Diffusive hydrodynamics from the symmetric exclusion process

A finite connected graph has a spanning tree: start with one vertex and, while the current vertex set is not $V$, add an edge from that set to its complement together with the new endpoint. Such an edge exists because otherwise no path could join the current set to its complement. Finiteness makes this process terminate; each added edge attaches one new vertex, so the resulting spanning subgraph is connected and has no cycle.

We prove by induction on the number of vertices that the $k$-element subsets of any finite tree are connected for every admissible $k$. The assertion holds for a one-vertex tree, since its $0$- and $1$-element subset graphs each have one vertex. For a tree $T$ with at least two vertices, choose a simple path of maximum length; either endpoint is a leaf, because an additional neighbor outside the path would extend it, while an additional neighbor on the path would create a cycle. Let $v$ be a leaf. Removing a leaf leaves a connected acyclic graph, hence a smaller tree: the unique path between two remaining vertices cannot have the leaf as an interior vertex. If two configurations $A,B$ have the same occupancy at $v$, removing $v$ leaves two subsets of the same cardinality in $T-v$; by induction a sequence of legal moves there connects them, and the same moves in $T$ keep the occupancy at $v$ fixed.

Suppose $v\in A$ and $v\mathrm{\notin }B$. Since $k<m$, choose a vacant vertex $w\mathrm{\notin }A$ and let $v=v_0,v_1,\dots ,v_r=w$ be the unique simple path in $T$ from $v$ to $w$. Existence follows from connectedness, and two distinct simple paths would contain a cycle, so the path is unique. Let $j$ be the first index with $v_j\mathrm{\notin }A$; then $v_0,\dots ,v_{j-1}$ are occupied. Move the particle at $v_{j-1}$ to the vacancy $v_j$, then the particle at $v_{j-2}$ to the new vacancy $v_{j-1}$, and continue until the particle at $v_0$ has moved to $v_1$. These are legal edge moves and leave $v$ vacant. The resulting configuration and $B$, restricted to $T-v$, both have $k$ particles; induction connects them without using $v$.

Suppose $v\mathrm{\notin }A$ and $v\in B$. Since $k>0$, choose an occupied vertex $w\in A$ and use the unique path $v=v_0,\dots ,v_r=w$. Let $j$ be the first index with $v_j\in A$; all earlier vertices are vacant. Move the particle from $v_j$ to $v_{j-1}$, then from $v_{j-1}$ to $v_{j-2}$, and continue until $v_0=v$ is occupied. The moves are legal along the path. The resulting configuration and $B$, restricted to $T-v$, both have $k-1$ particles; induction connects them without using $v$. This proves the claim for trees, including the cases $k=0$ and $k=m$ already covered by their singleton configuration graphs.

Apply step 1.1 to $G$ and let $T$ be a spanning tree. Every move along an edge of $T$ is also an allowed move in $G$. steps 3.1 and 3.2 therefore connect any two $k$-element subsets using edges of $G$, so ${\mathcal{C}}_k(G)$ is connected.

Since $0\le {\rho }_0\le 1$ and $dx({\mathbb{T}}^d)=1$, ${\rho }_0\in L^1$ and $\mathrm{\mid }{\widehat{\rho }}_0(k)\mathrm{\mid }\le 1$ for every $k$. For any smooth $G$ and integer $m\ge 1$, periodic integration by parts gives $(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^m\widehat{G}(k)=\widehat{(1-\mathrm{\Delta }{)}^{m}G}(k)$, hence $\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }\le \mathrm{\parallel }(1-\mathrm{\Delta }{)}^{m}G{\mathrm{\parallel }}_{L^1}(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}$. If $q\ge 0$ is fixed, choose $m>(d+q)\mathrm{/}2$; grouping lattice points by $r=\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}$ gives at most $(2r+1{)}^d-(2r-1{)}^d\le C_d{r}^{d-1}$ points in shell $r$. The shell contribution to ${\sum }_k(1+\mathrm{\mid }k\mathrm{\mid }{)}^q\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }$ is at most $C{r}^{d-1+q-2m}$. Grouping shells into dyadic blocks $2^j\le r<2^{j+1}$ bounds the $j$th block by $C^{\mathrm{\prime }}{2}^{j(d+q-2m)}$, a convergent geometric series since $2m>d+q$. Thus ${\sum }_k(1+\mathrm{\mid }k\mathrm{\mid }{)}^q\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }<\mathrm{\infty }$.

