Constructive quantum field theory — Lecture 5 (notes)

Introduction

At the end of Lecture 4 we had, at each lattice cutoff \(N\), the exact system (Theorem 11.6)

where \(Y\) is the approximate forecast built from the explicit action \(\bar{\Gamma}_q\), \(Z = \psi - Y\) is its error, and \(\psi = Y + Z\) is the exact forecast, a martingale. The inputs from Lecture 4 are:

1. Global coercivity (Lemma 13.1): \(H_q (y)\) is comparable to \((K_q + \gamma y^2)^{- 1}\) and \(\|H_q (y)\|_{p \to p} \le (1 + \varepsilon_q) / q\), \(\varepsilon_q = C (g_q + g_q^2)\).

2. The source (Proposition 13.2): \(b = b^{\text{b}} + b^{\text{l}}\) with \(\|b^{\text{b}}_q \|_{\infty} \le C \gamma^{7 / 2} q^{- 5 / 2}\) and \(\|b^{\text{l}}_q (y)\|_2^2 \le C \gamma^4 q^{- 5} \|t_q (y)\|_6^6\), where \(t_q (y) = y \mathbb{1}_{|y| > r_q}\).

3. The coordinate equation (13.5): \(K_q (Y_q + \varrho_q (Y_q)) + \frac{\gamma}{3} Y_q^3 = K_q \Phi_q = P_q\), with a small Lipschitz correction, \(\|D \varrho_q \|_{p \to p} \le \varepsilon_q\).

The goal is to show that \(Z\) is small and that \(Y\) rarely visits large fields, uniformly in \(N\). This is UV stability. Neither equation can be estimated alone: the forward equation contains \(Z\), and the source of the backward equation is controlled by the large fields of \(Y\).

How a priori estimates for FBSDEs are usually obtained.

• the backward equation is linear in \(Z\), given \(Y\), so it can be solved by a conditional representation formula with an explicit propagator;

• the forward equation is, after subtracting the Gaussian part, a cubic elliptic equation at each resolution, which is coercive and can be estimated pathwise;

• the two estimates are linked through two weighted energies whose weights are matched.

14The energies

14.1Moments at fixed cutoff

All estimates below are first proved on a finite interval \([q_0, T]\) and at fixed \(N\). There, everything is finite.

Lemma 14.1. (Preliminary moments) At fixed \(N\) and \(T\), the maxima over \([q_0, T]\) of \(P, Y, \psi, Z\) have all moments, and so does \(\int_{q_0}^T (\| \text{Cov}_q \|_{\text{HS}}^2 +\| \Xi_q \|_{\text{HS}}^2) \mathrm{d} q\). Consequently the energies \(E_T\), \(J_T\) and \(G_T\) defined below have all moments at fixed \(N, T\).

Proof. \(P\) has all maximal moments, since \(\nu_N\) has all moments. Global convexity, \(D^2 \bar{\Psi}_q \ge cq\) and \(D \bar{\Psi}_q (0) = 0\), gives \(\|Y_q \| \le C \|P_q \| / q\). Doob's inequality for the martingale \(\psi_q =\mathbb{E}_q \Phi_{\infty}\) gives its maximal moments, and \(Z = \psi - Y\). For the integrands, \(\| \text{Cov}_q \|_{\text{HS}} \le \text{TrCov}_q \le \mathbb{E}_q \| \Phi_{\infty} \|^2\), and \(\|H_q (Y_q)\|_{\text{HS}} \le C \sqrt{\dim H_N} / q\). The energies involve, besides \(Y\) and \(Z\), the Gaussian field \(X\) and \(\varrho_q (Y_q)\); finite-dimensional norm equivalence, \(\| \varrho_q (y)\| \le \varepsilon_q \|y\|\) and Gaussian moments of \(X\) give their moments.\(\Box\)

These constants depend on \(N\) and \(T\). They are used only to justify the absorptions below, never in the final bounds.

