Time penalisation moves a doubling-variables maximum away from the terminal boundary

The subsolution penalty. Let $\varphi \in C^1(Z)$ and suppose ${\tilde{u}}_{\eta }-\varphi$ has a local maximum at $z_0=(x_0,t_0)\in Z$. Then $u-\psi$ has a local maximum at $z_0$ for $\psi (z):=\varphi (z)+\eta \mathrm{/}(T-t)$, which is $C^1$ on $Z$ with ${\psi }_t={\varphi }_t+\eta \mathrm{/}(T-t{)}^2$ and $D\psi =D\varphi$ by [F2]; the subsolution inequality [F1] gives ${\varphi }_t(z_0)+\eta \mathrm{/}(T-t_0{)}^2+H(x_0,t_0,D\varphi (z_0))\le 0$, that is ${\varphi }_t+H(x_0,t_0,D\varphi (z_0))\le -\eta \mathrm{/}(T-t_0{)}^2<0$.

The supersolution penalty and the uniform terminal limits. If ${\tilde{v}}_{{\eta }^{\mathrm{\prime }}}-\varphi$ has a local minimum at $z_0\in Z$, then $v-\psi$ has a local minimum at $z_0$ for $\psi :=\varphi -{\eta }^{\mathrm{\prime }}\mathrm{/}(T-t)$, $C^1$ with ${\psi }_t={\varphi }_t-{\eta }^{\mathrm{\prime }}\mathrm{/}(T-t{)}^2$; the supersolution inequality [F1] gives ${\varphi }_t+H(x_0,t_0,D\varphi (z_0))\ge {\eta }^{\mathrm{\prime }}\mathrm{/}(T-t_0{)}^2>0$. For the terminal limits, ${\tilde{u}}_{\eta }(x,t)\le \sup u-\eta \mathrm{/}(T-t)$ and ${\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(x,t)\ge \inf v+{\eta }^{\mathrm{\prime }}\mathrm{/}(T-t)$ for every $x$, and the right-hand sides are independent of $x$ and tend to $-\mathrm{\infty }$, respectively $+\mathrm{\infty }$, as $t\uparrow T$.

Existence and finiteness of the maximum of ${\mathrm{\Phi }}_{\alpha ,\rho }$. On the product space ${\mathbb{R}}^n\times {\mathbb{R}}^n\times [0,T{]}^2$ the function ${\mathrm{\Phi }}_{\alpha ,\rho }$ is upper semicontinuous: it is built from the upper semicontinuous ${\tilde{u}}_{\eta }$, the function $-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}$, which is upper semicontinuous because $v$ is lower semicontinuous, and continuous terms, and at a sequence with $t\to T$ or $s\to T$ it tends to $-\mathrm{\infty }$ uniformly, since ${\tilde{u}}_{\eta }(x,t)\le \sup u-\eta \mathrm{/}(T-t)$ and $-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(y,s)\le -\inf v-{\eta }^{\mathrm{\prime }}\mathrm{/}(T-s)$; hence it takes the value $-\mathrm{\infty }$ on the terminal faces in the upper-semicontinuous sense fixed in [F3]. It is bounded above by $\sup u-\inf v<\mathrm{\infty }$, and its value at any diagonal point $(0,0,\tau ,\tau )$ with $0<\tau <T$ is finite, so $M:=\sup {\mathrm{\Phi }}_{\alpha ,\rho }\in \mathbb{R}$. For each $k\ge 1$ the set $A_k:=\{(\text{x,y,t,s}):{\mathrm{\Phi }}_{\alpha ,\rho }\ge M-1\mathrm{/}k\}$ is nonempty by the definition of $M$, closed by upper semicontinuity [F3], bounded because $\rho (\mathrm{\mid }x{\mathrm{\mid }}^2+\mathrm{\mid }y{\mathrm{\mid }}^2)\le \sup u-\inf v-M+1\mathrm{/}k\le \sup u-\inf v-M+1$ on $A_k$, and disjoint from the terminal faces because $M-1\mathrm{/}k$ is finite there; hence each $A_k$ is a compact subset of $A_1$ by [F4]. The family $\{A_k{\}}_{k\ge 1}$ is nested, so it has the finite intersection property, and [F5] applied in the compact set $A_1$ gives a point of ${\bigcap }_k{A}_k$, at which ${\mathrm{\Phi }}_{\alpha ,\rho }\ge M-1\mathrm{/}k$ for every $k$, hence ${\mathrm{\Phi }}_{\alpha ,\rho }\ge M$; since $M$ is an upper bound, ${\mathrm{\Phi }}_{\alpha ,\rho }=M$ there. Thus the maximum is attained and equals the finite number $M$, and every maximiser has $t,s<T$ because the terminal faces carry the value $-\mathrm{\infty }$.

Conclusion. Parts (1) and (2) of the statement are steps 1.1 and 1.2, and part (3) is step 1.3, whose construction nowhere selects a sequence or a point: the maximiser is obtained from the finite intersection property, which the cited lemma proves choice-free. Nothing in the argument rules out a maximiser with $t=0$ or $s=0$, since only the terminal faces $t=T$, $s=T$ carry the value $-\mathrm{\infty }$.