{aligned} \int _{\mathbb {R}^{N}}e^{\delta x^2} e^{-\frac{1}{2}x^{T}Ax-U(x)}\,\,\mathrm {d}x < +\infty \end{aligned}$$

for some $\delta > 0$. Now, for any polynomial P, there exists a constant C such that for all $A'$ in a sufficiently small neighborhood of A,

$$\begin{aligned} P(x) e^{-\frac{1}{2}x^{T}A'x-U(x)} \leqq C e^{\delta x^2} e^{-\frac{1}{2}x^{T}Ax-U(x)}. \end{aligned}$$

Since derivatives of all orders of the integrand in (2.2) are of the form

$$\begin{aligned} P(x) e^{-\frac{1}{2}x^{T}Ax-U(x)}, \end{aligned}$$

differentiation under the integral is justified, and the smoothness result follows.

1.5 Lemma 3.4

Proof

First assume $A \in \mathrm {dom}\,\Omega$, so $Z[A]<+\infty$. Let $\mu \in \mathcal {M}_2$ and define $f(x):=\frac{1}{2}x^{T}Ax+U(x)$. For any f such that $e^{-f}$ is integrable, define $\nu _f$ to be the probability measure with density proportional to $e^{-f}$. Then, provided that $\mu \ll \lambda$,

$$\begin{aligned} {\begin{matrix} \int f \,\,\mathrm {d}\mu -H(\mu ) &{} = \Omega [A] -\int \log \left( \frac{1}{Z[A]} e^{-f} \right) \,\,\mathrm {d}\mu -H(\mu ) \\ &{} = \Omega [A] + \int \log \left( \frac{\,\mathrm {d}\mu }{\,\mathrm {d}\lambda }\right) -\log \left( \frac{\,\mathrm {d}\nu _{f}}{\,\mathrm {d}\lambda } \right) \,\,\mathrm {d}\mu \\ &{} = \Omega [A] + \int \log \frac{\,\mathrm {d}\mu }{\,\mathrm {d}\nu _f} \,\,\mathrm {d}\mu \\ &{} = \Omega [A] - H_{\nu _f}(\mu ) \geqq \Omega [A]. \end{matrix}} \end{aligned}$$

(C.1)

Since $\mu \in \mathcal {M}_2$, we have $H(\mu ) < +\infty$ as discussed in Remark 3.5. Careful observation reveals that manipulations are valid in the sense of the extended real numbers even when $\int f\,\,\mathrm {d}\mu = +\infty$. Moreover, $\mu \not \ll \lambda$ if and only if $\mu \not \ll \nu _f$, in which case both sides of (C.1) are $+\infty$. Therefore (C.1) holds for all $\mu \in \mathcal {M}_2$.

For $A\in \mathrm {dom}\,\Omega$, (C.1) establishes the ‘$\leqq$’ direction of (3.3). For $A\notin \mathrm {dom}\,\Omega$, $\Omega [A]=-\infty$, so this direction is immediate.

Next suppose that $A \in \mathrm {dom}\,\Omega$. Since $\mathrm {dom}\,\Omega$ is open, it follows that $\nu _f \in \mathcal {M}_2$. From (C.1) and the inequality $-H_{\nu _f}(\mu )\geqq 0$ (which holds with equality if and only if $\mu = \nu _f$), it follows that (3.3) holds. Moreover, that the infimum in (3.3) is uniquely attained at $\mu = \nu _f$, that is, at $\mathrm{{d}}\mu (x)=\frac{1}{Z[A]}e^{-\frac{1}{2}x^{T}Ax-U(x)}\,\,\mathrm {d}x$.

1.6 Lemma 3.7

Proof

By definition $\mathcal {F}[G]=-\infty$ whenever $G\in \mathcal {S}^{N}\backslash \mathcal {S}_{+}^{N}$. Now we show that also $\mathcal {F}[G]=-\infty$ for G on the boundary $\partial \mathcal {S}_{+}^{N}$. This follows from the fact that for such G, any $\mu \in \mathcal {G}^{-1}(G)$ is supported on a subspace of ${\mathbb {R}^{N}}$ of positive codimension, that is, not absolutely continuous with respect to the Lebesgue measure, and therefore $H(\mu )=-\infty$. Moreover, since such $\mu$ is in $\mathcal {M}_2$, we have (via the weak growth condition) that $\int U\,\,\mathrm {d}\mu \in (-\infty ,\infty ]$, so the expression within the supremum of (3.2) is $-\infty$ for all $\mu \in \mathcal {G}^{-1}(G)$.