14.2Subtracting the Gaussian part

The Gaussian part of the forecast is the free field at resolution \(q\) driven by the same innovation,

This is the formula (4.3) of Lecture 3, but \(W\) is the innovation of the interacting problem; \(X\) is a Gaussian process under the interacting measure. Since \(\mathrm{d} (K_q X_q) = X_q \hspace{0.17em} \mathrm{d} q + \mathrm{d} W_q\),

On the other hand \(P_q = W_q + \int_{q_0}^q \psi_s \mathrm{d} s = W_q + \int_{q_0}^q (Y_s + Z_s) \mathrm{d} s\). Insert this into the coordinate equation (C) and subtract (14.2). With

the noise \(W\) cancels (\(V_q\) is not to be confused with the potentials \(V_t\), \(V_N\) of Lectures 3–4) and we obtain the identity

This is the equation we estimate. At each \(q\) it is the cubic elliptic equation \((q - \Delta) V + \frac{\gamma}{3} (V + X - \varrho)^3 = f\), a nonlinear Helmholtz equation in which the Gaussian field \(X\) enters, up to the small correction \(\varrho\), only as a shift inside the cube, and the right side has a memory. Three remarks:

• the memory term has coefficient one: \(\int_{q_0}^q V_s \mathrm{d} s\) is of the same size as the mass term \(qV_q\). It cannot be treated as a forcing; it involves \(V\) itself, and the estimate must absorb it;

• the only trace of the backward equation is \(F^Z = \int Z\);

• the identity is pathwise: no expectation has been taken.

14.3Thresholds and energies

Define the signed excess of \(V\) over three quarters of the threshold, the large-field part of \(Y\), and the large Gaussian values,

all pointwise in space, with \(r_q = (q / \gamma)^{1 / 2}\). The three random energies are

Only thresholded sixth powers are integrated over all scales: \(\mathbb{E} \|X_q \|_6^6 \approx q^{3 / 2}\) for \(2 q_0 \le q \lesssim N^2\), so the unthresholded Gaussian contribution to \(J\) would grow like \(\gamma N^2\), not uniformly in \(N\). The weights are matched to each other and to the source bounds:

14.4Two Hardy inequalities

Lemma 14.2. (Hardy) For \(f\) on \([q_0, \infty)\) with values in a Banach space, extended by zero to \((0, q_0)\),

Proof. Write \(q^{- 1} \int_0^q f_s \mathrm{d} s = \int_0^1 f_{q \theta} \mathrm{d} \theta\) and apply Minkowski's inequality. For the first bound, the substitution \(u = q \theta\) gives \(\|f_{\cdot \theta} \|_{L^4 (q^{- 1 / 2} \mathrm{d} q)} = \theta^{- 1 / 8} \|f\|_{L^4 (q^{- 1 / 2} \mathrm{d} q)}\), and \(\int_0^1 \theta^{- 1 / 8} \mathrm{d} \theta = \frac{8}{7}\). For the second, the left side is \(\|q^{- 1} \int_0^q f\|^2_{L^2 (q^{1 / 2} \mathrm{d} q)}\), \(\|f_{\cdot \theta} \|_{L^2 (q^{1 / 2} \mathrm{d} q)} = \theta^{- 3 / 4} \|f\|_{L^2 (q^{1 / 2} \mathrm{d} q)}\), and \((\int_0^1 \theta^{- 3 / 4} \mathrm{d} \theta)^2 = 16\).\(\Box\)

The first inequality will estimate the memory term of (14.4). Its constant \(\frac{8}{7}\) is larger than one. Coercivity from the mass term alone, which has coefficient one, would therefore not absorb the memory. The missing margin will come from the cubic term: at large fields it adds \(\frac{1}{4} q |w|^4\) to the mass, and \(1 + \frac{1}{4} - \frac{8}{7} = \frac{3}{28} > 0\).

Exercises

15Forward coercivity

15.1The scalar model

On one site the Laplacian disappears. We also drop the small Lipschitz correction \(\varrho\) (restored in Proposition 15.3) and allow an arbitrary continuous \(X\), thinking of one site of the lattice, where \(\text{Var} X_q \approx \sqrt{q}\). Then (14.4) becomes a scalar equation with memory. All the mechanisms of the forward estimate are already present, and the constants can be checked by hand.

