[proofplan] We subtract the two functions and reduce the nonlinear comparison to a linear parabolic inequality with a bounded zeroth-order coefficient. The local Lipschitz hypothesis gives a bounded coefficient on the compact range of $u$ and $v$. An exponential weight then makes the zeroth-order coefficient strictly positive, so a positive maximum inside the parabolic cylinder contradicts the elementary maximum principle argument at that maximum. Applying this on every truncated cylinder $[0,S] \times \overline{\Omega}$ with $S<T$ and passing to the limit gives the result up to time $T$. [/proofplan]

[step:Reduce the nonlinear inequality to a linear inequality for the difference]Define \begin{align*} w: [0,T] \times \overline{\Omega} \to \mathbb{R} \end{align*} by $w(t,x) := u(t,x) - v(t,x)$. Since $u,v \in C^{1,2}((0,T) \times \Omega) \cap C([0,T] \times \overline{\Omega})$, we have \begin{align*} w \in C^{1,2}((0,T) \times \Omega) \cap C([0,T] \times \overline{\Omega}). \end{align*} The assumed differential inequality gives, for every $(t,x) \in (0,T) \times \Omega$, \begin{align*} \partial_t w(t,x) - \Delta w(t,x) \leq f(u(t,x)) - f(v(t,x)). \end{align*} Because $[0,T] \times \overline{\Omega}$ is compact and $u,v$ are continuous, there exists $R>0$ such that \begin{align*} u([0,T] \times \overline{\Omega}) \cup v([0,T] \times \overline{\Omega}) \subset [-R,R]. \end{align*} Since $f$ is locally Lipschitz, there exists $L \geq 0$ such that \begin{align*} |f(a)-f(b)| \leq L |a-b| \quad \text{for all } a,b \in [-R,R]. \end{align*} Define \begin{align*} c: (0,T) \times \Omega \to \mathbb{R} \end{align*} as follows: if $u(t,x) \neq v(t,x)$, set \begin{align*} c(t,x) := \dfrac{f(u(t,x))-f(v(t,x))}{u(t,x)-v(t,x)}. \end{align*} If $u(t,x)=v(t,x)$, set \begin{align*} c(t,x) := 0. \end{align*} Then $|c(t,x)| \leq L$ for every $(t,x) \in (0,T) \times \Omega$, and \begin{align*} f(u(t,x)) - f(v(t,x)) = c(t,x)w(t,x) \end{align*} for every $(t,x) \in (0,T) \times \Omega$. Hence \begin{align*} \partial_t w(t,x) - \Delta w(t,x) - c(t,x)w(t,x) \leq 0 \end{align*} for every $(t,x) \in (0,T) \times \Omega$.[/step]

[guided]The only nonlinear term is $f(u)-f(v)$, so the purpose of this step is to rewrite that term as a bounded coefficient times $u-v$. Define the difference map \begin{align*} w: [0,T] \times \overline{\Omega} \to \mathbb{R} \end{align*} by $w(t,x) := u(t,x)-v(t,x)$. The regularity assumptions on $u$ and $v$ imply \begin{align*} w \in C^{1,2}((0,T) \times \Omega) \cap C([0,T] \times \overline{\Omega}). \end{align*} Subtracting the differential expressions in the hypothesis gives, pointwise on $(0,T) \times \Omega$, \begin{align*} \partial_t w(t,x) - \Delta w(t,x) \leq f(u(t,x)) - f(v(t,x)). \end{align*} Now we use local Lipschitz continuity of $f$ only on the values actually taken by $u$ and $v$. Since $[0,T] \times \overline{\Omega}$ is compact and $u,v$ are continuous, their ranges are compact subsets of $\mathbb{R}$. Therefore there exists $R>0$ such that both ranges lie in $[-R,R]$: \begin{align*} u([0,T] \times \overline{\Omega}) \cup v([0,T] \times \overline{\Omega}) \subset [-R,R]. \end{align*} The function $f$ is locally Lipschitz, so on this compact interval there is a Lipschitz constant $L \geq 0$ satisfying \begin{align*} |f(a)-f(b)| \leq L|a-b| \quad \text{for all } a,b \in [-R,R]. \end{align*} We now define the coefficient that linearizes the reaction term: \begin{align*} c: (0,T) \times \Omega \to \mathbb{R} \end{align*} as follows: if $u(t,x) \neq v(t,x)$, set \begin{align*} c(t,x) := \dfrac{f(u(t,x))-f(v(t,x))}{u(t,x)-v(t,x)}. \end{align*} If $u(t,x)=v(t,x)$, set \begin{align*} c(t,x) := 0. \end{align*} When $u(t,x) \neq v(t,x)$, the Lipschitz bound gives $|c(t,x)| \leq L$. When $u(t,x)=v(t,x)$, the definition gives $c(t,x)=0$, so the same bound holds. Also, if $u(t,x)=v(t,x)$ then both sides of \begin{align*} f(u(t,x))-f(v(t,x)) = c(t,x)w(t,x) \end{align*} are zero, while if $u(t,x) \neq v(t,x)$ this identity follows from the definition of $c$. Thus the nonlinear inequality becomes the linear inequality \begin{align*} \partial_t w(t,x) - \Delta w(t,x) - c(t,x)w(t,x) \leq 0 \end{align*} for every $(t,x) \in (0,T) \times \Omega$, with the important additional fact that $c$ is bounded.[/guided]