Meanwhile, for $G\in S_{++}^{N}$, one can see that $\mathcal {F}[G]>-\infty$ by considering $\mu$ to be mean-zero with a compactly supported smooth density, linearly transformed to have the appropriate covariance G. For such $\mu$, both terms in the supremum are finite.

Moreover, for $G\in S_{++}^{N}$ we also have that $\mathcal {F}[G] < +\infty$. Indeed, for $\mu \in \mathcal {G}^{-1}(G)$, by Lemma 2.8 we have $H(\mu ) \leqq \frac{1}{2}\log \left[ (2\pi e)^n \det G \right]$. Since $\int U\,\,\mathrm {d}\mu \geqq -C_U (1+\mathrm {Tr}\,G)$, we have a finite upper bound on the expression in the supremum in (3.2), which finishes the proof.

1.7 Lemma 3.8

Proof

Let $G_{1},G_{2}\in \mathcal {S}^N_{++}$, $\theta \in [0,1]$, and $\varepsilon >0$. Furthermore let $\mu _{1},\mu _{2}\in \mathcal {M}_2$ such that $\mu _{i}\in \mathcal {G}^{-1}(G_{i})$ and $\Psi [\mu _{i}]\geqq \mathcal {F}[G_{i}]-\varepsilon /2$. Then, noting that $\theta \mu _{1}+(1-\theta )\mu _{2}\in \mathcal {G}^{-1}\left( \theta G_{1}+(1-\theta )G_{2}\right)$, we observe

$$\begin{aligned} \mathcal {F}[\theta G_{1}+(1-\theta )G_{2}]= & {} \sup _{\mu \in \mathcal {G}^{-1}\left( \theta G_{1}+(1-\theta )G_{2}\right) }\Psi [\mu ]\\\geqq & {} \Psi \left[ \theta \mu _{1}+(1-\theta )\mu _{2}\right] \\\geqq & {} \theta \Psi [\mu _{1}]+(1-\theta )\Psi [\mu _{2}]\\\geqq & {} \theta \mathcal {F}[G_{1}]+(1-\theta )\mathcal {F}[G_{2}] - \varepsilon , \end{aligned}$$

where the penultimate step employs convexity of $\Psi$. Since $\varepsilon$ was arbitrary, we have established concavity.

The fact that $\mathcal {F}$ is proper follows from Lemma 3.7. Since $\mathcal {F}$ is concave, by Theorem A.9 it is continuous on $\mathrm {int}\,\mathrm {dom}\,\mathcal {F}$, which is in fact all of $\mathrm {dom}\,\mathcal {F}$ by the weak growth assumption. Thus we only need to check upper semi-continuity at points G outside of $\mathrm {dom}\,\mathcal {F}$. At $G \notin \overline{\mathrm {dom}\,\mathcal {F}} = \mathcal {S}^N_{+}$, upper semi-continuity is trivial because $\mathcal {F} \equiv -\infty$ on a neighborhood of G. Therefore let $G\in \partial \mathcal {S}^N_{++}$ and suppose that $G_k \in \mathcal {S}^N_{++}$ such that $G_k \rightarrow G$ as $k\rightarrow \infty$. We need to show that $\limsup _{k\rightarrow \infty } \mathcal {F}[G_k] = -\infty$. Throwing out all $G_k \notin \mathcal {S}^N_{++}$ from the sequence cannot increase the limit superior, so we can just assume that $G_k \in \mathcal {S}^N_{++}$ for all k. Since $G \in \partial \mathcal {S}^N_{++}$, we have $\det G = 0$, and therefore $\det G_k \rightarrow 0$. By Lemma 3.7 we have

$$\begin{aligned} \mathcal {F}[G_k] \leqq \frac{1}{2}\log \left[ (2\pi e)^n \det G_k \right] + C_U (1+\mathrm {Tr}\,G_k). \end{aligned}$$

Since the right-hand side of this inequality goes to $-\infty$ as $k\rightarrow \infty$, the proof is complete.

