Full support of the zero-temperature Sherrington-Kirkpatrick order parameter

Proof. For a nonnegative finite step coefficient and a smooth even convex one-Lipschitz terminal datum \psi, backward evolution across an interval with coefficient c is

These formulas preserve evenness, convexity, and the one-Lipschitz bound. Convexity in the first case follows either from Hölder’s inequality applied after a spatial translation, or by differentiating the logarithm: its second derivative is the tilted mean of the old second derivative plus c times the tilted variance of the old first derivative. The first derivative is a tilted mean of the old first derivative. The assertions in the case c=0 follow directly from convolution.

For these step solutions, Itô’s formula and completion of the square give the control representation

The supremum is over progressively measurable controls. Indeed, the drift of \Phi(s,Y_s)-\frac12\int_t^s\gamma a^2 is -\gamma(a-u_\gamma)^2/2 when \dd Y_s=\gamma(s)a_s\,\dd s+\dd B_s. Equality is attained by a_s=u_\gamma(s,Y_s); for the smooth finite-step problem the feedback equation has a unique strong solution. Consequently two such solutions obey

Here the supremum is over the whole time-space domain, even though the individual functions grow linearly. The bound follows by using the same controls on both sides of (2.4).

Nonnegative nondecreasing finite step functions are dense in \mathcal U for the L^1 norm. For example, truncate at a time increasing to one, approximate the resulting bounded monotone function by monotone step functions, and then let the truncation time tend to one. Since 0\leq M^{-1}\log(2\cosh(Mx))-|x|\leq M^{-1}\log2, (2.5) defines a unique uniform limit, with control representation (2.4) and terminal datum |x|. This agrees with the Parisi solution in the model definition. All the asserted shape properties pass to the limit. The zero control and Jensen’s inequality give the lower bound in (2.1). For the upper bound use the triangle inequality and |a|-a^2/2\leq1/2 in the control representation.

We give the smoothing estimates explicitly. This also establishes the uniformity needed below. On any strip [0,b] with b<1, monotonicity gives

Thus all coefficients in an L^1-bounded family have a common bound \Gamma there. Work first with the smooth-terminal step solutions and put U=\Phi_x. Differentiation of the mild equation gives

We always have \|U\|_\infty\leq1, independently of its terminal smoothing. The heat-kernel estimate \|\partial_xP_h f\|_\infty\leq C h^{-1/2}\|f\|_\infty implies

The constants do not involve derivatives of the terminal datum.

For completeness, the integral inequality in (2.8) gives a uniform bound away from b. The convolution of h^{-1/2} with itself equals \pi. Substituting the inequality once into its integral therefore reduces it to an ordinary backward Gronwall inequality, with forcing bounded by a constant times 1+(b-t)^{-1/2}. That forcing is integrable. The resulting estimate is uniform on every smaller strip.

The same argument bounds all higher derivatives. Suppose that derivatives of U through order m-1 are already bounded on a slightly larger strip. Differentiate (2.7) m times, placing one derivative on the heat kernel and the other m-1 on UU_x. Its highest-order term is U\partial_x^mU; every remaining product involves derivatives of orders at most m-1. The initial term is bounded by C(b-t)^{-1/2}\|\partial_x^{m-1}U(b)\|_\infty. Thus the same integral inequality applies to \|\partial_x^mU(t)\|_\infty, with an additional bounded forcing. Choose successively smaller strip endpoints in this induction. For every fixed m this proves the asserted bound, uniformly for the approximants.

Uniform convergence of the functions now gives convergence of all their spatial derivatives without a separate singular-kernel comparison. If a bounded C^2 function F on \R has bounded second derivative, Taylor’s formula gives, for every h>0,

Apply this to a difference of two approximating solutions, uniformly on a fixed smaller strip. The uniform second-derivative bound and uniform convergence imply convergence of the first derivatives. Apply (2.9) successively to the differences of the first, second, and later derivatives; the next two derivative bounds are available at every stage. This proves uniform convergence of every positive spatial derivative. Their limits are the spatial derivatives of the limiting function, by the fundamental theorem of calculus. Since the approximating derivatives are jointly continuous, so are their uniform limits.