[step:Record the nonpositive parabolic boundary values for the difference] For every $(t,x) \in (0,T) \times \partial\Omega$, the boundary hypothesis gives $w(t,x) \leq 0$. For every $x \in \Omega$, the initial hypothesis gives $w(0,x) \leq 0$. If $x \in \partial\Omega$, then continuity of $w$ on $[0,T] \times \overline{\Omega}$ and the lateral boundary inequality imply \begin{align*} w(0,x) = \lim_{t \downarrow 0} w(t,x) \leq 0. \end{align*} Therefore \begin{align*} w(0,x) \leq 0 \quad \text{for every } x \in \overline{\Omega}, \end{align*} and \begin{align*} w(t,x) \leq 0 \quad \text{for every } (t,x) \in (0,T) \times \partial\Omega. \end{align*} [/step]

[step:Apply an exponential weight to make the zeroth-order term positive] Choose $\lambda := L+1$. Define \begin{align*} z: [0,T] \times \overline{\Omega} \to \mathbb{R} \end{align*} by $z(t,x) := e^{-\lambda t}w(t,x)$. Then \begin{align*} z \in C^{1,2}((0,T) \times \Omega) \cap C([0,T] \times \overline{\Omega}). \end{align*} Since $w(t,x)=e^{\lambda t}z(t,x)$, direct differentiation gives \begin{align*} \partial_t z(t,x) - \Delta z(t,x) + (\lambda-c(t,x))z(t,x) = e^{-\lambda t}\bigl(\partial_t w(t,x)-\Delta w(t,x)-c(t,x)w(t,x)\bigr). \end{align*} Hence, for every $(t,x) \in (0,T) \times \Omega$, \begin{align*} \partial_t z(t,x) - \Delta z(t,x) + (\lambda-c(t,x))z(t,x) \leq 0. \end{align*} Moreover, because $|c(t,x)| \leq L$, we have \begin{align*} \lambda - c(t,x) \geq 1 \end{align*} for every $(t,x) \in (0,T) \times \Omega$. Since $e^{-\lambda t}>0$, the parabolic boundary inequalities for $w$ imply the same inequalities for $z$: \begin{align*} z(0,x) \leq 0 \quad \text{for every } x \in \overline{\Omega}, \end{align*} and \begin{align*} z(t,x) \leq 0 \quad \text{for every } (t,x) \in (0,T) \times \partial\Omega. \end{align*} [/step]

[step:Rule out a positive maximum on each truncated cylinder]Fix $S \in (0,T)$. We prove that $z \leq 0$ on $[0,S] \times \overline{\Omega}$. Suppose instead that \begin{align*} M := \max_{[0,S] \times \overline{\Omega}} z > 0. \end{align*} Since $z \leq 0$ on $\{0\} \times \overline{\Omega}$ and on $(0,S] \times \partial\Omega$, a point $(t_0,x_0)$ attaining this positive maximum satisfies \begin{align*} t_0 \in (0,S] \quad \text{and} \quad x_0 \in \Omega. \end{align*} At the spatial maximum point $x_0$ of the function $x \mapsto z(t_0,x)$, every coordinate second derivative satisfies $\partial_{x_i x_i} z(t_0,x_0) \leq 0$: this follows by applying the one-variable second-derivative maximum condition to the map $s \mapsto z(t_0,x_0+s e_i)$ for sufficiently small $s$, where $e_i \in \mathbb{R}^n$ is the $i$-th standard basis vector. Summing over $i \in \{1,\dots,n\}$ gives \begin{align*} \Delta z(t_0,x_0) \leq 0. \end{align*} At the time maximum point $t_0$ of the function $t \mapsto z(t,x_0)$ on $[0,S]$, we have \begin{align*} \partial_t z(t_0,x_0) \geq 0. \end{align*} If $t_0<S$, then the one-variable function $t \mapsto z(t,x_0)$ has an interior maximum at $t_0$, so its derivative there is zero. If $t_0=S$, it follows from the left-sided inequality \begin{align*} \frac{z(S,x_0)-z(S-h,x_0)}{h} \geq 0 \end{align*} for all sufficiently small $h>0$, together with differentiability in $t$ at $S$. Evaluating the differential inequality for $z$ at $(t_0,x_0)$ gives \begin{align*} 0 \geq \partial_t z(t_0,x_0)-\Delta z(t_0,x_0)+(\lambda-c(t_0,x_0))z(t_0,x_0). \end{align*} The three terms on the right satisfy \begin{align*} \partial_t z(t_0,x_0) \geq 0, \quad -\Delta z(t_0,x_0) \geq 0, \quad (\lambda-c(t_0,x_0))z(t_0,x_0) \geq z(t_0,x_0)>0. \end{align*} Thus the right-hand side is strictly positive, contradicting the displayed inequality. Therefore $z \leq 0$ on $[0,S] \times \overline{\Omega}$.[/step]