For each $t>0$ and each fixed derivative order, multiplication of the coefficients by any corresponding polynomial in $k$ still leaves an absolutely summable sequence, because $\mathrm{\mid }{\widehat{\rho }}_0(k)\mathrm{\mid }\le 1$ and the Gaussian factor $e^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}$ dominates every inverse power on lattice shells. Therefore the heat series and all fixed-order differentiated series converge uniformly on ${\mathbb{T}}^d$ for $t\ge \epsilon >0$. Pairing the $k$ and $-k$ terms shows ${\rho }_t$ is real. Termwise differentiation is valid on every such compact time interval; both ${\mathrm{\partial }}_t$ and $\mathrm{\Delta }$ multiply the $k$th term by $-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2$, so ${\mathrm{\partial }}_t{\rho }_t=\mathrm{\Delta }{\rho }_t$.

The decay estimate in step 1.1 makes the Fourier series for $G$ and for each fixed space or time derivative of $P_{t}G$ uniformly absolutely convergent even at $t=0$. By [F2], the series at $t=0$ equals $G$ in $L^2$; both are continuous, so equality almost everywhere implies equality everywhere. Uniform convergence of the differentiated series proves smoothness and uniform derivative bounds on $[0,T]\times {\mathbb{T}}^d$. Also $\mathrm{\parallel }{P}_{t}G-G{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sum }_k\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }\mathrm{\mid }{e}^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}-1\mathrm{\mid }\to 0$ by splitting into a finite set of modes and an absolutely summable tail. Termwise differentiation gives ${\mathrm{\partial }}_t{P}_{t}G=\mathrm{\Delta }{P}_{t}G$.

Absolute convergence permits termwise integration against ${\rho }_0\in L^1$. Reindexing $k\mapsto -k$ and using the even multiplier $e^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}$ gives $\int {\rho }_{t}G={\sum }_k{e}^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}{\widehat{\rho }}_0(k)\widehat{G}(-k)=\int {\rho }_0{P}_{t}G$. By step 2.2 this pairing tends to $\int {\rho }_{0}G$ as $t\downarrow 0$. Differentiating the pairing for $t>0$ and using $\mathrm{\Delta }{P}_{t}G=P_t\mathrm{\Delta }G$ yields $\frac{d}{dt}\int {\rho }_{t}G=\int {\rho }_t\mathrm{\Delta }G$; integration from $0$ proves the weak identity in [F3]. Thus the constructed family is a weak heat solution.