The argument applies also to an arbitrary L^1-convergent sequence in \mathcal U: first obtain these bounds for each member by step approximation, then use (2.5), the uniform local coefficient bound (2.6), and (2.9). Finally pass to the limit in the integral form of the differentiated equation on each compact strip. All its spatial factors converge uniformly, and its coefficient converges in L^1. This proves local absolute continuity in time and the asserted almost-everywhere equations.□

Proof. The drift is measurable in time, bounded by \gamma(t), and globally Lipschitz in space on every compact time strip by Lemma 2.1. Picard iteration gives a unique strong solution there. These solutions agree on overlaps and extend to time one because \int_0^1\gamma<\infty and Brownian paths are continuous.

For clarity, convergence needs no moment estimate for the approximating paths. Couple them using the same Brownian motion. The drift difference is bounded by a common spatial Lipschitz constant times |X_n-X|, plus |\gamma_n-\gamma|+\gamma_n\|u_n-u_\gamma\|_\infty. Gronwall’s inequality gives a deterministic bound for \sup_{t\leq T}|X_n(t)-X(t)| tending to zero. For step coefficients and smooth terminal data, Itô’s formula on each step cancels the drift of u_n(t,X_n(t)). The step-boundary values agree. Passing to the limit in the stochastic integral by its L^2 isometry proves (2.11). Its initial value is zero by evenness. The same isometry gives q(t)=\int_0^t\E[\Phi_{\gamma,xx}(s,X_s)^2]\,\dd s. The integrand is continuous by joint continuity, boundedness on compact strips, and continuity of X. This proves (2.12) and the stated convergence assertions.□

Proof. To preserve a positive plateau, truncate and discretize the coefficient separately to the left and right of a closed subinterval of I, retain the value c on that subinterval, and impose a positive lower cutoff tending to zero and smaller than c. All choices are monotone and converge in L^1. Increase the terminal smoothing parameter beyond the finite step bound. The convergence claims then follow from Lemmas 2.1 and 2.2. The products involved are uniformly bounded, and the spatial derivative bounds also give uniform continuity for evaluating them on the coupled paths.

For strict positivity, choose b\in I later than the time in question and use (2.3). The function h=\Phi_\gamma(b,\cdot) is smooth, convex, even, and differs from |x| by a bounded function. Its derivative is therefore not constant. Twice differentiating the logarithm in (2.3) gives

The tilted Gaussian law has a strictly positive density everywhere. The variance is strictly positive and the first term is nonnegative.

Inside I all functions are smooth in time as well as space, by the heat formula. Along X, the drifts of r, v, and w are respectively 0, -v^2, and -3vw, and their diffusion coefficients are v, w, and z. These identities follow by differentiating r_t=-r_{xx}/2-rr_x. Itô’s formula for v^2 and then for w^2-2v^3, with (2.12), proves (2.13)–(2.14). All differentiations under the expectation are justified by the bounded higher spatial derivatives on smaller intervals.

If c=0, the same argument gives

To see strictness, start the drift-free interval at any earlier time in I. Its Gaussian transition kernel implies that X_t has a strictly positive density on \R. The function \Phi_{xxx}(t,\cdot) cannot vanish identically: otherwise \Phi(t,\cdot) would be an even quadratic polynomial, whereas it is one-Lipschitz and differs from |x| by a bounded function.□

Proof. Here is a Gaussian estimate that also controls differentiation. Suppose q\in C^m(\mathbb R) satisfies, for 0\le j\le m,

For Y_x=\lambda x+b+\sqrt V Z, where Z is standard normal, 0<\lambda_*\le\lambda\le\lambda^*, |b|\le B, and 0\le V\le V_*,

uniformly in these parameters, for some \kappa'>0. Indeed, write \mu=\lambda x+b and take x large enough that \mu>0. On Y_x\ge\mu/2, use Ae^{-\kappa\mu/2}. On its complement, the global bound |q^{(j)}(y)|\le A e^{K|y|}, Cauchy–Schwarz, and \mathbb P(Y_x<\mu/2)\le e^{-\mu^2/(8V_*)} give a bound C\exp(K\mu-\mu^2/(16V_*)). This is smaller than an exponential in x; the case V=0 is immediate. Exponential integrability of the Gaussian justifies differentiation under the expectation. Adjusting the constant covers the remaining bounded range of x.

