Anderson reverse-time diffusion

For fixed $u<v$ and Borel $A$, the integrand $1_A(x)p_{T-v}(x)q(T-v,x;T-u,y)$ is nonnegative and jointly Borel in $(y,x)$. Tonelli's parameter-integral clause [F2] makes its integral Borel in $y$; division by the positive Borel function $p_{T-u}(y)$ preserves measurability. For each fixed $y$ the numerator defines a measure in $A$ by countable additivity of the Lebesgue integral.

Taking $A={\mathbb{R}}^d$ in the formula gives $K_{u,v}(y,{\mathbb{R}}^d)=p_{T-u}(y)\mathrm{/}{p}_{T-u}(y)=1$ by the marginal identity, so each $K_{u,v}(y,\cdot )$ is a probability measure. The identity kernel at $u=v$ is also a probability kernel.

If $u<v<w$, Tonelli and the two density ratios give ${\displaystyle \int K_{v,w}(x,A)K_{u,v}(y,dx)=\frac{{\int }_A{p}_{T-w}(z)\int q(T-w,z;T-v,x)q(T-v,x;T-u,y)dxdz}{p_{T-u}(y)}\mathrm{.}}$ The inner integral is $q(T-w,z;T-u,y)$ by forward Chapman–Kolmogorov, so the expression is $K_{u,w}(y,A)$. If two adjacent times coincide, composition with the identity kernel gives the same equality.

If $Q$ is a true continuous martingale and $\rho \le T$ is a stopping time, round $\rho$ up to finite-valued grid stopping times ${\rho }_m\downarrow \rho$. For fixed $s\le t$, include $s,t$ in the grid; the sampled process is a discrete martingale, so [F5] gives the event-integral identity for $Q_{t\wedge {\rho }_m}$ and $Q_{s\wedge {\rho }_m}$. The finite-valued stopping calculation identifies these stopped values as conditional expectations of $Q_T$. They are uniformly integrable: for any sub-$\sigma$-algebra $\mathcal{H}$, truncating $Q_T$ at $K\mathrm{/}2$ gives $E(\mathrm{\mid }E[Q_T\mid \mathcal{H}]\mathrm{\mid }{1}_{\mathrm{\mid }E[Q_T\mid \mathcal{H}]\mathrm{\mid }>K})\le 2E(\mathrm{\mid }{Q}_T\mathrm{\mid }{1}_{\mathrm{\mid }{Q}_T\mathrm{\mid }>K\mathrm{/}2})$. Continuity gives almost-sure convergence as $m\to \mathrm{\infty }$; dominated convergence on bounded truncations and this uniform tail bound pass the event identities to $Q^{\rho }$. Thus a bounded stop of a true continuous martingale is a true martingale. If $M$ is a local martingale with localizers ${\lambda }_k$, then $M^{\tau \wedge {\lambda }_k}=(M^{{\lambda }_k}{)}^{\tau }$ is a true martingale by this fact, so $({\lambda }_k)$ also localizes $M^{\tau }$. For each integer $n\ge 1$, let ${\tau }_n=\inf \{t\le T:\mathrm{\mid }{M}_t\mathrm{\mid }\ge n\}\wedge T$ and set $M^{(n)}=M^{{\tau }_n}$. The stopping-time definition and continuity make ${\tau }_n$ a stopping time, and the preceding argument shows $M^{(n)}$ is a local martingale. It is bounded by $n$, hence a true martingale: pass its event-integral identities along its localizing sequence by dominated convergence. Every continuous path is bounded on $[0,T]$, so ${\tau }_n=T$ eventually on each path. It suffices to construct the claimed limit for each bounded martingale $M^{(n)}$ and show the constructions agree on overlaps.