Lemma 15.1. (Scalar forward coercivity) Let \(\gamma > 0\), \(q_0 < T\), let \(X, f : [q_0, T] \to \mathbb{R}\) be continuous, and let \(v\) be continuous and solve

Define \(r_q\), \(w_q\), \(\xi_q\) as in (14.5), with \(V = v\). Then

Moreover the large-field part \(t_q = (v_q + X_q) \mathbb{1}_{|v_q + X_q | > r_q}\) satisfies \(|t_q | \le \frac{32}{7} (|w_q | + | \xi_q |)\).

Proof. Step 1: test with \(w^3\). Multiply (15.1) by \(w_q^3\). We estimate the four terms where \(w_q \ne 0\); there \(|v_q | = |w_q | + \frac{3}{4} r_q\) and \(v_q, w_q\) have the same sign.

Mass. \(qv_q w_q^3 = q|w_q |^4 + \frac{3}{4} qr_q |w_q |^3\).

Cube. \(|X_q | \le \frac{r_q}{32} + | \xi_q |\), so \(\text{sgn} (w_q) (v_q + X_q) \ge |w_q | + \frac{23}{32} r_q - | \xi_q | \ge |w_q | + \frac{11}{16} r_q - | \xi_q |\) (we give up \(\frac{1}{32} r_q\) to leave room, on the lattice, for the shift \(\| \varrho^0_q \|_{\infty} \le \varepsilon_q r_q\), \(\varepsilon_q \le \frac{1}{32}\)). Since the cube is increasing (also if the right side is negative),

by the elementary inequality (15.3) below and \(\gamma r_q^2 = q\). This is the large-field coercivity: at large fields the cube adds a sixth power and a quarter of the mass.

Memory. Since \(|v_s - w_s | \le \frac{3}{4} r_s\) and \(\int_{q_0}^q r_s \mathrm{d} s \le \frac{2}{3} qr_q\),

The first term, \(\frac{1}{2} qr_q |w_q |^3\), is absorbed by the second mass term \(\frac{3}{4} qr_q |w_q |^3\); the leftover \(\frac{1}{4} qr_q |w_q |^3 \ge 0\) is dropped here, and is used on the lattice to absorb the memory of \(\varrho^0\).

Forcing. By Young's inequality \(w_q^3 f_q \le \frac{\gamma}{48} w_q^6 + \frac{12}{\gamma} f_q^2\).

Step 2: integrate. Collecting, at every \(q\),

Multiply by \(q^{- 3 / 2}\) and integrate over \([q_0, T]\). With \(A = \int_{q_0}^T q^{- 1 / 2} w_q^4 \mathrm{d} q\), Hölder and the first Hardy inequality (with \(w\) extended by zero outside \([q_0, T]\)) give

Since \(\frac{5}{4} - \frac{8}{7} = \frac{3}{28}\), (15.2) follows. For the last claim, \(|v + X| \le |w| + \frac{3}{4} r + \frac{r}{32} + | \xi |\), so on \(\{|v + X| > r\}\) we have \(\frac{7}{32} |v + X| \le |v + X| - \frac{25}{32} r \le |w| + | \xi |\).\(\Box\)

Lemma 15.2. (Two elementary inequalities) For \(x, \rho \ge 0\), \((x - \rho)^3 \ge \frac{3}{4} x^3 - 65 \rho^3\). For \(u, \rho, r \ge 0\),

Proof. \(\frac{1}{4} x^3 + 64 \rho^3 - 3 x^2 \rho = \frac{1}{4} (x - 8 \rho)^2 (x + 4 \rho) \ge 0\), so \((x - \rho)^3 \ge x^3 - 3 x^2 \rho - \rho^3 \ge \frac{3}{4} x^3 - 65 \rho^3\). Apply it with \(x = u + \frac{11}{16} r\): \(\frac{1}{3} u^3 (x - \rho)^3 \ge \frac{1}{4} u^3 x^3 - \frac{65}{3} u^3 \rho^3\), and \(u^3 x^3 \ge u^6 + 3 (\frac{11}{16})^2 r^2 u^4\) with \(\frac{3}{4} (\frac{11}{16})^2 = \frac{363}{1024} > \frac{1}{4}\). Finally \(\frac{65}{3} u^3 \rho^3 \le \frac{1}{6} u^6 + \frac{4225}{6} \rho^6\), since \(\frac{1}{6} (u^3 - 65 \rho^3)^2 \ge 0\), and \(\frac{1}{4} - \frac{1}{6} = \frac{1}{12}\).\(\Box\)

What the scalar lemma says.