For backward evolution, suppose an even starting function l_0 has l_0(x)=cx+C+\rho(x) with differentiated exponential error. Set q(y)=e^{l_0(y)-cy-C}-1. This q obeys the hypotheses above: the positive tail follows by differentiating the exponential, and evenness of l_0 gives at most exponential growth for q and its derivatives on the negative tail. Gaussian tilting gives

Apply (3.6) uniformly for bounded \tau and take the logarithm. Its relative error tends uniformly to zero, so differentiating \log(1+\cdot) proves (3.3) with all the stated derivatives. The initial condition starts this induction since, for x>0,

At a backward boundary, l is multiplied by c_-/c_+>0, which preserves the required tail class. Finite induction proves the first claim.

Suppose at a positive time a that g(a,y)=e^{-y^2/(2a)+\beta y+C}(1+q(y)) has the required tail class. Evenness of g implies the same exponential-growth bound for the derivatives of q on the negative tail. Completing the square gives

On a fixed interval 0\le h\le H, the coefficient of x in the Gaussian mean is at least a/(a+H)>0. Thus (3.6) proves closure, uniformly including h=0. At a jump, write \Phi=x+D+\rho on the positive tail. Formula (3.2) replaces (\beta,C) by (\beta-\delta,C-\delta D) and replaces 1+q by (1+q)e^{-\delta\rho}, preserving the differentiated relative error. Finite induction proves (3.4). Logarithmic differentiation then gives (3.5).□

Proof. Evenness is preserved by both heat evolution and rescaling. If l_0''>0, the logarithm of its heat evolution obeys

where \nu is the Gaussian law of Y=x+\sqrt\tau Z tilted by e^{l_0(Y)}; at \tau=0 use l_0''(x)>0 directly. Since the terminal second derivative is cM\operatorname{sech}^2(Mx)>0 and backward rescalings are positive, this proves v>0. Thus r(x)>0 for x>0. All required derivatives are bounded on closed steps, by smoothness on compact sets and Lemma 3.1 at infinity.

Differentiating (3.1) twice gives

The terminal w=-2cM^2\operatorname{sech}^2(Mx)\tanh(Mx) is nonpositive for x>0. The lateral values are w(\tau,0)=0 and w(\tau,A)=o_A(1) uniformly on a closed step. Apply (3.7) to -w, with C=3\sup v, and let A\to\infty. Backward jumps multiply w by a positive constant, so the sign holds throughout.

Because r(x)=v(0)x+O(x^3) with v(0)>0, odd smoothness makes w/r smooth and even at zero. Direct differentiation of r_\tau=rH gives

At the terminal endpoint of the last step,

For (3.9), write v/r+r=x^{-1}+\eta(x), where \eta is smooth odd and \eta(x)=O(x) near zero. The radial lift \widehat H(y)=H(|y|) in \mathbb R^3 satisfies

Both the lifted function and vector field are smooth at zero: a smooth even function lifts smoothly, and \eta(x)/x is smooth even. Since H\to0 uniformly at infinity, comparison on balls followed by exhaustion gives H\le0.

For H_x, use intervals [\varepsilon,A]. The source wH is nonnegative, and

The lateral errors are H_x(\tau,\varepsilon)=O(\varepsilon), uniformly by smooth evenness, and H_x(\tau,A)=o_A(1) by the tails. Formula (3.7), with a bound independent of \varepsilon,A, proves H_x\ge0 on taking the two limits.

Finally, at a backward jump write r_{\rm new}=\alpha r_{\rm old}, where 0<\alpha\le1. Its derivatives rescale identically, and hence

These preserve both signs. Induction over the finitely many steps completes the proof.□

Proof. Evenness and positivity of g give odd smoothness of s. Within a step, the heat equation for g gives

On the first interval s=x/t, so the first two claims hold explicitly. At a jump of size \delta=c_+-c_-, (3.2) gives

Here \Phi_{xx}>0 and \Phi_{xxx}\le0 on the positive half-line by Lemma 3.2.

For subsequent intervals, put u=a-1/t. Equation (3.13) becomes the linear equation

For the fixed smooth solution, a is bounded on a closed step; this follows from (3.5) and compactness and does not presume its sign. Thus the zeroth-order coefficient is bounded above independently of a spatial truncation. Initially u\ge0, by the preceding step and (3.15); at both infinities u\to0 uniformly by Lemma 3.1. Comparison on [-A,A] and exhaustion prove a\ge1/t. Next apply comparison to -s_{xx} on [0,A] using (3.14). Its zeroth coefficient is -3a\le0, its initial values are nonnegative, and its boundary values tend to zero, at zero by oddness and at infinity by (3.5). This proves s_{xx}\le0.

