[proofplan] We prove the result by comparing $u$ with the explicit quadratic heat solution $\varepsilon(|x|^2 + 2nt)$. On each bounded parabolic cylinder, the difference satisfies the homogeneous [heat equation](/page/Heat%20Equation) and is non-positive on the parabolic boundary once the radius is chosen large enough relative to the global bound for $u$. A direct maximum principle argument on the bounded cylinder gives an upper bound for $u(t_0,x_0)$, and then letting $\varepsilon \downarrow 0$ gives $u(t_0,x_0) \leq 0$. Applying the same argument to $-u$ gives the reverse inequality. [/proofplan] [step:Construct a quadratic comparison function solving the heat equation] Since $u$ is bounded on $[0,T] \times \mathbb{R}^n$, define \begin{align*} M := \sup \{ |u(t,x)| : (t,x) \in [0,T] \times \mathbb{R}^n \}. \end{align*} Then $0 \leq M < \infty$. Fix $\varepsilon > 0$. Define \begin{align*} \phi_\varepsilon: [0,T] \times \mathbb{R}^n \to \mathbb{R}, \quad \phi_\varepsilon(t,x) := \varepsilon(|x|^2 + 2nt). \end{align*} For $x = (x_1,\dots,x_n)$, we have \begin{align*} \partial_t \phi_\varepsilon(t,x) = 2n\varepsilon. \end{align*} Also, since $\partial_{x_i}^2 |x|^2 = 2$ for each $i \in \{1,\dots,n\}$, \begin{align*} \Delta \phi_\varepsilon(t,x) = \varepsilon \sum_{i=1}^n \partial_{x_i}^2 |x|^2 = 2n\varepsilon. \end{align*} Therefore \begin{align*} \partial_t \phi_\varepsilon(t,x) - \Delta \phi_\varepsilon(t,x) = 0 \end{align*} for every $(t,x) \in (0,T] \times \mathbb{R}^n$. [/step] [step:Choose a bounded cylinder whose parabolic boundary controls $u - \phi_\varepsilon$] Fix an arbitrary point $(t_0,x_0) \in (0,T] \times \mathbb{R}^n$. Choose $R > |x_0|$ so large that \begin{align*} M \leq \varepsilon R^2. \end{align*} Define the open ball $B(0,R) := \{x \in \mathbb{R}^n : |x| < R\}$, its closed ball $\overline{B}(0,R) := \{x \in \mathbb{R}^n : |x| \leq R\}$, and its boundary $\partial B(0,R) := \{x \in \mathbb{R}^n : |x| = R\}$. Define the bounded parabolic cylinder \begin{align*} Q_R := (0,t_0] \times B(0,R) \end{align*} and define \begin{align*} v_\varepsilon: [0,t_0] \times \overline{B}(0,R) \to \mathbb{R}, \quad v_\varepsilon(t,x) := u(t,x) - \phi_\varepsilon(t,x). \end{align*} Then $v_\varepsilon$ is continuous on $[0,t_0] \times \overline{B}(0,R)$ and belongs to $C^{1,2}(Q_R)$. Since both $u$ and $\phi_\varepsilon$ solve the heat equation on $Q_R$, \begin{align*} \partial_t v_\varepsilon(t,x) - \Delta v_\varepsilon(t,x) = 0 \end{align*} for every $(t,x) \in Q_R$. On the initial face, for every $x \in \overline{B}(0,R)$, \begin{align*} v_\varepsilon(0,x) = u(0,x) - \varepsilon |x|^2 = -\varepsilon |x|^2 \leq 0. \end{align*} On the lateral boundary, for every $(t,x) \in [0,t_0] \times \partial B(0,R)$, the definition of $M$ and the choice of $R$ give \begin{align*} v_\varepsilon(t,x) \leq M - \varepsilon R^2 - 2n\varepsilon t \leq 0. \end{align*} [guided] The comparison function is chosen so that it grows like $\varepsilon |x|^2$ at spatial infinity but still solves the same heat equation as $u$. This lets us force the boundary values of $u - \phi_\varepsilon$ to be non-positive on a large ball. Fix $(t_0,x_0) \in (0,T] \times \mathbb{R}^n$. Since $M < \infty$, we can choose a radius $R > |x_0|$ satisfying \begin{align*} M \leq \varepsilon R^2. \end{align*} Define the open ball, closed ball, and boundary by \begin{align*} B(0,R) := \{x \in \mathbb{R}^n : |x| < R\}, \quad \overline{B}(0,R) := \{x \in \mathbb{R}^n : |x| \leq R\}, \quad \partial B(0,R) := \{x \in \mathbb{R}^n : |x| = R\}. \end{align*} The strict inequality $R > |x_0|$ ensures that $x_0$ lies inside the ball $B(0,R)$, so the point whose value we want to estimate belongs to the cylinder \begin{align*} Q_R := (0,t_0] \times B(0,R). \end{align*} Now define the comparison difference \begin{align*} v_\varepsilon: [0,t_0] \times \overline{B}(0,R) \to \mathbb{R}, \quad v_\varepsilon(t,x) := u(t,x) - \phi_\varepsilon(t,x). \end{align*} The function $v_\varepsilon$ is continuous on the compact set $[0,t_0] \times \overline{B}(0,R)$ because both $u$ and $\phi_\varepsilon$ are continuous there. It belongs to $C^{1,2}(Q_R)$ because both summands have that regularity in the open cylinder. Since each summand satisfies $\partial_t w - \Delta w = 0$ in $Q_R$, subtracting the two equations gives \begin{align*} \partial_t v_\varepsilon(t,x) - \Delta v_\varepsilon(t,x) = 0 \end{align*} for every $(t,x) \in Q_R$. We next check the parabolic boundary. At initial time, the hypothesis $u(0,x)=0$ gives \begin{align*} v_\varepsilon(0,x) = -\varepsilon |x|^2 \leq 0 \end{align*} for every $x \in \overline{B}(0,R)$. On the lateral boundary, $|x| = R$, and the global bound $u(t,x) \leq |u(t,x)| \leq M$ gives \begin{align*} v_\varepsilon(t,x) \leq M - \varepsilon(R^2 + 2nt). \end{align*} Since $M \leq \varepsilon R^2$ and $2n\varepsilon t \geq 0$, this implies \begin{align*} v_\varepsilon(t,x) \leq 0 \end{align*} for every $(t,x) \in [0,t_0] \times \partial B(0,R)$. Thus $v_\varepsilon$ is non-positive on the full parabolic boundary relevant for forward heat flow. [/guided] [/step] [step:Use the bounded-cylinder maximum argument to bound $u(t_0,x_0)$ from above] We claim that $v_\varepsilon \leq 0$ on $[0,t_0] \times \overline{B}(0,R)$. Fix $\delta > 0$ and define \begin{align*} w_{\varepsilon,\delta}: [0,t_0] \times \overline{B}(0,R) \to \mathbb{R}, \quad w_{\varepsilon,\delta}(t,x) := v_\varepsilon(t,x) - \delta t. \end{align*} Then $w_{\varepsilon,\delta}$ is continuous on $[0,t_0] \times \overline{B}(0,R)$ and belongs to $C^{1,2}(Q_R)$, and \begin{align*} \partial_t w_{\varepsilon,\delta}(t,x) - \Delta w_{\varepsilon,\delta}(t,x) = -\delta < 0 \end{align*} for every $(t,x) \in Q_R$. Because $[0,t_0] \times \overline{B}(0,R)$ is compact, $w_{\varepsilon,\delta}$ attains a maximum at some point $(s,y)$ in that compact set. This maximum cannot occur at a point of $Q_R$ with positive value: if $(s,y) \in Q_R$ were an interior spatial maximum at a positive time, then $y \in B(0,R)$ and $s \in (0,t_0]$. The spatial maximum condition gives $\nabla w_{\varepsilon,\delta}(s,y)=0$ and $\Delta w_{\varepsilon,\delta}(s,y) \leq 0$. The time maximum condition on the interval $[0,t_0]$ gives the left-time derivative inequality $\partial_t w_{\varepsilon,\delta}(s,y) \geq 0$; when $s=t_0$ this is the derivative from times $r<t_0$, and it equals $\partial_t w_{\varepsilon,\delta}(t_0,y)$ because $w_{\varepsilon,\delta}$ is $C^1$ in time on $(0,t_0]$. Hence \begin{align*} \partial_t w_{\varepsilon,\delta}(s,y) - \Delta w_{\varepsilon,\delta}(s,y) \geq 0, \end{align*} contradicting $\partial_t w_{\varepsilon,\delta} - \Delta w_{\varepsilon,\delta} = -\delta$. Therefore the maximum of $w_{\varepsilon,\delta}$ is attained on the initial face or the lateral boundary. On both parts, $v_\varepsilon \leq 0$, and since $-\delta t \leq 0$, we have $w_{\varepsilon,\delta} \leq 0$. Hence \begin{align*} w_{\varepsilon,\delta}(t,x) \leq 0 \end{align*} for every $(t,x) \in [0,t_0] \times \overline{B}(0,R)$. Letting $\delta \downarrow 0$ gives \begin{align*} v_\varepsilon(t,x) \leq 0 \end{align*} for every $(t,x) \in [0,t_0] \times \overline{B}(0,R)$. Since $x_0 \in B(0,R)$, evaluating at $(t_0,x_0)$ gives \begin{align*} u(t_0,x_0) \leq \phi_\varepsilon(t_0,x_0) = \varepsilon(|x_0|^2 + 2nt_0). \end{align*} [/step] [step:Let the barrier vanish and repeat the argument for $-u$] The preceding estimate holds for every $\varepsilon > 0$. Since $|x_0|^2 + 2nt_0$ is fixed, letting $\varepsilon \downarrow 0$ yields \begin{align*} u(t_0,x_0) \leq 0. \end{align*} Now define \begin{align*} \tilde{u}: [0,T] \times \mathbb{R}^n \to \mathbb{R}, \quad \tilde{u}(t,x) := -u(t,x). \end{align*} Then $\tilde{u}$ is continuous on $[0,T] \times \mathbb{R}^n$, belongs to $C^{1,2}((0,T] \times \mathbb{R}^n)$, satisfies \begin{align*} \partial_t \tilde{u}(t,x) - \Delta \tilde{u}(t,x) = 0 \end{align*} for every $(t,x) \in (0,T] \times \mathbb{R}^n$, has initial data $\tilde{u}(0,x)=0$, and is bounded by the same constant $M$. Applying the already proved upper-bound argument to $\tilde{u}$ gives \begin{align*} -u(t_0,x_0) = \tilde{u}(t_0,x_0) \leq 0. \end{align*} Thus \begin{align*} u(t_0,x_0) \geq 0. \end{align*} Combining the two inequalities gives $u(t_0,x_0)=0$. Since $(t_0,x_0) \in (0,T] \times \mathbb{R}^n$ was arbitrary, $u=0$ on $(0,T] \times \mathbb{R}^n$. The initial condition gives $u(0,x)=0$ for every $x \in \mathbb{R}^n$, so $u \equiv 0$ on $[0,T] \times \mathbb{R}^n$. [/step]