15.2The lattice version

Proposition 15.3. (Forward coercivity [FBSDE26]) For \(q_0\) large (depending on \(\gamma\)), at every cutoff \(N\) and every \(T\), pathwise,

with \(C\) independent of \(N\), \(T\) and \(p\).

Proof: the changes from the scalar case. Test (14.4) with \(w_q^3\) in \(L^2 (\Lambda_N)\) and proceed as in Lemma 15.1, with Hölder's inequality in space as well as in scale. Three new terms appear.

The Laplacian. \(\langle w^3, - \Delta_N V \rangle \ge 0\). Indeed \(w^3 = \phi' (V)\) for the convex function \(\phi (v) = \frac{1}{4} (|v| - \frac{3}{4} r)_+^4\), and for any nondecreasing \(\phi'\),

summing over unordered nearest-neighbour pairs. So the elliptic term only helps.

The coordinate correction. Split \(\varrho_q (y) = \varrho^0_q + \varrho^1_q\) according to \(|y| \le r_q\) or not (writing \(\varrho_q (y) = \textsf{L}_q (y) y\) with \(\textsf{L}\) the averaged derivative). Then \(\| \varrho^0_q \|_{\infty} \le \varepsilon_q r_q\) shifts the threshold by a small amount, and \(\| \varrho^1_q \|_j \le \varepsilon_q \|t_q \|_j\), \(j = 4, 6\). If \(\varepsilon_q \le \frac{1}{32}\), then \(|Y| \le |w| + \frac{13}{16} r_q + | \xi | + | \varrho^1 |\); on \(\{|Y| > r_q \}\) this gives \(\frac{3}{16} |Y| \le |w| + | \xi | + | \varrho^1 |\), so \(\|t_q \|_j \le \frac{16}{3} (\|w_q \|_j +\| \xi_q \|_j + \varepsilon_q \|t_q \|_j)\), and for small \(\varepsilon\)

This also supplies the \(t\) part of \(J_T\). These are operator bounds: \(\varrho\) is nonlocal, but small in every \(L^p\). In the cube the error is \(\rho = | \xi | + | \varrho^1 |\), and the cube error \(\frac{4225}{6} \gamma \| \rho \|_6^6\) is at most \(C \gamma (\| \xi_q \|_6^6 + \varepsilon_q^6 \|w_q \|_6^6)\): the \(w\) part is absorbed for small \(\varepsilon\), the \(\xi\) part goes into \(G_T\). The memory of \(\varrho^0\) costs at most \(\frac{2}{3} (\sup_s \varepsilon_s) \hspace{0.17em} qr_q \|w_q \|_3^3\) and is absorbed by the spare \(\frac{1}{4} qr_q \|w_q \|_3^3\) when \(\sup_s \varepsilon_s < \frac{3}{8}\) (automatic, since \(\varepsilon \le \frac{1}{32}\)); the memory of \(\varrho^1\) costs \(C \sup_q \varepsilon_q \hspace{0.17em} (A + G_T)\), with \(A = \int_{q_0}^T q^{- 1 / 2} \|w_q \|_4^4 \mathrm{d} q\), by Hölder, Hardy and Young, and is absorbed by the spare \(\frac{3}{28}\) if \(C \sup_q \varepsilon_q < \frac{3}{56}\); the subtraction is legitimate because \(A < \infty\) pathwise at fixed \(N, T\), by continuity on \([q_0, T]\). All are deterministic smallness conditions on \(g_{q_0}\), independent of \(p\).

The backward forcing. \(| \langle w^3, F^Z \rangle | \le \frac{\gamma}{48} \|w\|_6^6 + \frac{12}{\gamma} \|F^Z \|_2^2\), and by the second Hardy inequality \(\gamma^{- 1} \int q^{- 3 / 2} \|F^Z_q \|_2^2 \mathrm{d} q \le 16 \hspace{0.17em} E_T\).