For the ratio inequality, first note that s(x)>0 and r(x)>0 for x>0. On this half-line define

Then j=s\rho, so it suffices to prove \rho\ge0. On the first step, s=x/t and the concavity of r gives r(x)=\int_0^x v(y)\dd y\ge xv(x); hence \rho=1/x-v/r\ge0. Within any later step, forward time reverses (3.1), so r_t=-rH. Consequently

Also \tfrac12(d_r)_{xx}+d_r(d_r)_x =\tfrac12(w/r)_x=H_x-w. Subtracting gives

The source is nonnegative by the signs already proved, while (d_r)_x=w/r-v^2/r^2<0.

Here are the endpoint details for this last comparison. On a closed step away from time zero, v(t,0)>0 is continuous and bounded below, and a(t,0)\ge1/t>0. Odd Taylor expansions therefore give, uniformly,

At infinity, (3.3) and (3.5) give d_r=O(e^{-\kappa x}) and d_s=(x-\beta(t)t)^{-1}+O(e^{-\kappa' x}), so \rho\to0 uniformly. Use (3.7) on [\varepsilon,A], with upper zeroth-order bound zero, and let \varepsilon\downarrow0, A\to\infty. This proves \rho\ge0 provided it is nonnegative at the start of the step.

To verify that condition, write at a forward jump

Then direct cancellation gives

Since \rho_+=j_+/s_+ for x>0, its initial sign is preserved. This proves the ratio claim by induction. Smoothness of s/r at zero follows by factoring x from both odd functions; in particular j is even and j(0)=0. Reflection gives the assertion on the whole line.□

Proof. The signs of R,K,v,j,L follow from Lemmas 3.2 and 3.3; for K use the common sign of the odd functions r,s. Since H=v-b/2, we have B=-2H\ge0. On x>0, differentiation gives

All these expressions are even. The ratios w/r and s/r are smooth at zero since their numerators and denominator are odd and r_x(0)>0. The inequalities consequently extend there by continuity. For example b(0)=-z(0)/v(0), j(0)=0, and S(0)=0.

The derivative rules follow directly from the definitions. The only one involving the forward score is

which gives the last rule in (3.17).

The density has logarithmic derivative f_x/f=r-s. Therefore

By the differentiated tails, r is bounded and tends to c at the positive tail, v,w,z,b and their derivatives decay exponentially, s=x/t+O(1), j=1/t+O((1+x)e^{-\kappa x}), and K=O(1+x). Every polynomial F and its differentiated expression thus have at most polynomial growth. Moreover

These bounds prove absolute integrability and rFf\to0 at both infinities. Integrating the displayed derivative on [-A,A] and letting A\to\infty proves (3.18).□

Proof. Equation (4.8) reduces the assertion to the same identity with \mathcal W_0 and without the displayed L term. The construction of F_2 above cancels every term containing z. Expand the residual

using (3.17), substitute b=B+2v, and collect monomials in R,K,v,B,j. The coefficients are exactly those of Table 1, and every unlisted coefficient is zero. This proves the identity by rational arithmetic.□

Proof. Integrating (4.11) removes \mathcal W(F_2) by (4.3). Every remaining term on the right is nonnegative. Indeed, the prefactors of the five squares P_i are Rj,RB,Kv,vB,1; T=v(g_0R+hj)\geq0; and L(x_0v^2+hbv)\geq0. The table gives the stronger bound

Using Rb^2=w^2 proves (4.12).□

Proof. Suppose first c>0 and fix t\in I. Choose a closed neighborhood of t inside I. Lemma 2.3 supplies positive nondecreasing finite-step approximants \gamma_n, equal to c there, and terminal smoothings M_n^{-1}\log(2\cosh M_nx) with M_n\geq\|\gamma_n\|_\infty and M_n\to\infty. Write X^n for the associated diffusions and r_n=c\Phi_{n,x}, v_n=r_{n,x}, w_n=r_{n,xx}, z_n=r_{n,xxx} on that neighborhood.

