A proof for this equivalent version of the Infrared Bound/Gaussian Domination

Question

Consider the Ising Model in the $d$-dimensional discrete torus with side lengh $L$, denoted by $\mathbb{T}_L $, with nearest neighbors interaction (with interaction parameter $J$, no magnetic field, and inverse of temperature $\beta$

The Gaussian Domination Bound/ Infrared Bound states that for every non-zero $ p \in \mathbb{T}_L^*:=(\frac{2\pi}{L} \mathbb{Z}^d) \cap (-\pi,\pi]$, we have:

$$ \sum_{x \in \mathbb{T}_L} e^{ip \cdot x} \langle \sigma_0 \sigma_x \rangle_{L,\beta} \le \frac{1}{2\beta E(p)}, $$

where $\langle \cdot \rangle^{0}_{L,\beta} $ denotes the expected value of a random variable with respect to finite volume Gibbs Measure in $\mathbb{T}_L$; $i$ is the imaginary unit; $ip \cdot x$ denotes the inner product of $ip$ and $x$; and $$ E(p):= J \sum_{x: \|x\|_1=1} (1-e^{ip\cdot x}). $$

Multiple articles point out that the following statement is equivalent to that bound, $\forall (v_x) \in \mathbb{C}^{\mathbb{T}_L}$, we have $$ \sum_{x,y \in \mathbb{T}_L: \|x-y\|_1=1 }v_x \bar{v}_y \langle \sigma_x \sigma_y \rangle_{L,\beta} \le \frac{1}{2\beta}\sum_{x,y \in \mathbb{T}_L: \|x-y\|_1=1 } v_x \bar{v}_y G_L(x,y) + \frac{1}{L^d} \big| \sum_{x \in \mathbb{T}_L} v_x \big|^2 \sum_{x \in \mathbb{T}_L} \langle \sigma_0 \sigma_x \rangle_{L,\beta}, $$ where $$ G_L(x,y) := \sum_{p \in \mathbb{T}_L^* \backslash \{0\} } \frac{1}{L^d} \frac{e^{ip \cdot (x-y)}}{E(p)}. $$

Well, I tried to use the usual proof as in the Chapter 10 of the book Statistical Mechanics of Lattice Systems:a Concrete Mathematical Introduction. Following the proof in page 465 until the moment they choose the values of $\alpha_i$, I believe that this equivalence will come choosing new values for $\alpha_i$, but to this moment, I don't know what to choose.

Answer 1

So what you actually prove is that $$ \langle (\sigma (-\Delta h))^2 \rangle \le \frac{1}{2\beta} (-h\Delta h) $$ When you take $h=e^{ikx}$, your first statement pops up. However, for any $v$ with $\sum_x v_x =0$, we can always find $h$ such that $-\Delta h =v$ (the kernel of $-\Delta$ is all constant functions). Plug in $h$, and your second statement follows for "neutral" $v$. If $\sum_x v_x\ne0$, we can always define $u=v-1/L^d\sum _x v$ and use $u$ instead.