1.8 Lemma 3.9

Proof

Observe that (1) $\Omega$ and $\mathcal {F}$ are upper semi-continuous, proper concave functions (by Lemmas 3.2 and 3.8), (2) $\mathcal {F} = \Omega ^*$ and $\Omega = \mathcal {F}^*$, and (3) both $\mathrm {dom}\,\Omega$ and $\mathrm {dom}\,\mathcal {F} = \mathcal {S}^N_{++}$ are open. Then the strict concavity and differentiability of $\mathcal {F}$ on $\mathrm {dom}\,\mathcal {F} = \mathcal {S}^N_{++}$ follow directly from Theorem A.17.

Now we turn to proving $C^{\infty }$-smoothness. Though infinite-order differentiability is not typically discussed in convex analysis, it can be obtained from infinite-order differentiability and strict convexity of the convex conjugate via the implicit function theorem. Indeed, define the smooth function $h:\mathcal {S}^n_{++} \times \mathrm {dom}\,\Omega \rightarrow \mathcal {S}^n$ by

$$\begin{aligned} h(G,A)= \nabla \Omega [A] - G. \end{aligned}$$

Then $Dh=\left( \ -I_{\mathcal {S}^n}\ \big \vert \ \nabla ^2 \Omega \ \right)$, and since $\Omega$ is smooth and strictly concave, the right block is invertible for all A, G. Fix some $G'\in \mathcal {S}_{++}^n$, and let $A' = \nabla \mathcal {F}[G']\in \mathrm {dom}\,\Omega$, so $h(G',A') = 0$. Then the implicit function theorem gives the existence of a smooth function $\phi$ on a neighborhood $\mathcal {V}\subset \mathcal {S}^n_{++}$ of $G'$ such that $h(G,\phi (G))=0$ for all $G\in \mathcal {V}$. But this means precisely that $\phi = \nabla \mathcal {F}$, hence in particular $\nabla \mathcal {F}$ is smooth at $G'$.

1.9 Lemma 4.4

Proof

Write

$$\begin{aligned} Z[A,\varepsilon U] = \int e^{-\frac{1}{2} x^T A x - \varepsilon U(x)} \,\mathrm {d}x. \end{aligned}$$

We want to show that as $\varepsilon \rightarrow 0^+$, $Z[\,\cdot \,,\varepsilon U]$ epi-converges (see Definition A.19) to $Z[\,\cdot \,,\varepsilon U]$. If so, then $-\Omega [\,\cdot \,,\varepsilon U]$ epi-converges $-\Omega [\,\cdot \,,0]$, and Theorems A.21 and A.20 yield in particular that $\mathcal {F}[\,\cdot \,,\varepsilon U] \rightarrow \mathcal {F}[\,\cdot \,,0]$ pointwise on $\mathcal {S}^N_{++}$ as $\varepsilon \rightarrow 0^+$. Then by Theorem A.18 we have the pointwise convergence of the gradients on $\mathcal {S}^N_{++}$, that is, $A[G,\varepsilon U] \rightarrow A[G,0] = G^{-1}$ as $\varepsilon \rightarrow 0^+$ for $G\in \mathcal {S}^N_{++}$.

Thus it remains to show that $Z[\,\cdot \,,\varepsilon U]$ epi-converges to $Z[\,\cdot \,,\varepsilon U]$. The first of the conditions in Definition A.19 follows immediately from Fatou’s lemma, so we need only show that for any $A \in \mathcal {S}^N$, there exists a sequence $A_\varepsilon \rightarrow A$ such that

$$\begin{aligned} \limsup _{\varepsilon \rightarrow 0^+} Z [A_\varepsilon , \varepsilon U ] \leqq Z_\varepsilon [A, 0 ] \end{aligned}$$

In particular, it suffices to show that

$$\begin{aligned} \limsup _{\varepsilon \rightarrow 0^+} Z [A, \varepsilon U ] \leqq Z_\varepsilon [A, 0]. \end{aligned}$$

(C.2)

For $A \notin \mathcal {S}^N_{++}$, the righthand side is $+\infty$, so the inequality holds trivially.