Fix one such bounded martingale $M$, with $\mathrm{\mid }{M}_t\mathrm{\mid }\le n$, and a deterministic partition $\pi =(t_0,\dots ,t_N)$ of $[0,T]$. For coefficient bookkeeping put $H_s^{\pi }=M_{t_k}$ for $t_k\le s<t_{k+1}$ and $H_T^{\pi }=M_{t_{N-1}}$, and set ${\displaystyle I_t^{\pi }=\sum _{k=0}^{N-1}{M}_{t_k}(M_{t\wedge t_{k+1}}-M_{t\wedge t_k}),Q_t^{\pi }=\sum _{t_{k+1}\le t}(M_{t_{k+1}}-M_{t_k}{)}^2+(M_t-M_{t_k}{)}^2}$ when $t_k\le t<t_{k+1}$, with the last partial term omitted at $t=T$. The process $I^{\pi }$ is a continuous martingale because it is a finite sum of bounded, left-measurable multiples of martingale increments. Orthogonality [F1] gives $E\mathrm{\mid }{I}_T^{\pi }{\mathrm{\mid }}^2={\sum }_{k}E[M_{t_k}^2({\mathrm{\Delta }}_{k}M{)}^2]\le n^{2}E(M_T-M_0{)}^2\le n^4$. Directly expanding each square yields the exact identity $M_t^2-M_0^2=2I_t^{\pi }+Q_t^{\pi }$. Hence $\mathrm{\parallel }{Q}_T^{\pi }{\mathrm{\parallel }}_2\le 3n^2$, uniformly over all partitions.

For two partitions $\pi ,\rho$, take their common refinement $\eta$. Refinement does not change a step sum with a fixed coefficient on each coarse interval, so ${\displaystyle E\mathrm{\mid }{I}_T^{\pi }-I_T^{\rho }{\mathrm{\mid }}^2=E\sum _{[s,t]\in \eta }(H_s^{\pi }-H_s^{\rho }{)}^2(M_t-M_s{)}^2\le E\left[R_{\pi ,\rho }^2{Q}_T^{\eta }\right],}$ where $R_{\pi ,\rho }={\sup }_s\mathrm{\mid }{H}_s^{\pi }-H_s^{\rho }\mathrm{\mid }$. Uniform continuity [F3] gives $R_{\pi ,\rho }\to 0$ almost surely as both meshes tend to zero, and $R_{\pi ,\rho }\le 2n$. Cauchy–Schwarz [F4] and dominated convergence [F2] therefore make the last expectation tend to zero, using the uniform $L^2$ bound on $Q_T^{\eta }$ from step 1.2. Thus the terminal sums $I_T^{\pi }$ are Cauchy in $L^2$; completeness [F2] gives their $L^2$ limit, and Doob's inequality [F2] gives the same Cauchy property for ${\sup }_{t\le T}\mathrm{\mid }{I}_t^{\pi }-I_t^{\rho }\mathrm{\mid }$; for the continuous-time supremum, apply the discrete inequality on increasing finite rational grids and pass to their union by monotone convergence [F2]. Choose a subsequence whose successive expected squared sup errors are at most $2^{-4k}$. Markov's inequality and the union bound imply that this subsequence converges uniformly almost surely to a process $I$ with almost-sure continuous paths; define its value at each time as the finite pointwise limit of those sums, and zero if that limit does not exist. Since the approximating sums are continuous and adapted, this rule is adapted and progressively measurable, so its stopped values are measurable. The full family converges uniformly in probability to $I$. For fixed subsequence index $m$, the squared path-sup errors from its $m$th term to later terms converge almost surely to ${\sup }_{t\le T}\mathrm{\mid }{I}_t^{{\pi }_{n_m}}-I_t{\mathrm{\mid }}^2$. Applying monotone convergence to the infima of these tails gives Fatou's inequality; the $L^2$-Cauchy bound then shows these errors tend to zero in $L^2$. The event-integral martingale identities therefore pass to the $L^1$ limit.

The exact identity in step 1.2 shows that the partial-increment sums $Q^{\pi }$ converge uniformly in probability to $Q_t=M_t^2-M_0^2-2I_t$. A subsequence converges uniformly almost surely, so $Q$ is continuous. The step-endpoint sum differs from $Q^{\pi }$ by at most the maximal squared oscillation of $M$ over a partition interval; [F3] makes this tend to zero almost surely. Step-endpoint sums are nondecreasing. A uniformly convergent almost-sure subsequence of those sums therefore shows that $Q$ is increasing. Thus both covariation conventions converge to $Q$ along every vanishing-mesh sequence. Since $2I=M^2-M_0^2-Q$ is a martingale, $M^2-Q$ is a martingale; uniqueness of the partition-limit object follows from uniqueness of limits in probability.