Gaussian tails. \(X_q (x)\) is centred Gaussian with variance \(\int_{q_0}^q a_2 \le C \sqrt{q}\), while the threshold is \(r_q / 32 = \frac{1}{32} (q / \gamma)^{1 / 2}\). By Minkowski's inequality in \(x\) and translation invariance of \(\text{Law} (X_q)\), \(| \| \xi_q \|_j^j |_{L^p (\mathbb{P})} \le | | \xi_q (0) |^j |_{L^p (\mathbb{P})}\), and by Hölder, for \(j = 4, 6\),

and integrating in \(q\) with \(u = \sqrt{q} / (p \gamma)\) gives \(\|G_T \|_p \le C \gamma^3 p^5\).\(\Box\)

No information on the interacting field has been used, apart from the equation itself: the Gaussian tails are tails of the free field \(X\), which is Gaussian under the interacting measure because it is built from the innovation.

Exercises

16Backward feedback and closure

16.1The backward representation

The backward equation in (1) is linear in \(Z\). Its propagator is the adapted operator solution of

By (A), \(\| \mathscr{P}_{q, s} \|_{2 \to 2} \le \exp \int_q^s \frac{1 + \varepsilon_u}{u} \mathrm{d} u \le L_0 \hspace{0.17em} \frac{s}{q}\), with \(L_0 = \exp (\int_{q_0}^{\infty} \varepsilon_u \frac{\mathrm{d} u}{u}) \le \exp (C (g_{q_0} + g_{q_0}^2))\).

Lemma 16.1. (Conditional representation) For \(q_0 \le q \le T\), at fixed \(N\),

Proof. By the product rule and (1), \(\mathrm{d}_s (\mathscr{P}_{q, s} Z_s) = \mathscr{P}_{q, s} H_s Z_s \mathrm{d} s - \mathscr{P}_{q, s} (b_s + H_s Z_s) \mathrm{d} s + \mathscr{P}_{q, s} \Xi_s \mathrm{d} W_s = - \mathscr{P}_{q, s} b_s \mathrm{d} s + \mathscr{P}_{q, s} \Xi_s \mathrm{d} W_s\). The stochastic integral is a true martingale by Lemma 14.1 and the deterministic bound on \(\mathscr{P}\). Integrate from \(q\) to \(T\) and take \(\mathbb{E}_q\).\(\Box\)

So \(Z\) is the expected source accumulated on the unresolved scales, transported back to \(q\), plus the boundary error. This is the conditional-expectation form of a backward equation, as in \(F_t =\mathbb{E}_t [F_T]\) for the force (Theorem 5.4); here the source is the defect.

16.2Three contributions

Split \(Z = Z^{\text{t}} + Z^{\text{b}} + Z^{\text{l}}\) according to the terminal term and the two parts of the source. We estimate each in the scalar manner: norms of the propagator times norms of the source.

Bounded fields. By (B),

This is where the two correction steps of Lecture 4 are used: with one step the source would be \(O (s^{- 2})\) and the first integral would diverge logarithmically.

Boundary. \(q \|Z^{\text{t}}_q \|_2 \le L_0 \hspace{0.17em} \mathbb{E}_q \|TZ_T \|_2\), so, with \(\delta_{N, T, p} := \|TZ_T \|^2_{L^{2 p} (\mathbb{P}; L^2)}\),

At fixed \(N\), \(\delta_{N, T, p} \to 0\) as \(T \to \infty\) (Lemma 17.1).

Large fields. Let \(A^{\text{l}} := \int_{q_0}^T s^{7 / 2} \|b^{\text{l}}_s (Y_s)\|_2^2 \mathrm{d} s \le C \gamma^4 \int s^{- 3 / 2} \|t_s \|_6^6 \le C \gamma^3 J_T\) by (B). By Cauchy–Schwarz in \(s\),

so that, by Jensen's inequality, \(\|Z^{\text{l}}_q \|_2^2 \le 2 L_0^2 q^{- 5 / 2} \hspace{0.17em} \mathbb{E}_q A^{\text{l}}\). Integrating, and using that \(\mathbb{E}_q\) is a contraction on \(L^p (\mathbb{P})\),

This is how the clipping error is controlled. The large-field source \(b^{\text{l}}\) is not small; but its squared \(L^2\) norm is controlled by the sixth power of the large-field part, \(\|b^{\text{l}}_q \|_2^2 \le C \gamma^4 q^{- 5} \|t_q \|_6^6\) (B), and after transport by the propagator and integration with the energy weight it contributes to the backward energy only through \(J_T\), with the small factor \(\gamma^2 / q_0\).

