[proofplan] We prove the subsolution assertion first. A [test function](/page/Test%20Function) touching $u$ from above is modified by a positive quadratic term, which makes the contact strict on a compact neighbourhood contained in $Q$. Local [uniform convergence](/page/Uniform%20Convergence) then produces nearby maximum points for the approximating functions $u_k$ against the strictified test function, and these points converge to the original contact point. Applying the viscosity inequality for $u_k$ at those points and passing to the limit using the locally uniform convergence of both $u_k$ and $F_k$ gives the desired inequality for $u$. The supersolution case is the same argument with a negative quadratic perturbation and local minima. [/proofplan]

[step:Strictify an upper contact point inside a compact neighbourhood] Assume that each $u_k$ is a viscosity subsolution. Fix a point $(t_0,x_0) \in Q$ and a test function $\phi \in C^1(Q)$ such that $u-\phi$ has a local maximum at $(t_0,x_0)$. Choose $r>0$ such that the compact Euclidean ball \begin{align*} K := \overline{B}_{\mathbb{R}^{1+n}}((t_0,x_0),r) \end{align*} is contained in $Q$ and such that \begin{align*} u(t,x)-\phi(t,x) \le u(t_0,x_0)-\phi(t_0,x_0) \end{align*} for every $(t,x) \in K$. Fix $\alpha>0$. Define the strictified test function \begin{align*} \phi_\alpha: Q &\to \mathbb{R} \end{align*} by \begin{align*} \phi_\alpha(t,x) := \phi(t,x)+\alpha\bigl((t-t_0)^2+|x-x_0|^2\bigr). \end{align*} Then $\phi_\alpha \in C^1(Q)$, and $u-\phi_\alpha$ has a strict maximum over $K$ at $(t_0,x_0)$. Indeed, for every $(t,x) \in K$ with $(t,x) \ne (t_0,x_0)$, \begin{align*} u(t,x)-\phi_\alpha(t,x) \le u(t_0,x_0)-\phi(t_0,x_0)-\alpha\bigl((t-t_0)^2+|x-x_0|^2\bigr) < u(t_0,x_0)-\phi_\alpha(t_0,x_0). \end{align*} Let $\partial K$ denote the boundary of $K$ in the Euclidean space $\mathbb{R}^{1+n}$. In particular, because $\partial K$ is compact and does not contain $(t_0,x_0)$, the positive number \begin{align*} \eta_\alpha := u(t_0,x_0)-\phi_\alpha(t_0,x_0)-\max_{(t,x)\in \partial K}\bigl(u(t,x)-\phi_\alpha(t,x)\bigr) \end{align*} is well-defined and satisfies $\eta_\alpha>0$. [/step]

[step:Find nearby maximum points for the approximating subsolutions]For each $k \in \mathbb{N}$, define the [continuous function](/page/Continuous%20Function) \begin{align*} w_{k,\alpha}: K &\to \mathbb{R} \end{align*} by \begin{align*} w_{k,\alpha}(t,x) := u_k(t,x)-\phi_\alpha(t,x). \end{align*} Since $K$ is compact and $w_{k,\alpha}$ is continuous, the extreme value theorem implies that $w_{k,\alpha}$ attains a maximum on $K$; choose one maximizer and denote it by $(t_k,x_k) \in K$. The sequence $(t_k,x_k)$ converges to $(t_0,x_0)$. To prove this, let $(t_{k_j},x_{k_j})$ be any convergent subsequence, and write its limit as $(\bar{t},\bar{x}) \in K$. Local uniform convergence of $u_k$ to $u$ on $K$ gives uniform convergence of $w_{k,\alpha}$ to $u-\phi_\alpha$ on $K$. Since $(t_{k_j},x_{k_j})$ maximizes $w_{k_j,\alpha}$ on $K$, \begin{align*} w_{k_j,\alpha}(t_{k_j},x_{k_j}) \ge w_{k_j,\alpha}(t_0,x_0). \end{align*} Passing to the limit along the subsequence gives \begin{align*} u(\bar{t},\bar{x})-\phi_\alpha(\bar{t},\bar{x}) \ge u(t_0,x_0)-\phi_\alpha(t_0,x_0). \end{align*} Since $u-\phi_\alpha$ has a strict maximum over $K$ at $(t_0,x_0)$, this forces $(\bar{t},\bar{x})=(t_0,x_0)$. Every convergent subsequence has the same limit, and compactness of $K$ therefore implies $(t_k,x_k)\to(t_0,x_0)$. Since $(t_0,x_0)$ lies in the interior of $K$, the convergence just proved implies that $(t_k,x_k)$ lies in the interior of $K$ for all sufficiently large $k$. For such $k$, $u_k-\phi_\alpha$ has a local maximum at $(t_k,x_k)$ in $Q$.[/step]