Apply the scalar result to $M^{(n)}$ for each $n$. For $m\ge n$, the partial-increment sums for $M^{(m)}$ evaluated at $t\wedge {\tau }_n$ equal those for $M^{(n)}$ evaluated at $t$, pathwise. Their uniform probability limits therefore satisfy $[M^{(m)}{]}^{{\tau }_n}=[M^{(n)}]$. The limits paste to an adapted continuous increasing $[M]$, since ${\tau }_n=T$ eventually almost surely; alternatively its adapted values are the pointwise limits of a uniformly almost-surely convergent subsequence of the original partition sums, with zero assigned when a limit is not finite. The stopped identity $(M^2-[M]{)}^{{\tau }_n}=(M^{(n)}{)}^2-[M^{(n)}]$ is a true martingale by step 3.1, so $M^2-[M]$ is local. For vector $M$, choose localizing sequences for $M^i$ and $M^j$ and stop both at their minimum ${\sigma }_n$. The bounded-stopping fact in step 1.1 makes the two stopped coordinates true martingales, so $M^i+M^j$ and $M^i-M^j$ are local martingales. Apply the scalar result to them and set ${\displaystyle [M^i,M^j]=\frac{1}{4}([M^i+M^j]-[M^i-M^j])\mathrm{.}}$ The identity $4xy=(x+y{)}^2-(x-y{)}^2$ makes this the limit of cross-increment sums. Taking the same linear combination of the two square-compensated local martingales gives $M^i{M}^j-[M^i,M^j]$ local martingale.

Let $H$ be a bounded adapted process with everywhere-continuous paths and $N$ a bounded continuous martingale with $N_0=0$. For each partition $\pi$ use $H_s^{\pi }=H_{t_k}$ for $t_k\le s<t_{k+1}$ (with the last coefficient at $T$), and set ${\displaystyle J_t^{\pi }=\sum _k{H}_{t_k}(N_{t\wedge t_{k+1}}-N_{t\wedge t_k})\mathrm{.}}$ On a common refinement $\eta$, martingale orthogonality [F1] gives ${\displaystyle E\mathrm{\mid }{J}_T^{\pi }-J_T^{\rho }{\mathrm{\mid }}^2\le E[\underset{s}{\sup }\mathrm{\mid }{H}_s^{\pi }-H_s^{\rho }{\mathrm{\mid }}^2{Q}_T^{\eta }(N)]\mathrm{.}}$ The first factor tends to zero in $L^2$ by boundedness, continuity of $H$, and dominated convergence [F2]; the second has uniformly bounded $L^2$ norm by steps 1.2–2.1 applied to $N$. Cauchy–Schwarz [F4] makes the terminal sums Cauchy in $L^2$, whose limit exists by completeness [F2]. Doob's inequality [F2], applied on increasing finite rational grids and then passed to their union by monotone convergence [F2], makes the process sums Cauchy in $L^2$ of the path supremum. A subsequence with summable squared-sup errors converges uniformly almost surely; define the limit at each time where finite and zero otherwise, giving an adapted progressively measurable version with almost-sure continuous paths. Fatou gives $L^2$ convergence of the full process sums to it, so the event-integral martingale identities pass to the $L^1$ limit. This defines $\int HdN$ as the left-step limit. Localization extends the construction to continuous local martingales and locally bounded continuous adapted integrands.