Thus assume $A \in \mathcal {S}^N_{++}$. By the weak growth condition, we can write $U(x) = \widetilde{U}(x) - \lambda - \lambda \Vert x\Vert ^2$, where $C>0$ and $\widetilde{U} \geqq 0$. Then

$$\begin{aligned} Z[A,\varepsilon U] = \int e^{\varepsilon \lambda } e^{-\frac{1}{2} x^T (A - \varepsilon \lambda ) x - \varepsilon \widetilde{U}(x)} \,\mathrm {d}x \leqq \int e^{\varepsilon \lambda } e^{-\frac{1}{2} x^T (A - \varepsilon \lambda ) x} \,\mathrm {d}x, \end{aligned}$$

and evidently the righthand side converges to Z[A, 0] by dominated convergence.

1.10 Lemma 4.5

Proof

Let $G\in \mathcal {S}^N_{++}$. Recall Eq. (C.2) from the proof of Lemma 4.4. From this inequality, it follows that there exists $\tau > 0$ and an open neighborhood $\mathcal {N}$ of $G^{-1}$ in $\mathcal {S}^N_{++}$ such that $A \in \mathrm {dom}\,\Omega [\,\cdot \,,\varepsilon U]$ for all $(\varepsilon , A) \in (0,\tau ) \times \mathcal {N}$.

Now consider $\hat{\varepsilon } > 0$ sufficiently small so that $\hat{\varepsilon } < \tau$ and $\hat{A} := A_G(\hat{\varepsilon }) \in \mathcal {N}$ (possible by Lemma 4.4). Define the smooth function $h:(0,\tau ) \times \mathcal {N} \rightarrow \mathcal {S}^N$ by

$$\begin{aligned} h(\varepsilon ,A)= \nabla _A \Omega [A,\varepsilon U] - G. \end{aligned}$$

Then $Dh(\varepsilon ,A) =\left( \ * \ \big \vert \ \nabla ^2_A \Omega [A,\varepsilon U] \ \right)$, and since $\Omega [\,\cdot \,,\varepsilon U]$ is smooth and strictly concave, the right block is invertible for all $\varepsilon , A$. Moreover, we have $h(\hat{\varepsilon },\hat{A}) = 0$ by construction. Then the implicit function theorem gives the existence of a smooth function $\phi$ on a neighborhood I of $\hat{\varepsilon }$ such that $h(\varepsilon ,\phi (\varepsilon ))=0$ for all $\varepsilon \in I$, but this means precisely that $\phi = A_G$. The implicit function theorem then also says that

$$\begin{aligned} A_G '(\varepsilon ) = -(\nabla ^2_A \Omega [A_G(\varepsilon ),\varepsilon U])^{-1} \frac{\partial h}{\partial \varepsilon }(\varepsilon , A_G(\varepsilon )) \end{aligned}$$

(C.3)

for all $\varepsilon \in I$, where $A_G'$ denotes the ordinary derivative of the function $A_G$ of a single variable. In particular Eq. (C.3) holds at $\varepsilon = \hat{\varepsilon }$, but since $\hat{\varepsilon }$ was arbitrary (beyond being taken sufficiently small), it follows that Eq. (C.3) simply holds for all $\varepsilon > 0$ sufficiently small.

We want to show that all derivatives of $A_G:(0,\infty ) \rightarrow \mathcal {S}^N$ extend continuously to $[0,\infty )$. Starting with $A_G'$, we can examine these functions by taking further derivatives on the righthand side of Eq. (C.3). The result will be an expression involving integrals of the form

$$\begin{aligned} \int P(x,U(x))\, e^{-\frac{1}{2} x^T A_G(\varepsilon ) x - \varepsilon U(x)} \, \,\mathrm {d}x, \end{aligned}$$

where P is some polynomial, and it suffices to show that such integrals converge to their desired limits

$$\begin{aligned} \int P(x,U(x))\, e^{-\frac{1}{2} x^T G^{-1} x} \, \,\mathrm {d}x. \end{aligned}$$