[guided]We want to show that $z$ cannot become positive. Fix a time $S \in (0,T)$ and work on the compact cylinder $[0,S] \times \overline{\Omega}$. Because $z$ is continuous there, it attains a maximum. Suppose that this maximum is positive: \begin{align*} M := \max_{[0,S] \times \overline{\Omega}} z > 0. \end{align*} Let $(t_0,x_0) \in [0,S] \times \overline{\Omega}$ be a point where $z(t_0,x_0)=M$. The maximum point cannot lie on the initial face, because $z(0,x) \leq 0$ for every $x \in \overline{\Omega}$. It also cannot lie on the lateral boundary, because $z(t,x) \leq 0$ for every $(t,x) \in (0,S] \times \partial\Omega$. Since the maximum is positive, we must have \begin{align*} t_0 \in (0,S] \quad \text{and} \quad x_0 \in \Omega. \end{align*} Now we use the elementary calculus information available at a maximum. First, $x_0$ is an interior maximum point of the spatial function $x \mapsto z(t_0,x)$. Since $z$ is twice continuously differentiable in the spatial variables on $(0,T) \times \Omega$, we may restrict to each coordinate line through $x_0$. For each $i \in \{1,\dots,n\}$, let $e_i \in \mathbb{R}^n$ be the $i$-th standard basis vector. For sufficiently small $s \in \mathbb{R}$, the point $x_0+s e_i$ lies in $\Omega$, and the one-variable function $s \mapsto z(t_0,x_0+s e_i)$ has a maximum at $s=0$. Therefore its second derivative at $0$ is nonpositive, so $\partial_{x_i x_i}z(t_0,x_0) \leq 0$. Summing these inequalities over $i$ gives \begin{align*} \Delta z(t_0,x_0) \leq 0. \end{align*} Equivalently, $-\Delta z(t_0,x_0) \geq 0$. Second, $t_0$ is a maximum point of the one-variable function $t \mapsto z(t,x_0)$ on $[0,S]$. If $t_0<S$, then this one-variable function has an interior maximum at $t_0$, so its derivative there is zero and $\partial_t z(t_0,x_0)=0$. If $t_0=S$, then for every sufficiently small $h>0$, \begin{align*} z(S,x_0)-z(S-h,x_0) \geq 0. \end{align*} Dividing by $h>0$ and using differentiability in $t$ at $S$ gives \begin{align*} \partial_t z(S,x_0) \geq 0. \end{align*} In both cases, \begin{align*} \partial_t z(t_0,x_0) \geq 0. \end{align*} We now evaluate the differential inequality \begin{align*} \partial_t z(t,x)-\Delta z(t,x)+(\lambda-c(t,x))z(t,x) \leq 0 \end{align*} at the point $(t_0,x_0)$. This is legitimate because $(t_0,x_0) \in (0,T) \times \Omega$. We obtain \begin{align*} 0 \geq \partial_t z(t_0,x_0)-\Delta z(t_0,x_0)+(\lambda-c(t_0,x_0))z(t_0,x_0). \end{align*} But each contribution on the right is nonnegative, and the zeroth-order contribution is strictly positive. Indeed, \begin{align*} \partial_t z(t_0,x_0) \geq 0, \end{align*} \begin{align*} -\Delta z(t_0,x_0) \geq 0, \end{align*} and, since $\lambda-c(t_0,x_0) \geq 1$ and $z(t_0,x_0)=M>0$, \begin{align*} (\lambda-c(t_0,x_0))z(t_0,x_0) \geq z(t_0,x_0)>0. \end{align*} Therefore the right-hand side is strictly positive, contradicting the fact that it is at most zero. Hence no positive maximum exists on $[0,S] \times \overline{\Omega}$, and so \begin{align*} z(t,x) \leq 0 \quad \text{for every } (t,x) \in [0,S] \times \overline{\Omega}. \end{align*}[/guided]

[step:Pass from truncated cylinders to the full time interval] The previous step shows that for every $S \in (0,T)$, \begin{align*} z(t,x) \leq 0 \quad \text{for every } (t,x) \in [0,S] \times \overline{\Omega}. \end{align*} Thus $z(t,x) \leq 0$ for every $(t,x) \in [0,T) \times \overline{\Omega}$. Since $z$ is continuous on $[0,T] \times \overline{\Omega}$, taking $t \uparrow T$ gives \begin{align*} z(T,x) \leq 0 \quad \text{for every } x \in \overline{\Omega}. \end{align*} Therefore $z \leq 0$ on $[0,T] \times \overline{\Omega}$. Since $w(t,x)=e^{\lambda t}z(t,x)$ and $e^{\lambda t}>0$, it follows that \begin{align*} w(t,x) \leq 0 \quad \text{for every } (t,x) \in [0,T] \times \overline{\Omega}. \end{align*} By the definition $w=u-v$, this is exactly \begin{align*} u(t,x) \leq v(t,x) \quad \text{for every } (t,x) \in [0,T] \times \overline{\Omega}. \end{align*} This proves the [comparison principle](/theorems/4870). [/step]