Lemmas 3.2, 3.3, and 3.4 verify all sign, integrability, and boundary hypotheses of Proposition 4.2. Therefore

The derivatives through order four converge uniformly in space and are uniformly bounded, by Lemma 2.1. The coupled paths X_t^n converge, and the next spatial derivative bounds ensure uniform continuity when evaluating at these paths. Thus each expectation in the display converges to its counterpart for \gamma. In particular no limit of s_n, j_n, or w_n/r_n is needed.

Equation (2.14) identifies the resulting left side with c^2q'''(t). Lemma 2.3 also gives v(t,x)=c\Phi_{\gamma,xx}(t,x)>0 for every x, so its fourth moment is strictly positive. This proves (5.1) at every interior time, hence strict convexity of q'. The assertion for c=0 is the last part of Lemma 2.3; it is not obtained by dividing by c.□

Proof. Put \beta=\gamma+\epsilon\eta, u=\Phi_{\gamma,x}, \widetilde u=\Phi_{\beta,x}, and D=\Phi_\beta-\Phi_\gamma. Subtracting the two equations gives

Let Y^\epsilon_0=0 and

This diffusion exists uniquely by Lemma 2.1; its drift is bounded by the integrable function \beta. Itô’s formula on [0,T] yields the following identity. One may justify it without a classical time derivative by approximating \gamma and \beta separately by nondecreasing step functions, using the same smooth terminal datum for the pair. The two solutions, their spatial derivatives, and the associated linear diffusions converge on [0,T] by Lemma 2.1 and the same Gronwall estimate as in Lemma 2.2. The source converges in L^1, since the gradients are bounded by one. Passing to the limit in their classical identities gives

The time-local form of (2.5) gives \|D(T,\cdot)\|_\infty \leq\frac32\int_T^1|\beta-\gamma|. Letting T\uparrow1 therefore proves the exact identity

The coefficients \beta have uniformly bounded L^1 norms for 0\leq\epsilon\leq\epsilon_0, and converge to \gamma in L^1. Lemma 2.1 gives uniform convergence \widetilde u\to u and common spatial Lipschitz bounds on every compact strip. Coupling the diffusions with the same Brownian motion and using Gronwall’s inequality gives Y^\epsilon\to X uniformly there. The integrand in (6.2) consequently converges for each t<1 to \eta(t)q(t) and is bounded in absolute value by |\eta(t)|. Dominated convergence and differentiation of the linear penalty term in \mathcal P prove (6.1).□

Proof. For s<1, the direction \eta=\mathbf1_{[s,1)} adds an atom to \mu and is admissible. Minimality and Lemma 6.1 give G(s)\geq0. If K\subset[0,1) is compact, let \nu=\mu|_K and take \eta_K(t)=-\nu([0,t]). This bounded direction is admissible for 0\leq\epsilon\leq1, since it replaces \mu by \mu-\epsilon\nu. Fubini’s theorem gives

The interchange is legitimate because |q(t)-t|\leq1 and \nu(K)<\infty. Since G\geq0, it vanishes \mu-almost everywhere on every such K. Its continuity then gives G=0 on the relative support S.

By Lemma 2.2, G'(s)=s-q(s) and G''(s)=1-q'(s) on compact strips. Every interior support point is a minimum of G, proving (6.4). If 0\in S, use G(0)=0 and q(0)=0 to write

Nonnegativity for h>0 proves the asserted one-sided condition.□

Proof. The endpoint conclusion is recorded by Auffinger, Chen, and Zeng [4, Remark 7]. We give the quantitative argument needed here, including arbitrary positive constant plateaus. The boundary-layer method is also related to the zero-terminal-segment argument in El Alaoui, Montanari, and Sellke [9, Lemma 6.12 and the proof of Theorem 5 in the cited preprint]; their positivity-support statement concerns a different admissible class. Suppose such a plateau exists. We show that for some C>0,

Fix s\in(a,1), and let \rho_s denote the law of X_s. Choose R<\infty with p_0=\rho_s([-R,R])>0; no density assumption on \rho_s is needed.

When c>0, set h(t,y)=\E\exp(c|y+\sqrt{1-t}Z|), with Z standard normal. The heat formula gives \Phi(t,y)=c^{-1}\log h(t,y) and drift c\Phi_x=h_x/h. Its transition density on the plateau is

Indeed, the heat-semigroup identity makes h(t,B_t)/h(s,B_s) a positive mean-one martingale when Brownian motion is started at (s,x). Its stochastic logarithm has the bounded coefficient h_x/h=c\Phi_x, so change of measure gives precisely this drift and kernel. For c=0 use the same kernel formula with h\equiv1.