The argument is by dominated convergence. First observe that from the at-most-exponential growth assumption (Assumption 2.5), it follows that there exist $a,b>0$ such that $\vert P(x,U(x)) \vert \leqq a e^{b \Vert x\Vert }$ for all x. As in the proof of Lemma 4.4, write $U(x) = \widetilde{U}(x) - \lambda - \lambda \Vert x\Vert ^2$, where $C>0$ and $\widetilde{U} \geqq 0$. Then

$$\begin{aligned} \vert P(x,U(x))\, e^{-\frac{1}{2} x^T A_G(\varepsilon ) x - \varepsilon U(x)} \vert\leqq & {} \vert P(x,U(x)) \vert \, e^{\varepsilon \lambda } e^{-\frac{1}{2} x^T (A_G(\varepsilon ) - \varepsilon \lambda ) x - \varepsilon \widetilde{U}(x)} \\\leqq & {} a e^{b \Vert x \Vert } e^{\varepsilon \lambda } e^{-\frac{1}{2} x^T (A_G(\varepsilon ) - \varepsilon \lambda ) x}. \end{aligned}$$

Then for all $\varepsilon > 0$ small enough such that $\varepsilon < 1$ and $A_G(\varepsilon ) - \varepsilon \lambda \succ \frac{1}{2} G^{-1}$, we see that the absolute value of the integrand is bounded uniformly by

$$\begin{aligned} a e^{b \Vert x \Vert } e^{ \lambda } e^{-\frac{1}{4} x^T G^{-1} x}, \end{aligned}$$

which is integrable. This completes the dominated convergence argument, and we conclude that all derivatives of $A_G$ extend continuously to $[0,\infty )$.

Next we aim to use the preceding to show that all derivatives of $\Phi _G$ and $\Sigma _G$ also extend continuously to $[0,\infty )$.

To this end, recall the Dyson equation

$$\begin{aligned} \Sigma _G = A_G - G^{-1}, \end{aligned}$$

which requires that the desired extension property of $\Sigma _G$ is equivalent to that of $A_G$, which we have already proved.

Now for any $\varepsilon >0$, we have

$$\begin{aligned} \Phi _G (\varepsilon )= & {} 2\mathcal {F}[G,\varepsilon U] - \mathrm {Tr}\log G - N\log (2\pi e) \\= & {} \mathrm {Tr}[A_G (\varepsilon ) G] -2\Omega [A_G (\varepsilon ),\varepsilon U] - \mathrm {Tr}\log G - N\log (2\pi e) \end{aligned}$$

by Legendre duality, from which it follows from our extension property for $A_G$, together with the arguments used to establish it, that all derivatives of $\Phi _G$ extend continuously to $[0,\infty )$.

1.11 Lemma 4.12

Proof

Based on Eqs. (4.6) and (4.7), we want to show that $G[A^{(M)}(\varepsilon ), U_\varepsilon ^{(M)}] \sim G[A^{(M)}(\varepsilon ) , \varepsilon U]$. As a first step, we aim to show that $Z[A^{(M)}(\varepsilon ), U_\varepsilon ^{(M)}] \sim Z[A^{(M)}(\varepsilon ) , \varepsilon U]$. Indeed, we can write

$$\begin{aligned}&Z[A^{(M)}(\varepsilon ) , \varepsilon U] - Z[A^{(M)}(\varepsilon ), U_\varepsilon ^{(M)}] \nonumber \\&\quad \quad \quad =\ \ \int e^{-\frac{1}{2} x^T A^{(M)}(\varepsilon ) x - \varepsilon U(x)} \left( 1 - e^{- \frac{1}{2} x^T \left[ \Sigma _G(\varepsilon ) - \Sigma _G^{(\leqq M)}(\varepsilon ) \right] x} \right) \,\mathrm {d}x. \end{aligned}$$

(C.4)

We can choose C such that

$$\begin{aligned} -C \varepsilon ^{M+1} \preceq \Sigma _G(\varepsilon ) - \Sigma _G^{(\leqq M)}(\varepsilon ) \preceq C \varepsilon ^{M+1} \end{aligned}$$

for all $\varepsilon >0$ sufficiently small.

Now let $R(\varepsilon ) = \varepsilon ^{-p/2}$ for $p\in (0,1)$. We split the integral in (C.4) into a part over $B_{R(\varepsilon )}(0)$ and another part over the complement. The integrand is dominated by $e^{- \delta x^T x}$ for some $\delta$ uniform in $\varepsilon$, the integral of which over the complement of $B_{R(\varepsilon )}(0)$ decays super-algebraically as $\varepsilon \rightarrow 0$, so we can neglect this contribution.