Let $X=M+A$ and $Z=N+C$ as in the statement. First stop $M,N$ at common bounded hitting localizers from step 1.1; step 4.2 then defines their mutual integrals, and removing those stops gives the localized integrals. For a continuous finite-variation path $A$ and a continuous path $N$, summation by parts gives ${\sum }_k{A}_{t_k}{\mathrm{\Delta }}_{k}N=A_t{N}_t-A_0{N}_0-{\sum }_k{N}_{t_{k+1}}{\mathrm{\Delta }}_{k}A$, so these left sums converge pathwise to $A_t{N}_t-A_0{N}_0-{\int }_0^t{N}_{s}d{A}_s$; this defines $\int AdN$ even if $A$ is not adapted. The analogous observation applies to $C$ and $M$. On every partition, including a final partial interval, the exact identity $X_t{Z}_t-X_0{Z}_0={\sum }_k{X}_{t_k}{\mathrm{\Delta }}_{k}Z+{\sum }_k{Z}_{t_k}{\mathrm{\Delta }}_{k}X+{\sum }_k{\mathrm{\Delta }}_{k}X{\mathrm{\Delta }}_{k}Z$ holds. The first two sums converge by the preceding constructions and ordinary Lebesgue–Stieltjes integration. Each cross term containing a finite-variation increment vanishes, for example $\mathrm{\mid }{\sum }_k{\mathrm{\Delta }}_{k}M{\mathrm{\Delta }}_{k}C\mathrm{\mid }\le {\max }_k\mathrm{\mid }{\mathrm{\Delta }}_{k}M\mathrm{\mid }\mathrm{Var}(C)\to 0$; the same estimate is uniform in terminal time. The remaining cross sums converge uniformly in probability to $[M,N]$ by polarization in step 4.1. The limiting identity holds simultaneously at rational times and hence, by continuity, on the whole horizon. This proves the product formula without requiring adaptedness of $A,C$.

If $V$ is a continuous finite-variation local martingale with the stated representative convention, put $U=V-V_0$. Along every vanishing-mesh partition, its partial square sums are bounded by the maximal path oscillation times ${\mathrm{Var}}_{[0,T]}(V)$, which tends to zero almost surely. Thus $[U]=0$ by uniqueness of the limit in step 4.1, and $U^2$ is a nonnegative local martingale started at zero. For its localizers ${\lambda }_k$, the true stopped martingales satisfy $E{U}_{t\wedge {\lambda }_k}^2=0$; hence their values are zero almost surely at each rational $t$ and at $T$. Continuity and ${\lambda }_k\uparrow T$ imply $U\equiv 0$ on a common full event. Therefore $V$ is constant.

For each fixed $a,c$, the change $b=d+c$ is the translation in [F1]; the full map $S$ is invertible linear with determinant one. Let $E$ be a Lebesgue null set outside which the given factorization holds. Applying [F3] to $1_E$ shows that $S^{-1}E$ is null, so $q(S(a,d,c))=p_A(a)f(d)p_C(c)$ almost everywhere. For a nonnegative Borel test $\mathrm{\Phi }$, the function $(a,b,c)\mapsto \mathrm{\Phi }(a,b-c,c)q(a,b,c)$ is Lebesgue measurable. Change of variables [F3] gives $E\mathrm{\Phi }(A,B-C,C)=\int \mathrm{\Phi }(a,d,c)q(S(a,d,c))dadddc=\int \mathrm{\Phi }(a,d,c)p_A(a)f(d)p_C(c)dadddc$. In this last integral the factors depend on separate coordinates, and $\mathrm{\Phi }$ is measurable for the product Borel sigma-algebra. The integrand is therefore measurable for the product of the factor Lebesgue sigma-algebras; the completed-product identification in [F3] identifies its Euclidean integral with the product integral to which [F2] applies. This establishes the transformed joint density without assuming product measurability of $f(b-c)$.

Integrating this joint density over $c$ and using $\int p_C(c)dc=1$, Tonelli [F2] yields $p_{A,D}(a,d)=p_A(a)f(d)$. Its marginals are $p_A$ and $f$, so the product identity proves independence and identifies the density of $D$.