Put L=1-s and \delta=L/2. Uniformly for s+\delta\leq t<1, |x|\leq R, and |y|\leq1,

while h(t,y)\geq1 and h(s,x)\leq e^{cR}\E e^{c\sqrt L|Z|}=:M. Consequently the mixture of (6.6) against \rho_s is a density f_t satisfying

For \epsilon=1-t and |y|\leq\sqrt\epsilon, differentiation of the terminal heat formula gives, also when c=0,

For \epsilon\leq1 the denominator is at most D=e^c\E e^{c|Z|}<\infty. The unnormalized mass of each sign is at least p_*:=\Pr(Z\geq1)>0, because |y/\sqrt\epsilon|\leq1 and the exponential tilt is at least one. Each sign thus has probability at least \theta=p_*/D\in(0,1/2) under the normalized tilt. Hence |u(t,y)|\leq1-2\theta and 1-u(t,y)^2\geq4\theta(1-\theta)=:\kappa>0. Integrating this bound over [-\sqrt\epsilon,\sqrt\epsilon] and using (6.7) proves (6.5) with C=2m\kappa.

It follows that q(t)-t\leq (1-t)-C\sqrt{1-t}<0 sufficiently near one. Thus G(s')<0 for s' sufficiently near one, contrary to Lemma 6.2. This also covers c=0.□

Proof of Theorem 1.1. Let S be the support in Lemma 6.2. Since S is relatively closed in [0,1), a nonempty component of its complement is an initial, interior, or terminal interval. On each such interval \gamma is constant. A terminal component is impossible by Lemma 6.3; that lemma also excludes empty support, for which \gamma=0.

Consider a remaining gap with endpoints 0\leq a<b<1 and with a,b\in S. By Lemma 6.2, q(a)=a, q(b)=b, and q'(a),q'(b)\leq1, including the one-sided interpretation at a=0. If its constant coefficient is positive, Proposition 5.1 makes q' strictly convex inside the gap. By continuity up to its endpoints,

Integrating contradicts q(b)-q(a)=b-a, as illustrated by the positive area deficit in Figure 1. If the constant coefficient is zero, the strict increase in Proposition 5.1 instead gives q'(t)<q'(b)\leq1, with the same contradiction.

It remains only to consider an initial component [0,b) when 0\notin S and b\in S. Here \gamma=0 on the gap. Strict increase gives q'(t)<q'(b)\leq1 for 0<t<b, whereas q(0)=0 and q(b)=b. Integration is again a contradiction. Thus S=[0,1). The contact identities give q(t)=t for 0<t<1, and the identity at zero follows from symmetry. Differentiating this identity using Lemma 2.2 gives \E\Phi_{\gamma,xx}(t,X_t)^2=1, including the right derivative at zero. Finally, every nonempty interval (a,b) with 0\leq a<b<1 has positive \mu mass, so \gamma(b)-\gamma(a)=\mu((a,b])>0.□

Proof. Write f_k=\partial_x^k\Phi_\gamma for k\geq1, and evaluate these functions along X unless otherwise indicated. All spatial derivatives are bounded on each compact time strip by Lemma 2.1. Differentiating the PDE and applying Itô’s formula gives

The sum is empty for k=1. For general integrable \gamma, this identity follows by the finite-step approximation and Itô isometry used in Lemma 2.2. On a fixed strip the spatial factors converge uniformly along the coupled paths, the time coefficients converge in L^1, and the stochastic integrals converge in L^2.

For a=f_2 and b=f_3, the identity is \dd a=-\gamma a^2\,\dd t+b\,\dd B_t. Hence

The left side is zero by Theorem 1.1. Convexity gives a\geq0, and the same theorem gives \E a^2=1, so \E a^3\geq(\E a^2)^{3/2}=1. The ratio on the right of (6.8) is therefore well defined and continuous on every compact time strip. Its almost-everywhere equality with \gamma extends to every time by right continuity. In particular \gamma is continuous. At time zero, evenness gives b(0,0)=0, so the formula yields \gamma(0)=0.

