Comparison for periodic diagonal controlled-diffusion HJB equations

For any fixed lifts, a lattice minimizer exists: only finitely many integer vectors can give a norm no larger than the norm at $k=0$. If a minimizing difference $r$ has $\mathrm{\mid }r\mathrm{\mid }=D(x,y)<1\mathrm{/}2$, then for every nonzero $k\in {\mathbb{Z}}^d$, $\mathrm{\mid }k\mathrm{\mid }\ge 1$ and $r\cdot k\ge -\mathrm{\mid }r\mathrm{\mid }\mathrm{\mid }k\mathrm{\mid }$: for $r\mathrm{\ne }0$ this follows by expanding $\mid \mathrm{\mid }k\mathrm{\mid }r+\mathrm{\mid }r\mathrm{\mid }k{\mid }^2\ge 0$, and for $r=0$ it is immediate. Consequently $\mathrm{\mid }r+k{\mathrm{\mid }}^2-\mathrm{\mid }r{\mathrm{\mid }}^2=\mathrm{\mid }k{\mathrm{\mid }}^2+2r\cdot k\ge \mathrm{\mid }k\mathrm{\mid }(\mathrm{\mid }k\mathrm{\mid }-2\mathrm{\mid }r\mathrm{\mid })>0.$ Thus that minimizing difference is unique. Choose continuous local lifts near $x,y$ and keep the minimizing lattice vector fixed; their resulting difference $r^{\mathrm{\prime }}$ stays of norm less than $1\mathrm{/}2$ in a sufficiently small neighborhood. The same inequality for every nonzero $k$ shows that this difference remains the unique minimizer throughout that neighborhood. Hence there $D^2=\mathrm{\mid }{r}^{\mathrm{\prime }}{\mathrm{\mid }}^2$, a smooth quadratic in the local coordinates, justifying [F5].

Define $u(x,0)$ as the raw upper trace and $v(x,0)$ as the raw lower trace. Boundedness makes them finite. For example, if $x_k\to x$ and $u(x_k,0)$ approaches its limsup, the raw upper-trace definition and [A1] let us select interior $(y_k,t_k)$ with $\mathrm{\mid }{y}_k-x_k\mathrm{\mid }+t_k<1\mathrm{/}k$ and $u(y_k,t_k)>u(x_k,0)-1\mathrm{/}k$. Thus $y_k\to x$, $t_k\to 0$, and ${\mathrm{lim\hspace{0.17em}sup}}_{k}u(x_k,0)\le u(x,0)$. The lower-trace argument is dual. For a sequence with positive times tending to zero, the raw trace definition gives the USC/LSC inequalities directly; splitting a mixed sequence into positive-time and initial-edge subsequences proves the extensions are USC/LSC on the closed initial edge. The assumed order gives $u(x,0)\le v(x,0)$. Assume for contradiction that $u(x_{\ast },t_{\ast })-v(x_{\ast },t_{\ast })>0$ at some $0<t_{\ast }<T$. Choose $S$ with $t_{\ast }<S<T$, then choose $\eta ,\gamma >0$ sufficiently small that $m:=u(x_{\ast },t_{\ast })-v(x_{\ast },t_{\ast })-\eta t_{\ast }-\gamma \mathrm{/}(S-t_{\ast })>0.$ Put $P(t)=\eta t+\gamma \mathrm{/}(S-t)$. If $u(x,t)-v(y,s)-P(t)\ge 0$, boundedness bounds $P(t)$ above by $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $t\le S-c$ for some $c>0$.

Using [F5], define $M_{\epsilon }={\max }_{x,y\in {\mathbb{T}}^d,0\le t<S}\{u(x,t)-v(y,t)-P(t)-D(x,y{)}^2\mathrm{/}(2\epsilon )\}\mathrm{.}$ The maximum exists by step 1.2 and [F4]: the expression is USC, and its positive superlevel sets stay in a compact time interval bounded away from $S$. Also $M_{\epsilon }\ge m$. Write $M_0={\max }_{x,0\le t<S}(u(x,t)-v(x,t)-P(t))\ge m$, also attained. At any maximizing triple $(x_{\epsilon },y_{\epsilon },t_{\epsilon })$, boundedness gives $D(x_{\epsilon },y_{\epsilon }{)}^2=O(\epsilon )$. Compactness and semicontinuity imply $M_{\epsilon }\to M_0$: for every sequence $\epsilon \downarrow 0$, a subsequence of maximizers tends to $(x,x,t)$, and the limsup of the unpenalized expression is at most $M_0$, whereas $M_{\epsilon }\ge M_0$. Moreover $0\le D(x_{\epsilon },y_{\epsilon }{)}^2\mathrm{/}(4\epsilon )\le M_{2\epsilon }-M_{\epsilon }⟶0,$ by evaluating the $2\epsilon$ maximum at an $\epsilon$ maximizer. These conclusions hold uniformly over maximizing triples, by the same sequential compactness argument. No sequence of these maximizers can have $t_{\epsilon }\to 0$: then $x_{\epsilon },y_{\epsilon }\to x$ along a subsequence and semicontinuity and the initial order give limsup of their maximum values at most $-P(0)<0$, contradicting $M_{\epsilon }\ge m$.

