Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers

Properties of $L$. Fix $v\in {\mathbb{R}}^n$. Superlinearity of $H$ gives $R>0$ with $H(p)\ge 2\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }p\mathrm{\mid }$ for $\mathrm{\mid }p\mathrm{\mid }\ge R$, so $p\cdot v-H(p)\le -\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }p\mathrm{\mid }\le 0$ there; on the closed ball $\bar{B}(0,R)$, which is closed and bounded and hence compact by [F4], the continuous function $p\mapsto p\cdot v-H(p)$ (continuity of $H$ is [F3]) attains a maximum $M_v<\mathrm{\infty }$ by [F4]. Hence $p\cdot v-H(p)\le \max \{M_v,0\}<\mathrm{\infty }$ for every $p$, while $L(v)\ge -H(0)$ because $p=0$ is admissible; by the least-upper-bound property of [F2], $L(v)$ is a real number. Convexity of $L$: for fixed $p$ the map $v\mapsto p\cdot v-H(p)$ is affine, and a pointwise supremum of affine functions is convex; $L$ is real-valued, so it is finite and convex on the open convex set ${\mathbb{R}}^n$ and therefore continuous by [F3]. Superlinearity: fix $a>0$; for $v\mathrm{\ne }0$ the admissible test point $p:=av\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$ gives $L(v)\ge a\mathrm{\mid }v\mathrm{\mid }-H(av\mathrm{/}\mathrm{\mid }v\mathrm{\mid })\ge a\mathrm{\mid }v\mathrm{\mid }-{\max }_{\mathrm{\mid }q\mathrm{\mid }\le a}H(q)$, where the maximum is finite by [F3] and [F4]; dividing by $\mathrm{\mid }v\mathrm{\mid }$ and letting $\mathrm{\mid }v\mathrm{\mid }\to \mathrm{\infty }$ gives ${\mathrm{lim\hspace{0.17em}inf}}_{\mathrm{\mid }v\mathrm{\mid }\to \mathrm{\infty }}L(v)\mathrm{/}\mathrm{\mid }v\mathrm{\mid }\ge a$, and since $a>0$ was arbitrary, $L(v)\mathrm{/}\mathrm{\mid }v\mathrm{\mid }\to +\mathrm{\infty }$.

Attainment and localisation. Fix $x$, $t>0$ and $\delta >0$, and put $\phi (y):=u_0(y)+tL((x-y)\mathrm{/}t)$, so that $Q_t{u}_0(x)=\inf \phi$ by [F1]. The competitor $y=x$ gives $Q_t{u}_0(x)\le \phi (x)=u_0(x)+tL(0)\le \mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}+tL(0)=:A<\mathrm{\infty }$, where $L(0)\in \mathbb{R}$. Also every term is at least $\inf u_0-tH(0)>-\mathrm{\infty }$ by [F2], so $Q_t{u}_0(x)\in \mathbb{R}$. Let $y$ satisfy $\phi (y)\le Q_t{u}_0(x)+\delta$; then $tL\left(\frac{x-y}{t}\right)\le A+\delta -\inf u_0\le 2\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}+tL(0)+\delta =:C\mathrm{.}$ Superlinearity of $L$ from step 1.1 gives $R>0$ such that $L(w)>C\mathrm{/}t$ whenever $\mathrm{\mid }w\mathrm{\mid }\ge R$. If $\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}t\ge R$, then $tL((x-y)\mathrm{/}t)>C$, contradicting the preceding bound. Hence $\mathrm{\mid }y-x\mathrm{\mid }<tR$, so every $\delta$-near-minimiser lies in the closed ball $\bar{B}(x,\rho )$ with $\rho :=tR<\mathrm{\infty }$; the radius depends only on $t,\delta ,\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$ and the fixed $H$ (and may harmlessly be viewed as a function of $x$ as in the statement). For attainment, let $S:=\{y\in {\mathbb{R}}^n:\phi (y)\le Q_t{u}_0(x)+1\}$. By the definition of the finite infimum, $S$ is nonempty. It is closed by continuity of $\phi$ and bounded by the localisation just proved with $\delta =1$, so [F4] makes it compact. The continuous function $\phi$ attains a minimum on $S$ at some $y_{\ast }$. This minimum equals the global infimum: it is at least $Q_t{u}_0(x)$, and for every $0<\epsilon <1$ the infimum property gives $y$ with $\phi (y)<Q_t{u}_0(x)+\epsilon$, which lies in $S$, so the minimum is at most $Q_t{u}_0(x)+\epsilon$. Thus $\phi (y_{\ast })=Q_t{u}_0(x)$ and the infimum is attained, without selecting a sequence.

Real-valuedness of $Q_t{u}_0$. For $t>0$ each term satisfies $u_0(y)+tL((x-y)\mathrm{/}t)\ge {\inf }_{{\mathbb{R}}^n}{u}_0-tH(0)>-\mathrm{\infty }$ by the estimate $L\ge -H(0)$ of step 1.1, and the value at $y=x$ is finite; hence $Q_t{u}_0(x)\in \mathbb{R}$ by [F1]. For $t=0$ this is $Q_0{u}_0=u_0$, real-valued by hypothesis.

Conclusion. Part (1) is step 1.1, part (2) is step 2.1, and part (3) is step 3.1.