To bootstrap regularity without assuming time derivatives of \gamma, consider all polynomial moments of the spatial derivatives at once. For a polynomial P=P(y_1,\ldots,y_m), define

In formal variables y_1,y_2,\ldots, set

These are polynomials in finitely many variables, with D_1=0. Itô’s product rule and (6.10) give

Every such moment is finite and continuous on each compact strip. Since \gamma is continuous, (6.11) makes every M_P continuously differentiable; the ratio (6.8) then makes \gamma continuously differentiable. More generally, if \gamma and all these moments are C^k, the integrand in (6.11) is C^k for every P, making all moments C^{k+1}. The same ratio makes \gamma C^{k+1}. Induction proves smoothness, including right derivatives at zero.

Smoothness and \gamma(0)=0 identify its Stieltjes measure with \gamma'(t)\,\dd t, and monotonicity gives \gamma'\geq0. It remains to prove strict positivity away from zero. Set d=f_4. The polynomial-moment identities give

Differentiating \E b^2=2\gamma\E a^3 therefore yields

All factors in this identity are evaluated at (t,X_t).

Fix t\in(0,1) and put c=\gamma(t). Full support implies c>0. We next apply the integrated coercivity estimate at this fixed time, although \gamma need not be constant on a neighborhood of t. Choose \delta_n\downarrow0 with [t-\delta_n,t+\delta_n]\subset(0,1). Replace \gamma on [t-\delta_n,t+\delta_n) by the value c. This preserves monotonicity, since \gamma(s)\leq c for s<t and \gamma(s)\geq c for s>t, and the change tends to zero in L^1. Outside that interval, truncate at a height greater than c, impose a positive lower cutoff less than c, and approximate separately on the left and right by nondecreasing finite step functions with values at most c and at least c, respectively. Let the truncation height tend to infinity, the lower cutoff tend to zero, and the discretization error tend to zero. The resulting positive bounded nondecreasing coefficients \gamma_n converge to \gamma in L^1 and equal c on a neighborhood of t for every n.

Use terminal data M_n^{-1}\log(2\cosh(M_nx)), with M_n\geq\|\gamma_n\|_\infty and M_n\to\infty, and let a_n,b_n,d_n and X^n denote the corresponding spatial derivatives and diffusion. At time t, the scaled derivatives in Proposition 4.2 are v_n=ca_n, w_n=cb_n, and z_n=cd_n. Thus that proposition and Lemmas 3.2–3.4 give

The derivative bounds and convergence on a fixed compact strip from Lemmas 2.1 and 2.2 do not depend on the widths \delta_n of the artificial plateaus. They therefore allow passage to the limit in every expectation in the last display. Combine the limiting inequality with (6.12) and c=\gamma(t) to obtain (6.9). Its right side is strictly positive because c>0 and \E a^4\geq(\E a^2)^2=1. The proof asserts only \gamma'(0)\geq0 at the left endpoint.□

Proof. Lemma 2.2 gives the bounded martingale u(t,X_t)=\int_0^t a(s,X_s)\,\dd B_s. By Theorem 1.1, its second moment is t and \E a(t,X_t)^2=1. Martingale convergence therefore gives U_1 with |U_1|\leq1 and \E U_1^2=1, hence U_1^2=1 almost surely. The diffusion converges to X_1 almost surely and in L^2, because its drift has deterministic integrable bound \gamma.

The uniform terminal bound (2.1) and convexity show that u(t,x_t)\to\operatorname{sgn}(x) whenever x_t\to x\ne0. Indeed, for a fixed small h>0, the derivative of a convex function at x_t lies between its left and right difference quotients of step h; uniform convergence to |\cdot| makes both limits equal to the sign when the interval stays away from zero. Consequently U_1X_1=|X_1| almost surely, with no need to exclude X_1=0.

Itô’s formula along the optimal diffusion, justified on compact strips as in Lemma 2.2, gives

Let t\uparrow1. The terminal bound and L^1 convergence of X_t identify the left side with \E|X_1|. Substitute \E u(s,X_s)^2=s and the definition of \mathcal P to obtain the first line of (7.1). Since X_1=B_1+\int_0^1\gamma(s)u(s,X_s)\,\dd s, martingality and Fubini’s theorem yield

All terms are integrable because |u|,|U_1|\leq1 and \gamma\in L^1. Finally the L^2 stochastic-integral identity at time one implies \E[U_1B_1]=\int_0^1\E a(s,X_s)\,\dd s. The integral is absolutely bounded by one using \E|a(s,X_s)|\leq(\E a(s,X_s)^2)^{1/2}=1.□