For fixed sufficiently small $\epsilon$, introduce independent times and maximize ${\mathrm{\Phi }}_{\epsilon ,\delta }(x,t,y,s)=u(x,t)-v(y,s)-P(t)-D(x,y{)}^2\mathrm{/}(2\epsilon )-(t-s{)}^2\mathrm{/}(2\delta )$ over $x,y\in {\mathbb{T}}^d$, $0\le t<S$, $0\le s\le S$. Maxima exist by step 1.2 and [F4] and are at least $M_{\epsilon }\ge m$. Boundedness gives $\mathrm{\mid }t-s\mathrm{\mid }=O(\sqrt{\delta })$ and $t\le S-c$ uniformly. For fixed $\epsilon$, compactness shows that as $\delta \downarrow 0$ every limit of maximizing quadruples is a synchronized maximizer of $M_{\epsilon }$: the limsup of the expression without the temporal penalty is at most $M_{\epsilon }$, while the maximum is at least $M_{\epsilon }$. Hence the temporal penalty tends to zero too. By step 2.1 and [F5], for all sufficiently small $\epsilon$ these limiting times are positive and the spatial distance is less than $1\mathrm{/}2$. Consequently for each such $\epsilon$ and all sufficiently small $\delta$, both times are strictly in $(0,S)$ and $D(x,y{)}^2=\mathrm{\mid }r{\mathrm{\mid }}^2$ is smooth near the maximizer in local lifts with $r=x-y$ and $\mathrm{\mid }r\mathrm{\mid }=D(x,y)$. Thus all contacts used next are interior in the original cylinder.

At such a maximizer apply [F2] to $u(t,x)$ and $-v(s,y)$ with $\varphi (t,x,s,y)=\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(2\epsilon )+(t-s{)}^2\mathrm{/}(2\delta )+P(t),\mu =\epsilon \mathrm{.}$ It gives a closed upper jet of $u$ with gradient $(a,p)$ and Hessian $H$, and a closed lower jet of $v$ with gradient $(b,p)$ and Hessian $K$, where $p=r\mathrm{/}\epsilon ,a=(t-s)\mathrm{/}\delta +P^{\mathrm{\prime }}(t),b=(t-s)\mathrm{/}\delta ,a-b=P^{\mathrm{\prime }}(t)\ge \eta \mathrm{.}$ Negating the jet of $-v$ gives the lower jet of $v$ directly from its defining expansion and witnessing sequence. Let $X,Y$ be the spatial principal blocks of $H,K$. The full Hessian of $\varphi$ has no space-time mixed entries. Its spatial block is ${\epsilon }^{-1}B$ with $B=\left(\begin{array}{cc}I& -I\\ -I& I\end{array}\right)$ and $B^2=2B$. Restricting the upper matrix inequality in [F2] to vectors with zero temporal components therefore gives $\left(\begin{array}{cc}X& 0\\ 0& -Y\end{array}\right)\le (3\mathrm{/}\epsilon )B\mathrm{.}$ The temporal Hessian, including $P^{\mathrm{\prime }\mathrm{\prime }}(t)$ and the potentially large $1\mathrm{/}\delta$, contributes nothing to this spatial restriction.

For each control $c$ write $G_c(x,t,p,X)={\sum }_j({\alpha }_j^c(x,t)X_{jj}+b_j^c(x,t)p_j)+f^c(x,t)$. By [F3], $a\le {\max }_c{G}_c(x,t,p,X)$ and $b\ge {\max }_c{G}_c(y,s,p,Y)$. Finiteness gives $\eta \le a-b\le {\max }_c\{G_c(x,t,p,X)-G_c(y,s,p,Y)\}\mathrm{.}$ For ${\sigma }_j^c=\sqrt{{\alpha }_j^c}$, rationalization and [F1] give $\mathrm{\mid }{\sigma }_j^c(x,t)-{\sigma }_j^c(y,s)\mathrm{\mid }\le \frac{L}{2\sqrt{{\alpha }_{\ast }}}(\mathrm{\mid }r\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid })\mathrm{.}$ Evaluate the spatial matrix inequality of step 4.1 on each vector $({\sigma }_j^c(x,t)e_j,{\sigma }_j^c(y,s)e_j)$ and sum over $j$ to obtain ${\sum }_j({\alpha }_j^c(x,t)X_{jj}-{\alpha }_j^c(y,s)Y_{jj})\le \frac{3}{\epsilon }{\sum }_j\mathrm{\mid }{\sigma }_j^c(x,t)-{\sigma }_j^c(y,s){\mathrm{\mid }}^2\le \frac{3d{L}^2}{4{\alpha }_{\ast }\epsilon }(\mathrm{\mid }r\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid }{)}^2\mathrm{.}$ The drift difference is bounded by $C(\mathrm{\mid }r\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid })\mathrm{\mid }r\mathrm{\mid }\mathrm{/}\epsilon$, and the forcing difference by $L(\mathrm{\mid }r\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid })$, uniformly over the finite controls, using [F1]. Hence a constant $C$ independent of $\epsilon ,\delta$ satisfies $\eta \le C\{\frac{(\mathrm{\mid }r\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid }{)}^2}{\epsilon }+\frac{(\mathrm{\mid }r\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid })\mathrm{\mid }r\mathrm{\mid }}{\epsilon }+\mathrm{\mid }r\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid }\}\mathrm{.}$

First let $\delta \downarrow 0$ for fixed $\epsilon$, selecting a convergent subsequence of maximizing quadruples by [A1] and [F4]. Since $\epsilon$ is fixed, every term containing $\mathrm{\mid }t-s\mathrm{\mid }\mathrm{/}\epsilon$ vanishes. The remaining spatial pair is a maximizer of $M_{\epsilon }$, so step 5.1 gives $\eta \le C^{\mathrm{\prime }}(D(x_{\epsilon },y_{\epsilon }{)}^2\mathrm{/}\epsilon +D(x_{\epsilon },y_{\epsilon }))$. Now let $\epsilon \downarrow 0$. Both terms vanish by step 2.1, contradicting $\eta >0$. The assumed positive gap is impossible, so $u\le v$ at every interior point. The initial-edge assertion is its assumed order.