[guided]The point of the strictification is to prevent the maximum points of the approximating functions from drifting to another point where the original contact was flat. We now make that compactness argument explicit. For each $k \in \mathbb{N}$, we consider the map \begin{align*} w_{k,\alpha}: K &\to \mathbb{R} \end{align*} defined by \begin{align*} w_{k,\alpha}(t,x) := u_k(t,x)-\phi_\alpha(t,x). \end{align*} The domain $K$ is compact and $w_{k,\alpha}$ is continuous, so the extreme value theorem gives a point $(t_k,x_k)\in K$ where $w_{k,\alpha}$ attains its maximum on $K$. We claim that these maximizers converge to the original point $(t_0,x_0)$. Take an arbitrary convergent subsequence $(t_{k_j},x_{k_j})$ and call its limit $(\bar{t},\bar{x})\in K$. Such subsequences exist because $K$ is compact. Since $u_k\to u$ locally uniformly on $Q$, and $K\subset Q$ is compact, the convergence is uniform on $K$. Therefore $w_{k,\alpha}=u_k-\phi_\alpha$ converges uniformly on $K$ to $u-\phi_\alpha$. The maximizing property gives \begin{align*} u_{k_j}(t_{k_j},x_{k_j})-\phi_\alpha(t_{k_j},x_{k_j}) \ge u_{k_j}(t_0,x_0)-\phi_\alpha(t_0,x_0). \end{align*} The left-hand side converges to $u(\bar{t},\bar{x})-\phi_\alpha(\bar{t},\bar{x})$, using uniform convergence of $u_{k_j}$ to $u$ on $K$ and continuity of $\phi_\alpha$. The right-hand side converges to $u(t_0,x_0)-\phi_\alpha(t_0,x_0)$. Hence \begin{align*} u(\bar{t},\bar{x})-\phi_\alpha(\bar{t},\bar{x}) \ge u(t_0,x_0)-\phi_\alpha(t_0,x_0). \end{align*} But the previous step proved that $u-\phi_\alpha$ has a strict maximum on $K$ at exactly $(t_0,x_0)$. Thus $(\bar{t},\bar{x})=(t_0,x_0)$. Because every convergent subsequence of $(t_k,x_k)$ has the same limit and the whole sequence lies in the compact set $K$, the full sequence converges: \begin{align*} (t_k,x_k)\to(t_0,x_0). \end{align*} Finally, $(t_0,x_0)$ is an interior point of $K$. Therefore, for all sufficiently large $k$, the maximizing point $(t_k,x_k)$ also lies in the interior of $K$. Since it maximizes $u_k-\phi_\alpha$ on $K$, it is in particular a local maximum point in $Q$, so the viscosity subsolution inequality for $u_k$ is applicable there.[/guided]