If $T=0$, set $B_0=0$; the integral, bracket and finite-horizon Brownian claims are immediate. Henceforth suppose $T>0$. Form left-step approximations to $\sigma$ on deterministic dyadic partitions, retaining ${\sigma }_0$ at time zero. They are predictable by [F1]. Take their pointwise limit where it exists as a finite invertible matrix with inverse norm at most $C$, and use $C^{-1}{I}_d$ otherwise; necessarily $C>0$. The resulting predictable $\widehat{\sigma }$ agrees with $\sigma$ at all times on its full continuity event. Continuity on the compact horizon gives ${\int }_0^T\mathrm{\parallel }{\widehat{\sigma }}_s{\mathrm{\parallel }}_F^{2}ds<\mathrm{\infty }$ almost surely, and its bracket identity is the given one. Thus [F9] supplies a Brownian motion $B=\int {\widehat{\sigma }}^{-1}dM$ adapted to the original filtration, with bracket $t{I}_d$. This invocation makes no raw stopped evaluation of an unnormalized almost-surely continuous process.

This integral also has the left-step construction required in the Statement. On the augmented, normalized stopped intervals of [F9], put $H={\sigma }^{-1}$ and use its bounded deterministic left-step approximations. They converge uniformly on each continuous path and are bounded by $C$. The bracket-energy isometry and maximal estimate proved in [F9], together with dominated convergence [F7] and the integrable stopped bracket trace, give localized $L^2$ convergence of their integrals to the same $B$. Equality with the predictable reconstruction follows from uniqueness in that energy completion. Its original-adapted continuous version and Brownian increment identities are supplied by [F9]; the bracket is $[B^i,B^j{]}_t={\delta }_{ij}t$. Hence every asserted conclusion holds on the given filtration.

For each $a\in \mathcal{C}$, choose unit eigenvectors $e_i(a)$ and eigenvalues $s_i(a)\ge 0$, so $a={\sum }_{i=1}^d{s}_i(a)e_i(a)e_i(a{)}^{\mathsf{T}}$. For $0<h<1$, define $L_h^{a}u(t,x)={\sum }_i\frac{s_i(a)}{h^2}(u(t,x+h{e}_i)-2u(t,x)+u(t,x-h{e}_i))+\frac{r(a)}{h}(u(t+h,x)-u(t,x))\mathrm{.}$ All off-center coefficients are nonnegative. Their sum is $q_h(a)=2\mathrm{tr}(a)\mathrm{/}{h}^2+r(a)\mathrm{/}h$.

We will use this elementary representation fact on a bounded open ball $B$: if a linear functional $T$ on compact smooth tests satisfies $\mathrm{\mid }T\eta \mathrm{\mid }\le C\mathrm{\parallel }\eta {\mathrm{\parallel }}_1$, it is integration against a function of absolute value at most $C$. Indeed, $\mathrm{\parallel }\eta {\mathrm{\parallel }}_1\le \mathrm{\mid }B{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }\eta {\mathrm{\parallel }}_2$ extends $T$ continuously from the dense test subspace to $L^2(B)$. Real $L^2$ duality represents it by $g\in L^2(B)$. The original $L^1$ estimate extends to all $L^2$ tests, because their $L^2$ approximations also converge in $L^1$. Test with $\mathrm{sgn}(g){\mathbf{1}}_{\{\mathrm{\mid }g\mathrm{\mid }>C+ϵ\}}$ to obtain a contradiction unless that set is null. Hence $\mathrm{\mid }g\mathrm{\mid }\le C$. The unique local representatives agree on overlaps.

Write $L_h^{a}u=M_h^{a}u-q_h(a)u$, where $M_h^a$ is its off-center positive operator, and define $P_{h}u(z)={\sup }_{a\in \mathcal{C}}\frac{M_h^{a}u(z)+k(a)f(z)}{\lambda +q_h(a)}\mathrm{.}$ Put $Q_h=2h^{-2}+h^{-1}$ and ${\gamma }_h=Q_h\mathrm{/}(\lambda +Q_h)<1$. Positivity and $M_h^{a}1=q_h(a)\le Q_h$ give $\mathrm{\parallel }{P}_{h}u-P_{h}w{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\gamma }_h\mathrm{\parallel }u-w{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ The same argument bounds $\sup (P_{h}u-P_{h}w{)}^{+}$ by ${\gamma }_h\sup (u-w{)}^{+}$.