Proposition 16.2. (Backward feedback) At every cutoff,

with \(C\) independent of \(N\), \(T\) and \(p\).

Proof. \(\|Z\|^2 \le 3 (\|Z^{\text{t}} \|^2 +\|Z^{\text{b}} \|^2 +\|Z^{\text{l}} \|^2)\) and the three estimates above.\(\Box\)

16.3Closing the loop

Theorem 16.3. (Cutoff-uniform energy estimate) There is \(q_0\), depending only on \(\gamma\), such that for every \(N\), \(T\) and \(p \ge 1\),

Proof. Let \(x = \|E_T \|_p\) and \(y = \|J_T \|_p\). By Propositions 15.3 and 16.2,

with \(A_0, B_0, B_1\) and the constant in \(\eta\) independent of \(N, T, p\), and bounded for all \(q_0\) beyond the thresholds already required. Choose \(q_0\) such that also \(A_0 \eta \le \frac{1}{2}\). Then \(y \le A_0 (C \gamma^3 p^5 + B_0 + B_1 \delta_{N, T, p}) + \frac{1}{2} y\). Since \(y < \infty\) by Lemma 14.1, it can be absorbed, and \(x\) follows.\(\Box\)

The structure of the argument is the point:

Forward coercivity estimates the large fields using the backward energy; the backward representation returns that energy with a small coefficient. Neither half closes on its own. The smallness is in the feedback, \(\gamma^2 / q_0\), not in the coupling of either equation, and it is independent of the moment order, so a single \(q_0\) closes the loop for every \(p\). The growth \(p^5\) is then inherited from the Gaussian tails \(G_T\), and gives stretched exponential moments \(\mathbb{E} \exp \{c (E_{\infty} + J_{\infty})^{1 / 5} \} < \infty\) once the boundary term is removed. Note that \(\gamma^2 / q_0 = g_{q_0}^2\).

16.4Consequences

The boundary term \(\delta_{N, T, p}\) is removed in Lecture 6, at each fixed \(N\), by letting \(T \to \infty\). Assuming this for the moment, we obtain

The second bound follows from the representation formula on \([q, \infty)\): the three contributions are, up to the factor \(3\), \(O (q^{- 3})\), \(0\) and, by the computation above, \(2 L_0^2 q^{- 5 / 2} \mathbb{E}A^{\text{l}}\). So the approximate forecast differs from the exact one by \(O (q^{- 5 / 4})\) in \(L^2 (\mathbb{P}; L^2)\) (root mean square), uniformly in the lattice. Since \(\|H_q \| \le C / q\), the coupling term \(H_q Z_q\) in the forward equation is \(O (q^{- 9 / 4})\) in \(L^2 (\mathbb{P}; L^2)\), integrable in \(q\): the forward equation is, up to an integrable drift, the explicit SDE \(\mathrm{d} Y = b (Y) \mathrm{d} q + H (Y) \mathrm{d} W\).

Remark 16.4. (Comparison with the variational method) In the Boué–Dupuis approach (Theorem 7.1) one estimates a drift \(u\) which is the costate up to \(\dot{Q}^{1 / 2}\), and the renormalisation is a change of variables \(u = - 4 \lambda \dot{Q}^{1 / 2} : X^3 : + r\) (in the normalisation \(\lambda \int : \varphi^4\): of Lecture 3) [BG20]. Here the explicit action \(\bar{\Gamma}\) plays the role of all such changes of variables at once, in the forecast coordinate, and \(Z\) plays the role of the remainder \(r\). The coercivity of the quartic term is used in both, but here it is used pathwise in the forward equation, while the remainder is estimated in expectation through the backward equation.

Exercises

References

[BG20]

N. Barashkov, M. Gubinelli. A variational method for \(\Phi^4_3\). Duke Math. J. 169 (2020), 3339–3415.

[FBSDE26]

M. Gubinelli. The massive \(\Phi^4_3\) field through forward–backward equations. Course notes, October 2026.

[MY99]

J. Ma, J. Yong. Forward-Backward Stochastic Differential Equations and their Applications. Lecture Notes in Math. 1702, Springer, 1999.