Parseval and the heat coefficients give $\mathrm{\parallel }{\rho }_t-{\rho }_0{\mathrm{\parallel }}_2^2={\sum }_k\mathrm{\mid }{\widehat{\rho }}_0(k){\mathrm{\mid }}^2\mathrm{\mid }{e}^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}-1{\mathrm{\mid }}^2\to 0$ as $t\downarrow 0$: choose a finite set whose tail in the convergent series ${\sum }_k\mathrm{\mid }{\widehat{\rho }}_0(k){\mathrm{\mid }}^2$ is small, then take the limit on that finite set. For $ϵ>0$ put $a_{ϵ}(z)=\frac{1}{2}(1-z\mathrm{/}\sqrt{z^2+{ϵ}^2})$ and ${\mathrm{\Phi }}_{ϵ}(z)={\int }_0^z{\int }_0^r{a}_{ϵ}(q)dqdr$. Then ${\mathrm{\Phi }}_{ϵ}$ is smooth and convex, ${\mathrm{\Phi }}_{ϵ}(0)={\mathrm{\Phi }}_{ϵ}^{\mathrm{\prime }}(0)=0$, and $0\le {\mathrm{\Phi }}_{ϵ}(z)\le z^2\mathrm{/}2$. On any interval $[a,b]\subset (0,T]$, the heat equation, chain rule, and periodic integration by parts give $\int {\mathrm{\Phi }}_{ϵ}({\rho }_b)-\int {\mathrm{\Phi }}_{ϵ}({\rho }_a)=-{\int }_a^b\int {\mathrm{\Phi }}_{ϵ}^{\mathrm{\prime }\mathrm{\prime }}({\rho }_s){\sum }_{i=1}^d\mathrm{\mid }{\mathrm{\partial }}_{x_i}{\rho }_s{\mathrm{\mid }}^{2}dxds\le 0$. For each real $z$, the defining double integral tends to $\frac{1}{2}(z^{-}{)}^2$, where $z^{-}:=\max \{-z,0\}$; dominated convergence on the bounded range of $\rho$ over $[a,b]\times {\mathbb{T}}^d$ gives the same monotonicity for $\mathrm{\parallel }{\rho }_t^{-}{\mathrm{\parallel }}_2^2$. Letting $a\downarrow 0$, the $L^2$ continuity just proved and ${\rho }_0\ge 0$ imply ${\rho }_t^{-}=0$. The assumed datum already gives the required bound when $T=0$.

For the upper bound put $b_{ϵ}(z)=\frac{1}{2}(1+(z-1)\mathrm{/}\sqrt{(z-1{)}^2+{ϵ}^2})$ and ${\mathrm{\Psi }}_{ϵ}(z)={\int }_1^z{\int }_1^r{b}_{ϵ}(q)dqdr$. This is smooth and convex with ${\mathrm{\Psi }}_{ϵ}(1)={\mathrm{\Psi }}_{ϵ}^{\mathrm{\prime }}(1)=0$ and $0\le {\mathrm{\Psi }}_{ϵ}(z)\le (z-1{)}^2\mathrm{/}2$. The same integration by parts gives $\int {\mathrm{\Psi }}_{ϵ}({\rho }_b)\le \int {\mathrm{\Psi }}_{ϵ}({\rho }_a)$ for $0<a<b$, and ${\mathrm{\Psi }}_{ϵ}(z)\to \frac{1}{2}((z-1{)}^{+}{)}^2$, where $r^{+}:=\max \{r,0\}$. Dominated convergence and $L^2$ continuity at zero, together with ${\rho }_0\le 1$, give $({\rho }_t-1{)}^{+}=0$. The zero Fourier coefficient is unchanged by the heat multiplier, so $\int {\rho }_t={\widehat{\rho }}_0(0)=\int {\rho }_0$. The assumed datum already gives the bound when $T=0$.

Let $(u_t)$ be any weak heat solution in the sense of [F3]. Testing its weak identity with $e_{-k}$ shows $g_k(t):={\widehat{u}}_t(k)$ is continuous and satisfies $g_k(t)={\widehat{\rho }}_0(k)-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{\int }_0^t{g}_k(s)ds$. Hence $g_k$ is differentiable and solves $g_k^{\mathrm{\prime }}=-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{g}_k$, so $g_k(t)=e^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}{\widehat{\rho }}_0(k)$, the coefficient of the constructed ${\rho }_t$. For each $t$, both functions are in $L^2$; equality of all Fourier coefficients and completeness in [F2] imply $u_t={\rho }_t$ almost everywhere. Thus the weak heat solution is unique.