Starting with $u_0=0$, iterate $u_{j+1}=P_h{u}_j$. The contraction bounds successive differences by a summable geometric sequence. Their sums converge uniformly to a bounded function $v_h$, since a uniformly Cauchy sequence of bounded real functions has a bounded pointwise limit and converges uniformly. The contraction estimate gives $P_h{v}_h=v_h$ and uniqueness among bounded fixed points. Multiplying the individual residuals by their denominators, which lie in $[\lambda ,\lambda +Q_h]$, shows that this is equivalent to ${\sup }_{a\in \mathcal{C}}\{L_h^a{v}_h-\lambda v_h+k(a)f\}=0.\text{(1)}$.

Let $\psi$ be a supersolution as in the Statement. The centered second difference is an average of the corresponding directional second derivative, and the forward time difference is an average of ${\mathrm{\partial }}_t\psi$. Bounded uniform continuity gives $E_h\to 0$ such that, uniformly in $z$ and $a$, $\mid L_h^a\psi -r(a){\mathrm{\partial }}_t\psi -a:D_x^2\psi \mid \le E_h\mathrm{.}$ For example, the sum of the spatial-Hessian and time-derivative moduli of continuity at scale $h$ suffices, since $r(a)+\mathrm{tr}a=1$. Therefore ${\mathrm{\Psi }}_h=\psi +E_h\mathrm{/}\lambda$ satisfies $P_h{\mathrm{\Psi }}_h\le {\mathrm{\Psi }}_h$.

AM–GM applied to $r(a),s_1(a),\dots ,s_d(a)$, whose sum is $1$, gives $0\le k(a)\le 1\mathrm{/}n$. Hence $P_h$ preserves $0\le u\le V$: at the upper bound its numerator is at most $q_h(a)V+\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}n=(\lambda +q_h(a))V$. Thus $0\le v_h\le V$.

If $u$ has Lipschitz constant at most $L$, each affine control expression in $P_{h}u$ has Lipschitz constant at most $\frac{q_h(a)L+L_f\mathrm{/}n}{\lambda +q_h(a)}\le L\mathrm{.}$ A supremum preserves this common bound. Iteration and uniform convergence therefore make every $v_h$ globally $L$-Lipschitz.

If $u(z)+N\mathrm{\mid }z{\mathrm{\mid }}^2$ is convex, every translate of $u$ has the same semiconvexity constant $N$. The numerator in each affine control expression has semiconvexity constant at most $q_h(a)N+N_f\mathrm{/}n$. After division this is at most $N$, since $N_f\mathrm{/}n=\lambda N$. Adding $N\mathrm{\mid }z{\mathrm{\mid }}^2$ makes each expression convex; the translation corrections to its quadratic term are affine plus constants. Their supremum is convex. Starting from $0$ and passing to the uniform limit proves that $v_h+N\mathrm{\mid }z{\mathrm{\mid }}^2$ is convex.

Take a sequence $h_j\downarrow 0$. Diagonally select a subsequence along which $v_{h_j}$ converges at every rational point. Boundedness gives the successive subsequences, and finite rational nets plus the common Lipschitz constant give uniform convergence on every compact set. Call the limit $v$. It satisfies $0\le v\le V$, the global $L$-Lipschitz bound, and convexity of $v+N\mathrm{\mid }z{\mathrm{\mid }}^2$.

The one-sided contraction estimate gives $\sup (v_h-{\mathrm{\Psi }}_h{)}^{+}\le {\gamma }_h\sup (v_h-{\mathrm{\Psi }}_h{)}^{+}\mathrm{.}$ The supremum is finite, so it vanishes. Pass along the chosen subsequence to obtain $v\le \psi$. This proves the comparison assertion for the same constructed limit.

Fix $a\in \mathcal{C}$ and a nonnegative compactly supported smooth test $\eta$. Equation (1) implies $L_h^a{v}_h-\lambda v_h+k(a)f\le 0$. Integrate against $\eta$ and transfer translations to $\eta$. Its adjoint centered differences converge uniformly, on a common compact support, to the spatial second derivatives; its adjoint forward difference converges to $-{\mathrm{\partial }}_t\eta$. Local uniform convergence of $v_h$ gives $\int v(a:D_x^2\eta -r(a){\mathrm{\partial }}_t\eta -\lambda \eta )+\int k(a)f\eta \le 0.\text{(2)}$ Thus each fixed-control inequality holds as a distribution.