Proof. Euler convergence and normalization. Use Brownian increments \sqrt\Delta Z_j=B_{t_j}-B_{t_{j-1}} to couple (7.2) to X up to time T. Lemma 2.1, on a slightly larger strip, bounds all spatial derivatives of u,a. Their almost-everywhere time derivatives given by the PDE are bounded there, so u,a are uniformly Lipschitz in time and space. The drift \gamma u is spatially Lipschitz with a common bound. For its time dependence, it is enough to use that \gamma is bounded and monotone. If \pi_N(s)=t_{j-1} on [t_{j-1},t_j), then

Also \E|X_s-X_{\pi_N(s)}|^2\leq C\Delta. Subtract the two integral drift equations at the grid points, use these bounds and the Lipschitz bounds, and apply discrete Gronwall. This gives

More explicitly, the L^2 norm of the sum of local drift errors is at most C\sqrt\Delta+C\Delta; the remaining errors satisfy e_j\leq C\Delta\sum_{i<j}e_i+o(1), which proves the claim.

Uniform convergence in (7.5), boundedness and Lipschitz continuity of a, and \E a(t,X_t)^2=1 imply \max_{j\leq N}|c_j-1|\to0. The g_j form an orthonormal family in L^2: they have unit second moment by definition and zero conditional mean given the previous increments. Itô isometry, (7.5), and the same continuity estimates give

Rounding and the two coefficient vectors. The independent random variable F_{\rm G}(Z_{N+1}) is uniform on (0,1). For any v\in[-1,1], the sign of v+2U-1, with U uniform on (0,1), has mean v. Hence \E[F_N\mid Z_1,\ldots,Z_N]=u(T,Y_N) exactly. This preserves every correlation with a function of the first N Gaussians. Bessel’s inequality applied to u(T,Y_N)-M_N against the two orthonormal families (g_j)_{j=1}^N and (Z_i)_{i=1}^N yields

For the second equality, terms of M_N before i are independent of Z_i, and those after i have zero conditional mean; the ith term contributes the stated expectation. The first equality follows from orthogonality of the g_j and M_N=\sum_j\sqrt\Delta\,g_j/c_j.

The shifted product. The first coefficient vector is asymptotically the constant vector with entries \sqrt\Delta, because c_j\to1 uniformly. The second has entries \sqrt\Delta times the expected Hessian at the preceding mesh point. Pairing A_j with B_{j+1} therefore selects the expected Hessian at t_j, which is a Riemann-sum value for the target integral. To justify the accumulated errors, the vectors in both displays have bounded \ell^2 norm. Cauchy–Schwarz shows that their errors contribute o(1) to the shifted scalar product with 1\leq j<N. The first display also gives A_N=\sqrt\Delta/c_N+o(1)=o(1). The remaining term therefore obeys |A_NB_{N+1}|\leq|A_N|=o(1), since |B_{N+1}|\leq1. Thus the sum in (7.4) differs by o(1) from

This converges to the stated integral by (7.5), uniform convergence of c_j to one, and continuity of t\mapsto\E a(t,X_t). Corollary 7.1 gives its limit as T\uparrow1.□

Proof. Theorem 1.1 gives a strictly increasing minimizer in \mathcal U, exactly the no-overlap-gap hypothesis in [9, Assumption 2] for \xi(t)=t^2/2. We apply their Corollary 2.2, with its operation count from Theorem 3.

To match their Gaussian quadratic input, form a symmetric matrix A with A_{ij}=J_{ij}/\sqrt n for i<j and independent diagonal entries A_{ii}=\sqrt{2/n}\,G_i, where the G_i are independent standard Gaussians. The quadratic extension \widehat H_n(x)=\frac12x^{\mathsf T}Ax has covariance (2n)^{-1}(\sigma^{\mathsf T}\tau)^2 on the hypercube, and

The second term is independent of \sigma, so every optimization gap is identical for H_n and \widehat H_n.

Evaluating \nabla\widehat H_n(x)=Ax costs O(n^2) operations. The part with distinct indices is H_n(x), and evaluating one coordinate of its gradient costs O(n). Thus the bound C(\epsilon)(\chi+n)+n\chi_1 in the cited theorem is C(\epsilon)n^2, after changing the constant. Its probability guarantee and the equality of the optimization gaps give the assertion.□