[proofplan] We first verify that the infimum defining $Q_t f$ is finite, attained, and produces a [continuous function](/page/Continuous%20Function). The essential algebraic input is the dynamic programming identity for the quadratic cost: splitting a time interval into two pieces gives exactly the same infimal convolution. That identity converts test-function contact inequalities into the viscosity subsolution and supersolution inequalities. Finally, [uniform continuity](/page/Uniform%20Continuity) of $f$ and the coercive quadratic penalty show that $Q_t f$ converges uniformly to $f$ as $t \downarrow 0$. [/proofplan] [step:Show that the infimal convolution is finite and has minimizers] Let $B := \|f\|_\infty$, so $|f(y)| \leq B$ for every $y \in \mathbb{R}^n$. Fix $t > 0$ and $x \in \mathbb{R}^n$. Define the objective function $\Phi_{t,x}: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} \Phi_{t,x}(y) := f(y) + \frac{|x-y|^2}{2t}. \end{align*} The function $\Phi_{t,x}$ is continuous because $f$ is uniformly continuous and the quadratic term is continuous. It is also coercive, since \begin{align*} \Phi_{t,x}(y) \geq -B + \frac{|x-y|^2}{2t}. \end{align*} Thus $\Phi_{t,x}(y) \to \infty$ as $|y| \to \infty$. By the Weierstrass theorem applied on a sufficiently large closed ball containing a sublevel set of $\Phi_{t,x}$, the infimum is attained. Hence $u(t,x)$ is a finite real number for every $(t,x) \in (0,\infty) \times \mathbb{R}^n$. [/step] [step:Prove continuity in space and time on positive time intervals] Let $\omega_f:[0,\infty) \to [0,\infty)$ denote a modulus of continuity for $f$, meaning that $\omega_f(\rho) \to 0$ as $\rho \downarrow 0$ and \begin{align*} |f(a)-f(b)| \leq \omega_f(|a-b|) \end{align*} for all $a,b \in \mathbb{R}^n$. Fix $t>0$ and $x,x' \in \mathbb{R}^n$. For each $y \in \mathbb{R}^n$, set $y' := y + x' - x$. Then $|x'-y'|=|x-y|$, and hence \begin{align*} u(t,x') \leq f(y') + \frac{|x-y|^2}{2t} \leq f(y) + \omega_f(|x'-x|) + \frac{|x-y|^2}{2t}. \end{align*} Taking the infimum over $y \in \mathbb{R}^n$ gives \begin{align*} u(t,x') - u(t,x) \leq \omega_f(|x'-x|). \end{align*} Interchanging $x$ and $x'$ gives \begin{align*} |u(t,x')-u(t,x)| \leq \omega_f(|x'-x|). \end{align*} Thus $x \mapsto u(t,x)$ is uniformly continuous, uniformly in $t>0$. Now fix $0<a<b<\infty$ and let $a \leq s<t \leq b$. Choosing the same spatial point $y$ in the formula defining $u(s,x)$ and $u(t,x)$ shows $u(t,x) \leq u(s,x)$, because the coefficient $1/(2t)$ is smaller than $1/(2s)$. Let $y_{t,x} \in \mathbb{R}^n$ be a minimizer for $u(t,x)$, whose existence was proved above. Since $|u|\leq B$, this minimizer satisfies \begin{align*} \frac{|x-y_{t,x}|^2}{2t} = u(t,x)-f(y_{t,x}) \leq 2B. \end{align*} Using $y_{t,x}$ as a competitor in the formula for $u(s,x)$ gives \begin{align*} u(s,x) \leq f(y_{t,x}) + \frac{|x-y_{t,x}|^2}{2s}. \end{align*} Subtracting this inequality from the equality defining $u(t,x)$ at $y_{t,x}$ yields \begin{align*} u(t,x)-u(s,x) \geq -\frac{t-s}{2st}|x-y_{t,x}|^2 \geq -\frac{2B}{a}(t-s). \end{align*} Therefore \begin{align*} |u(t,x)-u(s,x)| \leq \frac{2B}{a}|t-s| \end{align*} for all $x \in \mathbb{R}^n$ and all $s,t \in [a,b]$. Combining the spatial and temporal estimates proves that $u$ is continuous on $(0,\infty)\times\mathbb{R}^n$. [/step] [step:Establish the dynamic programming identity] Fix $0 < s < t$ and $x \in \mathbb{R}^n$. We prove \begin{align*} u(t,x) = \inf_{z \in \mathbb{R}^n}\left\{u(s,z) + \frac{|x-z|^2}{2(t-s)}\right\}. \end{align*} Using the definition of $u(s,z)$, the right-hand side equals \begin{align*} \inf_{z \in \mathbb{R}^n}\inf_{y \in \mathbb{R}^n}\left\{f(y) + \frac{|z-y|^2}{2s} + \frac{|x-z|^2}{2(t-s)}\right\}. \end{align*} For fixed $x,y \in \mathbb{R}^n$, define $G_{x,y}: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} G_{x,y}(z) := \frac{|z-y|^2}{2s} + \frac{|x-z|^2}{2(t-s)}. \end{align*} Completing the square gives \begin{align*} G_{x,y}(z) = \frac{|x-y|^2}{2t} + \frac{t}{2s(t-s)}\left|z-\frac{s}{t}x-\frac{t-s}{t}y\right|^2. \end{align*} Therefore \begin{align*} \inf_{z \in \mathbb{R}^n} G_{x,y}(z) = \frac{|x-y|^2}{2t}. \end{align*} Taking the infimum over $y \in \mathbb{R}^n$ yields the desired identity. [/step] [step:Verify the viscosity subsolution inequality from a local maximum test] Let $\varphi: (0,\infty) \times \mathbb{R}^n \to \mathbb{R}$ be a $C^1$ [test function](/page/Test%20Function). Suppose $u-\varphi$ has a local maximum at $(t_0,x_0) \in (0,\infty) \times \mathbb{R}^n$, and normalize by subtracting a constant from $\varphi$ so that \begin{align*} u(t_0,x_0) = \varphi(t_0,x_0). \end{align*} We must prove \begin{align*} \partial_t \varphi(t_0,x_0) + \frac{1}{2}|\nabla \varphi(t_0,x_0)|^2 \leq 0. \end{align*} Let $p := \nabla \varphi(t_0,x_0) \in \mathbb{R}^n$. For $h > 0$ sufficiently small, set \begin{align*} z_h := x_0 - hp. \end{align*} The dynamic programming identity with $s=t_0-h$ and $t=t_0$ gives \begin{align*} u(t_0,x_0) \leq u(t_0-h,z_h) + \frac{|x_0-z_h|^2}{2h}. \end{align*} Since $u-\varphi$ has a local maximum at $(t_0,x_0)$ and $(t_0-h,z_h) \to (t_0,x_0)$, for all sufficiently small $h$ we have \begin{align*} u(t_0-h,z_h)-u(t_0,x_0) \leq \varphi(t_0-h,z_h)-\varphi(t_0,x_0). \end{align*} Combining the two inequalities and using $|x_0-z_h|^2 = h^2|p|^2$ gives \begin{align*} 0 \leq \varphi(t_0-h,x_0-hp)-\varphi(t_0,x_0) + \frac{h}{2}|p|^2. \end{align*} Divide by $h>0$ and let $h \downarrow 0$. Since $\varphi$ is $C^1$, \begin{align*} 0 \leq -\partial_t \varphi(t_0,x_0) - p \cdot \nabla \varphi(t_0,x_0) + \frac{1}{2}|p|^2. \end{align*} Because $p=\nabla \varphi(t_0,x_0)$, this becomes \begin{align*} 0 \leq -\partial_t \varphi(t_0,x_0) - \frac{1}{2}|\nabla \varphi(t_0,x_0)|^2. \end{align*} This is precisely the viscosity subsolution inequality. [/step] [step:Verify the viscosity supersolution inequality from a local minimum test] Let $\varphi: (0,\infty) \times \mathbb{R}^n \to \mathbb{R}$ be a $C^1$ test function. Suppose $u-\varphi$ has a local minimum at $(t_0,x_0) \in (0,\infty) \times \mathbb{R}^n$, and normalize so that $u(t_0,x_0)=\varphi(t_0,x_0)$. We prove \begin{align*} \partial_t \varphi(t_0,x_0) + \frac{1}{2}|\nabla \varphi(t_0,x_0)|^2 \geq 0. \end{align*} For $h \in (0,t_0)$, the dynamic programming identity and continuity of $z \mapsto u(t_0-h,z)$ imply that the coercive function \begin{align*} z \mapsto u(t_0-h,z) + \frac{|x_0-z|^2}{2h} \end{align*} attains its minimum on $\mathbb{R}^n$. Therefore there is a point $z_h \in \mathbb{R}^n$ such that \begin{align*} u(t_0,x_0) = u(t_0-h,z_h) + \frac{|x_0-z_h|^2}{2h}. \end{align*} Since $|u| \leq B$, this equality implies \begin{align*} \frac{|x_0-z_h|^2}{2h} \leq 2B. \end{align*} Hence $z_h \to x_0$ as $h \downarrow 0$. For sufficiently small $h$, the point $(t_0-h,z_h)$ lies in the neighbourhood on which $u-\varphi$ has its local minimum, so \begin{align*} u(t_0-h,z_h)-u(t_0,x_0) \geq \varphi(t_0-h,z_h)-\varphi(t_0,x_0). \end{align*} Using the equality for $u(t_0,x_0)$ gives \begin{align*} 0 \geq \varphi(t_0-h,z_h)-\varphi(t_0,x_0) + \frac{|x_0-z_h|^2}{2h}. \end{align*} Define $r_h \in \mathbb{R}^n$ by \begin{align*} r_h := \frac{x_0-z_h}{h}. \end{align*} Then the last inequality becomes \begin{align*} 0 \geq \varphi(t_0-h,x_0-hr_h)-\varphi(t_0,x_0) + \frac{h}{2}|r_h|^2. \end{align*} We first prove that $(r_h)$ is bounded for sufficiently small $h$. Since $\varphi$ is $C^1$, there are constants $\rho>0$ and $M>0$ such that \begin{align*} |\varphi(t,x)-\varphi(t_0,x_0)| \leq M(|t-t_0|+|x-x_0|) \end{align*} whenever $|t-t_0|+|x-x_0|<\rho$. Since $(t_0-h,z_h)\to(t_0,x_0)$, this estimate applies for sufficiently small $h$. The preceding inequality then gives \begin{align*} 0 \geq -M h(1+|r_h|) + \frac{h}{2}|r_h|^2. \end{align*} Dividing by $h>0$ shows \begin{align*} \frac{1}{2}|r_h|^2 \leq M(1+|r_h|), \end{align*} so $(r_h)$ is bounded along small $h$. With this boundedness established, differentiability of $\varphi$ at $(t_0,x_0)$ gives \begin{align*} \varphi(t_0-h,x_0-hr_h)-\varphi(t_0,x_0) = -h\,\partial_t\varphi(t_0,x_0) - h\,\nabla\varphi(t_0,x_0)\cdot r_h + o(h). \end{align*} Choose a sequence $h_k \downarrow 0$ such that $r_{h_k} \to r \in \mathbb{R}^n$. Dividing the inequality by $h_k$ and letting $k \to \infty$ gives \begin{align*} 0 \geq -\partial_t\varphi(t_0,x_0) - \nabla\varphi(t_0,x_0)\cdot r + \frac{1}{2}|r|^2. \end{align*} For every $r \in \mathbb{R}^n$, the elementary inequality \begin{align*} -\nabla\varphi(t_0,x_0)\cdot r + \frac{1}{2}|r|^2 \geq -\frac{1}{2}|\nabla\varphi(t_0,x_0)|^2 \end{align*} follows by completing the square. Therefore \begin{align*} 0 \geq -\partial_t\varphi(t_0,x_0) - \frac{1}{2}|\nabla\varphi(t_0,x_0)|^2. \end{align*} This is the viscosity supersolution inequality. [guided] The supersolution direction is the subtler half because a lower test function must be compared with an actual minimizing trajectory in the dynamic programming formula. We begin exactly as in the terse proof. Let $\varphi: (0,\infty)\times \mathbb{R}^n \to \mathbb{R}$ be $C^1$, suppose $u-\varphi$ has a local minimum