Put $f_G(\eta )=N^{-d}{\sum }_x{\eta }_{x}G(x\mathrm{/}N)$. For the bond $\{x,x+e_i\}$, $f_G({\eta }^{x,x+e_i})-f_G(\eta )=N^{-d}({\eta }_{x+e_i}-{\eta }_x)(G(x\mathrm{/}N)-G((x+e_i)\mathrm{/}N))$. Summing its rate-$N^2$ contribution over all $x,i$ and reindexing the terms containing ${\eta }_{x+e_i}$ gives $L_N{f}_G(\eta )=N^{-d}{\sum }_x{\eta }_x{N}^2{\sum }_i[G((x+e_i)\mathrm{/}N)+G((x-e_i)\mathrm{/}N)-2G(x\mathrm{/}N)]=⟨{\pi }^N(\eta ),{\mathrm{\Delta }}_{N}G⟩$. Thus the finite-chain martingale in [F2], applied to $f_G$, is exactly $M_t^{N,G}$ and has bracket given by the sum of rates times squared jumps.

A bond with equal endpoint occupations has zero jump; otherwise its squared jump is $N^{-2d}(G((x+e_i)\mathrm{/}N)-G(x\mathrm{/}N){)}^2$. Substituting these jumps and rate $N^2$ into [F2] gives the displayed bracket. By the fundamental theorem of calculus along the coordinate segment, $\mathrm{\mid }G((x+e_i)\mathrm{/}N)-G(x\mathrm{/}N)\mathrm{\mid }\le N^{-1}\mathrm{\parallel }{\mathrm{\partial }}_{i}G{\mathrm{\parallel }}_{\mathrm{\infty }}$, while $({\eta }_{x+e_i}-{\eta }_x{)}^2\le 1$. There are $N^d$ bonds in each coordinate direction, so the bracket is increasing, continuous, predictable, and bounded by $T{N}^{-d}{\sum }_i\mathrm{\parallel }{\mathrm{\partial }}_{i}G{\mathrm{\parallel }}_{\mathrm{\infty }}^2$. The finite state space makes $f_G$ bounded, hence [F2] also gives square integrability.

For fixed $s$, reindexing the bond sum as in the model generator gives $L_N{F}_s^N(\eta )=⟨{\pi }^N(\eta ),{\mathrm{\Delta }}_{N}H(s,\cdot )⟩$; also ${\mathrm{\partial }}_s{F}_s^N(\eta )=⟨{\pi }^N(\eta ),{\mathrm{\partial }}_{s}H(s,\cdot )⟩$. Hence the compensating drift in the Statement is exactly $({\mathrm{\partial }}_s{F}_s^N+L_N{F}_s^N)({\eta }_s)$.

Fix $N$ and a state $i={\eta }_s$. By [F2], on an interval of length $h$, two or more potential uniformization events have probability $O_N(h^2)$, one actual transition $i\to j$ has probability $h{L}_N(i,j)+O_N(h^2)$, and no state change has probability $1-h{\lambda }_i+O_N(h^2)$, with ${\lambda }_i={\sum }_{j\mathrm{\ne }i}{L}_N(i,j)$. Taylor expansion is uniform over the finite state set and $s\in [0,T]$. If no state change occurs, then $F_{s+h}^N(i)-F_s^N(i)=h{\mathrm{\partial }}_s{F}_s^N(i)+O_N(h^2)$ and the compensating integral is $h({\mathrm{\partial }}_s{F}_s^N+L_N{F}_s^N)(i)+O_N(h^2)$, so the increment of $M^N$ is $-h{L}_N{F}_s^N(i)+O_N(h^2)$. A single transition $i\to j$ gives increment $F_s^N(j)-F_s^N(i)+O_N(h)$; summing its conditional mean over $j$ gives $h{L}_N{F}_s^N(i)+O_N(h^2)$, which cancels the no-transition term. Paths with at least two potential events contribute $O_N(h^2)$ because $M^N$ has bounded increments for fixed $N,T$. Therefore $\mathrm{\mid }\mathbb{E}[M_{s+h}^N-M_s^N\mid {\mathcal{F}}_s]\mathrm{\mid }\le C_N{h}^2$. In the conditional second moment, no transition contributes $O_N(h^2)$, one transition contributes $h{\sum }_{j\mathrm{\ne }i}{L}_N(i,j)(F_s^N(j)-F_s^N(i){)}^2+O_N(h^2)=h{\mathrm{\Gamma }}_{F_s^N}(i)+O_N(h^2)$, and two or more potential events contribute $O_N(h^2)$.