[step:Apply the subsolution inequality and pass to the limit] For all sufficiently large $k$, the viscosity subsolution property of $u_k$ applied to the test function $\phi_\alpha$ at $(t_k,x_k)$ gives \begin{align*} F_k(t_k,x_k,u_k(t_k,x_k),\partial_t\phi_\alpha(t_k,x_k),\nabla\phi_\alpha(t_k,x_k)) \le 0. \end{align*} We record the limits of the entries in this expression. Since $(t_k,x_k)\to(t_0,x_0)$ and $u_k\to u$ uniformly on $K$, \begin{align*} u_k(t_k,x_k)\to u(t_0,x_0). \end{align*} Since $\phi_\alpha \in C^1(Q)$, \begin{align*} \partial_t\phi_\alpha(t_k,x_k)\to \partial_t\phi_\alpha(t_0,x_0)=\partial_t\phi(t_0,x_0) \end{align*} and \begin{align*} \nabla\phi_\alpha(t_k,x_k)\to \nabla\phi_\alpha(t_0,x_0)=\nabla\phi(t_0,x_0). \end{align*} Define the compact set $C_\alpha \subset Q\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}^n$ to be the union of the [limit point](/page/Limit%20Point) \begin{align*} \{(t_0,x_0,u(t_0,x_0),\partial_t\phi(t_0,x_0),\nabla\phi(t_0,x_0))\} \end{align*} with all sufficiently large tuples \begin{align*} (t_k,x_k,u_k(t_k,x_k),\partial_t\phi_\alpha(t_k,x_k),\nabla\phi_\alpha(t_k,x_k)). \end{align*} Because the tuples converge to the displayed limit point, this set is compact. Local uniform convergence $F_k\to F$ therefore implies \begin{align*} F_k(t_k,x_k,u_k(t_k,x_k),\partial_t\phi_\alpha(t_k,x_k),\nabla\phi_\alpha(t_k,x_k))-F(t_k,x_k,u_k(t_k,x_k),\partial_t\phi_\alpha(t_k,x_k),\nabla\phi_\alpha(t_k,x_k))\to 0. \end{align*} Continuity of $F$ gives \begin{align*} F(t_k,x_k,u_k(t_k,x_k),\partial_t\phi_\alpha(t_k,x_k),\nabla\phi_\alpha(t_k,x_k))\to F(t_0,x_0,u(t_0,x_0),\partial_t\phi(t_0,x_0),\nabla\phi(t_0,x_0)). \end{align*} Passing to the limit in the subsolution inequality yields \begin{align*} F(t_0,x_0,u(t_0,x_0),\partial_t\phi(t_0,x_0),\nabla\phi(t_0,x_0)) \le 0. \end{align*} Since $(t_0,x_0)$ and $\phi$ were arbitrary, $u$ is a viscosity subsolution of $F(t,x,u,\partial_tu,\nabla u)=0$ on $Q$. [/step]

[step:Repeat the argument for lower contact points] Assume now that each $u_k$ is a viscosity supersolution. Fix $(t_0,x_0)\in Q$ and $\phi\in C^1(Q)$ such that $u-\phi$ has a local minimum at $(t_0,x_0)$. Choose the same kind of compact ball $K\subset Q$ on which \begin{align*} u(t,x)-\phi(t,x) \ge u(t_0,x_0)-\phi(t_0,x_0) \end{align*} for every $(t,x)\in K$. Fix $\alpha>0$ and define \begin{align*} \psi_\alpha: Q &\to \mathbb{R} \end{align*} by \begin{align*} \psi_\alpha(t,x) := \phi(t,x)-\alpha\bigl((t-t_0)^2+|x-x_0|^2\bigr). \end{align*} Then $u-\psi_\alpha$ has a strict minimum over $K$ at $(t_0,x_0)$. For each $k$, choose a minimizer $(s_k,y_k)\in K$ of $u_k-\psi_\alpha$ on $K$. The same compactness and uniform convergence argument, with maxima replaced by minima, gives \begin{align*} (s_k,y_k)\to(t_0,x_0). \end{align*} Thus, for all sufficiently large $k$, $(s_k,y_k)$ is an interior local minimum point of $u_k-\psi_\alpha$ in $Q$. Applying the viscosity supersolution inequality for $u_k$ gives \begin{align*} F_k(s_k,y_k,u_k(s_k,y_k),\partial_t\psi_\alpha(s_k,y_k),\nabla\psi_\alpha(s_k,y_k)) \ge 0. \end{align*} As before, \begin{align*} u_k(s_k,y_k)\to u(t_0,x_0), \end{align*} \begin{align*} \partial_t\psi_\alpha(s_k,y_k)\to \partial_t\psi_\alpha(t_0,x_0)=\partial_t\phi(t_0,x_0), \end{align*} and \begin{align*} \nabla\psi_\alpha(s_k,y_k)\to \nabla\psi_\alpha(t_0,x_0)=\nabla\phi(t_0,x_0). \end{align*} The tuples involved again lie eventually in a compact subset of $Q\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}^n$, so local uniform convergence $F_k\to F$ and continuity of $F$ allow passage to the limit. We obtain \begin{align*} F(t_0,x_0,u(t_0,x_0),\partial_t\phi(t_0,x_0),\nabla\phi(t_0,x_0)) \ge 0. \end{align*} Since the lower contact point and test function were arbitrary, $u$ is a viscosity supersolution of $F(t,x,u,\partial_tu,\nabla u)=0$ on $Q$. [/step]