Meanwhile, for $x\in B_{R(\varepsilon )}(0)$, we have

$$\begin{aligned} \left| x^T \left[ \Sigma _G(\varepsilon ) - \Sigma _G^{(\leqq M)}(\varepsilon ) \right] x \right| \leqq C \varepsilon ^{M+1-p}, \end{aligned}$$

hence there exists $C'$ such that

$$\begin{aligned} \left| 1 - e^{- \frac{1}{2} x^T \left[ \Sigma _G(\varepsilon ) - \Sigma _G^{(\leqq M)}(\varepsilon ) \right] x} \right| \leqq C' \varepsilon ^{M+1-p} \end{aligned}$$

for all $x\in B_{R(\varepsilon )}(0)$. Combining with (C.4) and dominated convergence, we have established $Z[A^{(M)}(\varepsilon ), U_\varepsilon ^{(M)}] \sim Z[A^{(M)}(\varepsilon ) , \varepsilon U]$.

This result, together, together with analogous arguments applied to integrals of the form

$$\begin{aligned} \int x_i x_j \,e^{-\frac{1}{2} x^T A^{(M)}(\varepsilon ) x - \varepsilon U(x)} \left( 1 - e^{- \frac{1}{2} x^T \left[ \Sigma _G(\varepsilon ) - \Sigma _G^{(\leqq M)}(\varepsilon ) \right] x} \right) \,\mathrm {d}x, \end{aligned}$$

yields $G[A^{(M)}(\varepsilon ), U_\varepsilon ^{(M)}] \sim G[A^{(M)}(\varepsilon ) , \varepsilon U]$.

1.12 Lemma 5.1

Proof

For convenience, we define

$$\begin{aligned} \mathcal {F}_{c} [G] := \sup _{\mu \in \mathcal {G}^{-1}(G)\cap \mathcal {M}_c} \left[ H(\mu ) - \int U\,\,\mathrm {d}\mu \right] . \end{aligned}$$

Evidently $\mathcal {F}_c \leqq \mathcal {F}$ and $\mathcal {F}_{c}[G] = -\infty$ if $G \notin \mathcal {S}^N_{++}$, so we can restrict attention to $G \in \mathcal {S}^N_{++}$.

Fix $\varepsilon > 0$. Let $G\in \mathcal {S}^N_{++}$, so $\mathcal {F}[G]$ is finite, and let $\mu \in \mathcal {M}_2$ such that

$$\begin{aligned} H(\mu ) - \int U \,\,\mathrm {d}\mu \geqq \mathcal {F}[G] - \varepsilon /2. \end{aligned}$$

In particular, $H(\mu ) \ne -\infty$, so $\mathrm{{d}}\mu = \rho \,\,\mathrm {d}x$ for some density $\rho$. Then consider the measure $\mu _R \in \mathcal {M}_c (R)$ given by density $\rho _R := Z_R ^{-1}\cdot \rho \cdot \chi _R$, where $\chi _R$ is the indicator function for $B_R (0)$ and $Z_R = \int _{B_R (0)} \rho \,\,\mathrm {d}x$. By monotone convergence, $Z_R \rightarrow 1$.

   Unfortunately we cannot expect $\mathcal {G}(\mu _R) = G$, but we do have $\mathcal {G}(\mu _R) \rightarrow G$ (following from dominated convergence, together with the finite second moments of $\mu$). We then want to modify $\mu _R$ (keeping its support compact) to construct a nearby measure with the correct second moments.

   To this end let $G_R = \tau _R [G - \mathcal {G}(\mu _R)] + \mathcal {G}(\mu _R)$, where $\tau _R > 1$ is chosen so that $\tau _R \rightarrow +\infty$ and the eigenvalues of $G_R$ remain uniformly bounded away from zero and infinity (possible since $\mathcal {G}(\mu _R) \rightarrow G$). Note that we have $G = \tau _R^{-1} G_R + (1 - \tau _R^{-1}) \mathcal {G}(\mu _R)$.