If $T=0$, the only time is $0$ and $M_0^N=0$. For $T>0$, fix $s<t$ and partition $[s,t]$ into $n$ equal intervals of length $h=(t-s)\mathrm{/}n$. Applying the conditional-mean estimate from step 2.1 on every subinterval, then the tower property, bounds $\mathrm{\mid }\mathbb{E}[M_t^N-M_s^N\mid {\mathcal{F}}_s]\mathrm{\mid }$ by $n{C}_N{h}^2\to 0$. Thus $M^N$ is an all-pairs martingale. For fixed $N$ its paths are bounded on $[0,T]$ because the state space is finite, $H$ and its time derivative are bounded, and the compensating integrand is bounded; hence it is square-integrable.

Set ${\gamma }_s(\eta )={\mathrm{\Gamma }}_{F_s^N}(\eta )$ and $A_t^N={\int }_0^t{\gamma }_s({\eta }_s)ds$. If $T=0$, both processes are zero. For $T>0$, the second-moment expansion in step 2.1 and continuity of ${\gamma }_s(i)$ in $s$ imply $\mathbb{E}[A_{s+h}^N-A_s^N\mid {\mathcal{F}}_s]=h{\gamma }_s({\eta }_s)+O_N(h^2)$: without a state change the integrand differs from ${\gamma }_s({\eta }_s)$ by $O_N(h)$, and a state change has probability $O_N(h)$ and changes the bounded integrand over an interval of length $h$. Hence $\mathbb{E}[(M_{s+h}^N-M_s^N{)}^2-(A_{s+h}^N-A_s^N)\mid {\mathcal{F}}_s]=O_N(h^2)$. For fixed $s<t$, partition $[s,t]$ into $n$ equal intervals and write ${\mathrm{\Delta }}_{j}M=M_{t_{j+1}}^N-M_{t_j}^N$, ${\mathrm{\Delta }}_{j}A=A_{t_{j+1}}^N-A_{t_j}^N$. Since $M^N$ is a martingale, $\mathbb{E}[M_{t_j}^N{\mathrm{\Delta }}_{j}M\mid {\mathcal{F}}_s]=0$ by the tower property. Expanding the square, conditioning each remainder ${\mathrm{\Delta }}_j{M}^2-{\mathrm{\Delta }}_{j}A$ first on ${\mathcal{F}}_{t_j}$, and using the bound just proved gives an absolute conditional expectation at most $n{C}_N{h}^2\to 0$. Therefore $(M_t^N{)}^2-A_t^N$ is an all-pairs martingale.

The integrand ${\gamma }_s({\eta }_s)$ is nonnegative, so $A^N$ is increasing; it is continuous and adapted. If $T=0$, the zero process is predictable. Otherwise, for each nested deterministic partition, the process equal to $A^N$ at the left endpoint on each interval $(a,b]$ and zero at time zero is predictable by [F4]; these processes converge pointwise to $A^N$ by continuity. Thus $A^N$ is predictable. At a bond $\{x,x+e_i\}$, the test-function jump is $N^{-d}({\eta }_{x+e_i}-{\eta }_x)(H(s,x\mathrm{/}N)-H(s,(x+e_i)\mathrm{/}N))$. Multiplying its square by the rate $N^2$ and summing over all bonds yields the displayed formula for ${\gamma }_s({\eta }_s)$. The fundamental theorem of calculus gives $\mathrm{\mid }H(s,(x+e_i)\mathrm{/}N)-H(s,x\mathrm{/}N)\mathrm{\mid }\le N^{-1}\mathrm{\parallel }{\mathrm{\partial }}_{i}H(s,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}$; using $({\eta }_{x+e_i}-{\eta }_x{)}^2\le 1$ and $N^d$ bonds per coordinate gives the stated bound for $A_T^N$.