at $(t_0,x_0)$, and normalize so that $u(t_0,x_0)=\varphi(t_0,x_0)$. We need to show \begin{align*} \partial_t \varphi(t_0,x_0) + \frac{1}{2}|\nabla \varphi(t_0,x_0)|^2 \geq 0. \end{align*} Fix $h \in (0,t_0)$. The dynamic programming identity says that $u(t_0,x_0)$ is obtained by minimizing over the previous spatial point $z$ at time $t_0-h$. The map $z \mapsto u(t_0-h,z)$ is continuous by the continuity step, and the added quadratic cost is coercive. Since $u$ is bounded by $B$, the whole objective is coercive. Hence the minimum is attained. Choose $z_h \in \mathbb{R}^n$ such that \begin{align*} u(t_0,x_0) = u(t_0-h,z_h) + \frac{|x_0-z_h|^2}{2h}. \end{align*} The boundedness of $f$ gives $|u| \leq B$, where $B=\|f\|_\infty$. Hence \begin{align*} \frac{|x_0-z_h|^2}{2h} = u(t_0,x_0)-u(t_0-h,z_h) \leq 2B. \end{align*} This estimate is important: it forces $z_h \to x_0$ as $h \downarrow 0$, so the minimizing point remains inside the neighbourhood where the test function touches $u$ from below. Because $u-\varphi$ has a local minimum at $(t_0,x_0)$, and because $(t_0-h,z_h) \to (t_0,x_0)$, for all sufficiently small $h$ we have \begin{align*} u(t_0-h,z_h)-u(t_0,x_0) \geq \varphi(t_0-h,z_h)-\varphi(t_0,x_0). \end{align*} Substituting the equality for $u(t_0,x_0)$ into this contact inequality gives \begin{align*} 0 \geq \varphi(t_0-h,z_h)-\varphi(t_0,x_0) + \frac{|x_0-z_h|^2}{2h}. \end{align*} Now introduce the velocity variable $r_h \in \mathbb{R}^n$ by \begin{align*} r_h := \frac{x_0-z_h}{h}. \end{align*} Then $z_h=x_0-hr_h$, and the inequality becomes \begin{align*} 0 \geq \varphi(t_0-h,x_0-hr_h)-\varphi(t_0,x_0) + \frac{h}{2}|r_h|^2. \end{align*} The point of this notation is that the quadratic cost becomes the kinetic term $\frac{h}{2}|r_h|^2$, while the test function can be expanded to first order. Before using a first-order expansion along this moving direction, we prove that the velocities $r_h$ are bounded. Since $\varphi$ is $C^1$, it is locally Lipschitz near $(t_0,x_0)$. Thus there are constants $\rho>0$ and $M>0$ such that \begin{align*} |\varphi(t,x)-\varphi(t_0,x_0)| \leq M(|t-t_0|+|x-x_0|) \end{align*} whenever $|t-t_0|+|x-x_0|<\rho$. Since $(t_0-h,z_h)\to(t_0,x_0)$, this applies for all sufficiently small $h$. Combining the local Lipschitz estimate with \begin{align*} 0 \geq \varphi(t_0-h,x_0-hr_h)-\varphi(t_0,x_0) + \frac{h}{2}|r_h|^2 \end{align*} gives \begin{align*} 0 \geq -Mh(1+|r_h|) + \frac{h}{2}|r_h|^2. \end{align*} After division by $h>0$, the quadratic inequality \begin{align*} \frac{1}{2}|r_h|^2 \leq M(1+|r_h|) \end{align*} shows that $(r_h)$ is bounded along small $h$. Now the differentiability expansion is legitimate in a uniform form along the selected sequence of directions: because $r_h$ is bounded, the increment $(-h,-hr_h)$ has size $O(h)$, and hence \begin{align*} \varphi(t_0-h,x_0-hr_h)-\varphi(t_0,x_0) = -h\,\partial_t\varphi(t_0,x_0) - h\,\nabla\varphi(t_0,x_0)\cdot r_h + o(h). \end{align*} Choose a sequence $h_k \downarrow 0$ such that $r_{h_k} \to r \in \mathbb{R}^n$. Divide the inequality by $h_k$ and pass to the limit. The differentiability remainder vanishes because the velocities are bounded, and we obtain \begin{align*} 0 \geq -\partial_t\varphi(t_0,x_0) - \nabla\varphi(t_0,x_0)\cdot r + \frac{1}{2}|r|^2. \end{align*} Finally, complete the square: \begin{align*} -\nabla\varphi(t_0,x_0)\cdot r + \frac{1}{2}|r|^2 = \frac{1}{2}|r-\nabla\varphi(t_0,x_0)|^2 - \frac{1}{2}|\nabla\varphi(t_0,x_0)|^2. \end{align*} The square is non-negative, so \begin{align*} -\nabla\varphi(t_0,x_0)\cdot r + \frac{1}{2}|r|^2 \geq -\frac{1}{2}|\nabla\varphi(t_0,x_0)|^2. \end{align*} Combining this lower bound with the limiting inequality gives \begin{align*} 0 \geq -\partial_t\varphi(t_0,x_0) - \frac{1}{2}|\nabla\varphi(t_0,x_0)|^2. \end{align*} Equivalently, \begin{align*} \partial_t\varphi(t_0,x_0) + \frac{1}{2}|\nabla\varphi(t_0,x_0)|^2 \geq 0. \end{align*} This proves the viscosity supersolution inequality. [/guided] [/step] [step:Prove uniform attainment of the initial condition] We prove \begin{align*} \lim_{t \downarrow 0}\sup_{x \in \mathbb{R}^n}|u(t,x)-f(x)| = 0. \end{align*} The upper bound follows by choosing $y=x$ in the infimum: \begin{align*} u(t,x) \leq f(x). \end{align*} Let $\varepsilon>0$. Since $f$ is uniformly continuous, choose $\delta>0$ such that $|f(y)-f(x)|<\varepsilon$ whenever $|x-y|<\delta$. If $|x-y|<\delta$, then \begin{align*} f(y)+\frac{|x-y|^2}{2t} \geq f(x)-\varepsilon. \end{align*} If $|x-y|\geq\delta$, then \begin{align*} f(y)+\frac{|x-y|^2}{2t} \geq -B+\frac{\delta^2}{2t}. \end{align*} Choose $t_\varepsilon>0$ such that $\delta^2/(2t) \geq 2B+\varepsilon$ whenever $0<t<t_\varepsilon$. Then for $0<t<t_\varepsilon$ and $|x-y|\geq\delta$, \begin{align*} f(y)+\frac{|x-y|^2}{2t} \geq B+\varepsilon \geq f(x)-\varepsilon. \end{align*} Thus every $y \in \mathbb{R}^n$ satisfies \begin{align*} f(y)+\frac{|x-y|^2}{2t} \geq f(x)-\varepsilon. \end{align*} Taking the infimum over $y$ gives \begin{align*} u(t,x) \geq f(x)-\varepsilon. \end{align*} Together with $u(t,x)\leq f(x)$, this yields \begin{align*} \sup_{x \in \mathbb{R}^n}|u(t,x)-f(x)| \leq \varepsilon \end{align*} for all $0<t<t_\varepsilon$. This proves the stated uniform initial convergence. [/step] [step:Conclude the Hamilton-Jacobi viscosity solution property] The continuity step proves that $u$ is continuous on $(0,\infty)\times\mathbb{R}^n$. The subsolution step proves that every $C^1$ test function touching $u$ from above satisfies \begin{align*} \partial_t \varphi + \frac{1}{2}|\nabla \varphi|^2 \leq 0 \end{align*} at the contact point. The supersolution step proves that every $C^1$ test function touching $u$ from below satisfies \begin{align*} \partial_t \varphi + \frac{1}{2}|\nabla \varphi|^2 \geq 0 \end{align*} at the contact point. Therefore $u$ is a viscosity solution of \begin{align*} \partial_t u + \frac{1}{2}|\nabla u|^2 = 0 \end{align*} on $(0,\infty)\times\mathbb{R}^n$. The previous step proves the initial condition in the uniform sense, completing the proof. [/step]