Choose a nonnegative smooth compactly supported unit-mass kernel $\rho$ and put ${\rho }_{ϵ}(z)={ϵ}^{-n}\rho (z\mathrm{/}ϵ)$, $v_{ϵ}=v\ast {\rho }_{ϵ}$. The common Lipschitz bound gives $\mathrm{\mid }{D}_z{v}_{ϵ}\mathrm{\mid }\le L$, and $v_{ϵ}\to v$ locally uniformly. Integration by parts bounds each distributional directional derivative by $L\mathrm{\parallel }\eta {\mathrm{\parallel }}_1$. Step 1.2 represents these by bounded weak derivatives. Using a countable dense set of unit directions gives $\mathrm{\mid }{D}_{z}v\mathrm{\mid }\le L$ almost everywhere.

Semiconvexity gives $D_z^2{v}_{ϵ}\ge -2NI$. For a unit spatial vector $e$, insert the rank-one control $a=e{e}^{\mathsf{T}}$ into (2). It has $r(a)=k(a)=0$, so $D_{ee}v-\lambda v\le 0$ as distributions. Convolution makes this $D_{ee}{v}_{ϵ}\le \lambda v_{ϵ}$ everywhere. Taking a countable dense set of $e$ and then continuity yields $-2NI\le D_x^2{v}_{ϵ}\le \lambda v_{ϵ}I\le \lambda VI\mathrm{.}\text{(3)}$.

Set $C=\max (2N,\lambda V)$. Every spatial Hessian entry in (3) has absolute value at most $C$. For a compact smooth test $\eta$, $\int v{\mathrm{\partial }}_{x_i{x}_j}\eta ={\lim }_{ϵ\downarrow 0}\int (D_{ij}{v}_{ϵ})\eta \mathrm{.}$ Step 1.2 therefore represents this distribution by an actual weak derivative $D_{ij}v\in L^{\mathrm{\infty }}$, with norm at most $C$. Passing the directional bounds through tests gives $-2NI\le D_x^{2}v\le \lambda vI$ almost everywhere. The control $a=0$ in (2) similarly gives ${\mathrm{\partial }}_{t}v\le \lambda v$.

Each individually existing derivative commutes with convolution by its weak-test identity: if $g=D^{\beta }v$, where $\beta$ is a first index or two spatial indices, test against $y\mapsto {\rho }_{ϵ}(z-y)$ to obtain $D^{\beta }(v\ast {\rho }_{ϵ})(z)=\int g(y){\rho }_{ϵ}(z-y)dy\mathrm{.}$ The factors $(-1{)}^{\mathrm{\mid }\beta \mathrm{\mid }}$ cancel. This argument uses only that particular weak derivative. All expressions in (2) now have bounded-function representatives. Discard the null sets for a countable dense subset of $\mathcal{C}$; continuity in $a$, including continuity of the determinant at singular matrices, extends the inequalities to every $a$ on one common full-measure set.

By semiconvexity and the local Alexandrov theorem, almost every $z_0$ has a direction-uniform expansion $v(z)=q(z)+o(\mathrm{\mid }z-z_0{\mathrm{\mid }}^2)$ for a quadratic polynomial $q$ with $q(z_0)=v(z_0)$. Require also that ball averages recover each bounded weak first and spatial second derivative $g$, and $g^2$, at $z_0$. These finitely many requirements remove only a null set. Expanding the square gives $\frac{1}{\mathrm{\mid }B(z_0,s)\mathrm{\mid }}{\int }_{B(z_0,s)}\mathrm{\mid }g-g(z_0){\mathrm{\mid }}^2⟶0,$ so Cauchy–Schwarz gives mean absolute convergence as well.