   Now let $\pi \in \mathcal {M}_2$ be any compactly supported measure with a density and finite entropy, and let $\pi _R = T_R \# \pi$, where $T_R$ is a linear transformation chosen so that $\mathcal {G}(\pi _R) = G_R$. Since the eigenvalues of $G_R$ are uniformly bounded away from zero and infinity, the $T_R$ can be chosen to have determinants uniformly bounded away from zero and infinity (which guarantees that that the $\vert H(\pi _R) \vert$ are uniformly bounded), and $\pi _R$ can be taken to have uniformly bounded support. Then finally we can define a measure $\nu _R := \tau _R^{-1} \pi _R + (1 - \tau _R^{-1}) \mu _R$, so $\mathcal {G}(\nu _R) = G$ and $\nu _R$ is compactly supported.

   For the proof it suffices to show that

$$\begin{aligned} H(\nu _R) - \int U \,\,\mathrm {d}\nu _R \rightarrow H(\mu ) - \int U \,\,\mathrm {d}\mu \end{aligned}$$

(C.5)

as $R \rightarrow \infty$.

By the weak growth condition (Definition 2.3), we can choose a constant C such that $\widetilde{U}$ defined by $\widetilde{U}(x) := C(1+\Vert x\Vert ^2) + U(x)$ satisfies $\widetilde{U}(x) \geqq \Vert x\Vert ^2$. Now

$$\begin{aligned} \int (1+\Vert x\Vert ^2) \,\,\mathrm {d}\mu _R \rightarrow \int (1+\Vert x\Vert ^2) \,\,\mathrm {d}\mu < +\infty \end{aligned}$$

by monotone convergence together with the fact that $Z_R \rightarrow 1$. Furthermore

$$\begin{aligned} \tau _R^{-1} \int (1+\Vert x\Vert ^2) \,\,\mathrm {d}\pi _R \rightarrow 0, \end{aligned}$$

so in fact

$$\begin{aligned} \int (1+\Vert x\Vert ^2) \,\,\mathrm {d}\nu _R \rightarrow \int (1+\Vert x\Vert ^2) \,\,\mathrm {d}\mu < +\infty \end{aligned}$$

Therefore, without loss of generality, we can prove C.5 under the assumption that $U(x) \geqq \Vert x\Vert ^2$. But then $\int U \,\,\mathrm {d}\mu _R \rightarrow \int U\,\,\mathrm {d}\mu$ by monotone convergence, and $\tau _R^{-1} \int U \,\,\mathrm {d}\pi _R \rightarrow 0$ since the $\pi _R$ have uniformly bounded support, so in fact $\int U\,\,\mathrm {d}\nu _R \rightarrow \int U\,\,\mathrm {d}\mu$.

      Then we need only show that $H(\nu _R) \rightarrow H(\mu )$. Here one verifies from the construction that $\nu _R$ converges weakly to $\mu$, and moreover the second moments of $\nu _R,\mu$ are uniformly bounded, so by Lemma 2.9, we have $\limsup _R H(\nu _R) \leqq H(\mu )$.

However, by the concavity of the entropy, we have $H(\nu _R) \geqq \tau _R^{-1} H(\pi _R) + (1-\tau _R^{-1}) H(\mu _R)$. Now recall that the $\vert H(\pi _R) \vert$ are uniformly bounded in R, so $\tau _R^{-1} H(\pi _R) \rightarrow 0$. Thus the statement $\liminf _R H(\nu _R) \geqq H(\mu )$ (and hence also $H(\nu _R) \rightarrow H(\mu )$) will follow if we can establish $H(\mu _R) \rightarrow H(\mu )$.

Now

$$\begin{aligned} H(\mu _R) = \log (Z_R) - Z_R^{-1}\int _{B_R (0)} \rho \,\log \rho \,\,\mathrm {d}x, \end{aligned}$$

but we know that $Z_R \rightarrow 1$, so we need only show that

$$\begin{aligned} \int _{B_R (0)} \rho \log \rho \,\,\mathrm {d}x \rightarrow \int \rho \log \rho \,\,\mathrm {d}x. \end{aligned}$$

From Lemma 2.8, the negative part of $\rho \log \rho$ is integrable. But then the fact that $H(\mu ) > -\infty$ precisely means that the positive part of $\rho \log \rho$ is integrable, that is, $\rho \log \rho$ is absolutely integrable. Then the desired fact follows from dominated convergence.