The first-order blowups $[v(z_0+sy)-v(z_0)]\mathrm{/}s$ converge locally uniformly to $Dq(z_0)\cdot y$. Their weak derivatives are $D_{z}v(z_0+sy)$, which converge locally in $L^1$ to the weak derivative values at $z_0$ by step 10.2. Distributional limits identify those values with $Dq(z_0)$. Similarly the second-order blowups $\frac{v(z_0+sy)-v(z_0)-sDq(z_0)\cdot y}{s^2}$ converge locally uniformly to $\frac{1}{2}{y}^{\mathsf{T}}{D}^{2}qy$, while their spatial second weak derivatives are $D_x^{2}v(z_0+sy)$ and converge locally in $L^1$ to $D_x^{2}v(z_0)$. Thus the spatial block of $D^{2}q$ equals that actual weak Hessian. These scaled weak identities follow directly by changing variables in the compact-test identities.

Fix $ϵ>0$ and set $q_{\pm }(z)=q(z)\pm ϵ\mathrm{\mid }z-z_0{\mathrm{\mid }}^2$. On a sufficiently small closed ball centered at $z_0$, the expansion makes $q_{+}$ a strict upper contact and $q_{-}$ a strict lower contact. Let $z_j$ maximize $v_{h_j}-q_{+}$ there. Local uniform convergence implies $z_j\to z_0$ and $m_j=(v_{h_j}-q_{+})(z_j)\to 0$. For large $j$, every stencil neighbor of $z_j$ remains in that ball. Nonnegative off-center weights then give $0\le {\sup }_a\{L_{h_j}^a{q}_{+}-\lambda q_{+}+k(a)f\}(z_j)-\lambda m_j$.

Centered spatial differences are exact on a quadratic polynomial; its forward time difference differs from its time derivative by $h_j{\mathrm{\partial }}_{tt}{q}_{+}\mathrm{/}2$. This error is uniform in the controls, since $0\le r(a)\le 1$. Bounded coefficients and uniform Lipschitz continuity of $f$ therefore allow passage in step 11.2 to ${\max }_a\{r(a){\mathrm{\partial }}_t{q}_{+}+a:D_x^2{q}_{+}-\lambda q_{+}+k(a)f\}(z_0)\ge 0$.

Minimize $v_{h_j}-q_{-}$ on the same ball. The identical positive-weight calculation reverses the inequality and gives the corresponding maximum for $q_{-}$ nonpositive in the limit. Let $ϵ\downarrow 0$ in both conclusions. Continuity of the finite-dimensional maximum gives zero for $q$. By step 11.1 its first and spatial second coefficients are the actual weak derivatives of $v$ at $z_0$. This proves the asserted Bellman equation outside a null set.

At a point where the equation and derivative bounds hold, put $S=\lambda v-{\mathrm{\partial }}_{t}v\ge 0$ and $H=\lambda vI-D_x^{2}v\ge 0$. The equation becomes ${\max }_{\begin{array}{c}r\ge 0,a\ge 0\\ r+\mathrm{tr}a=1\end{array}}\{-rS-a:H+(r\det a{)}^{1\mathrm{/}n}f\}=0.\text{(4)}$ In particular every normalized control satisfies $rS+a:H\ge (r\det a{)}^{1\mathrm{/}n}f$.

For $\delta >0$, let $S_{\delta }=S+\delta$, $H_{\delta }=H+\delta I$, and $c_{\delta }=(S_{\delta }^{-1}+\mathrm{tr}{H}_{\delta }^{-1}{)}^{-1},r_{\delta }=c_{\delta }\mathrm{/}{S}_{\delta },a_{\delta }=c_{\delta }{H}_{\delta }^{-1}\mathrm{.}$ This is a normalized control, and $r_{\delta }S+a_{\delta }:H\le n{c}_{\delta },(r_{\delta }\det a_{\delta }{)}^{1\mathrm{/}n}=\frac{c_{\delta }}{(S_{\delta }\det H_{\delta }{)}^{1\mathrm{/}n}}\mathrm{.}$ Insert it into the inequality from step 14.1 and cancel $c_{\delta }>0$ to get $f\le n(S_{\delta }\det H_{\delta }{)}^{1\mathrm{/}n}$. Sending $\delta \downarrow 0$ gives $f\le n(S\det H{)}^{1\mathrm{/}n}$, including $S=0$ or singular $H$.