[proofplan]
The proof is an integration-by-parts equivalence. If $u$ satisfies the pointwise conservation law, multiplying by a compactly supported [test function](/page/Test%20Function) and integrating over spacetime moves the derivatives from $u$ and $f(u)$ onto the test function, with no boundary terms because the test function vanishes near the parabolic boundary. Conversely, the weak identity can be integrated by parts in the reverse direction to show that the [continuous function](/page/Continuous%20Function) $\partial_t u+\operatorname{div}_x f(u)$ has zero integral against every compactly supported smooth test function. A local non-negative bump test function then forces this continuous function to vanish pointwise.
[/proofplan]
[step:Define the spacetime residual and verify its continuity]
Define the spacetime domain $Q := (0,T) \times U \subset \mathbb{R}^{n+1}$. Define the residual function $G: Q \to \mathbb{R}$ by
\begin{align*}
G(t,x) = \partial_t u(t,x) + \sum_{i=1}^n f_i'(u(t,x))\,\partial_{x_i}u(t,x).
\end{align*}
Since $u \in C^1(Q;\mathbb{R})$ and $f_i \in C^1(\mathbb{R};\mathbb{R})$ for each $i \in \{1,\dots,n\}$, each composed function $(t,x) \mapsto f_i'(u(t,x))\,\partial_{x_i}u(t,x)$ is continuous on $Q$. Hence $G \in C(Q;\mathbb{R})$. By definition, $u$ is a classical solution exactly when $G(t,x)=0$ for every $(t,x)\in Q$.
[/step]
[step:Derive the weak identity from the pointwise conservation law]
Assume that $u$ is a classical solution, so $G=0$ on $Q$. Let $\phi \in C_c^\infty(Q;\mathbb{R})$ be arbitrary, and let $K := \operatorname{supp}\phi \subset Q$. Since $K$ is compact in $Q$, choose numbers $a,b \in \mathbb{R}$ with $0<a<b<T$ and a bounded [open set](/page/Open%20Set) $W \subset U$ with compact closure in $U$ such that $K \subset (a,b)\times W$.
Multiplying $G=0$ by $\phi$ and integrating gives
\begin{align*}
0 = \int_0^{\,T} \int_U G(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Substituting the definition of $G$,
\begin{align*}
0 = \int_0^{\,T} \int_U \partial_t u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t) + \sum_{i=1}^n \int_0^{\,T} \int_U f_i'(u(t,x))\partial_{x_i}u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
For the time term, the compact support of $\phi$ in $(0,T)$ gives no boundary contribution, and ordinary [integration by parts](/theorems/210) in the variable $t$ gives
\begin{align*}
\int_0^{\,T} \int_U \partial_t u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
=
-\int_0^{\,T} \int_U u(t,x)\partial_t\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
For each $i \in \{1,\dots,n\}$, the chain rule gives
\begin{align*}
\partial_{x_i}\bigl(f_i(u(t,x))\bigr)=f_i'(u(t,x))\partial_{x_i}u(t,x).
\end{align*}
Since $\phi$ has compact support in $U$ with respect to the spatial variable, [integration by parts](/theorems/2098) in $x_i$ gives
\begin{align*}
\int_0^{\,T} \int_U f_i'(u(t,x))\partial_{x_i}u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
=
-\int_0^{\,T} \int_U f_i(u(t,x))\partial_{x_i}\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Combining the time and spatial identities yields
\begin{align*}
0 =
-\int_0^{\,T} \int_U \left(u(t,x)\partial_t\phi(t,x)+f(u(t,x))\cdot\nabla_x\phi(t,x)\right)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Multiplying by $-1$ proves the weak identity.
[guided]
Assume that $u$ is a classical solution. This means that the residual function $G: Q \to \mathbb{R}$ defined by
\begin{align*}
G(t,x)=\partial_t u(t,x)+\sum_{i=1}^n f_i'(u(t,x))\partial_{x_i}u(t,x)
\end{align*}
satisfies $G(t,x)=0$ for every $(t,x)\in Q$.
Let $\phi \in C_c^\infty(Q;\mathbb{R})$ be a test function. The compact support condition is the reason no boundary terms appear: there is a compact set $K:=\operatorname{supp}\phi\subset Q$, so $\phi$ vanishes outside $K$ and in particular vanishes near $t=0$, near $t=T$, and near the spatial boundary of $U$. Multiplying the pointwise equation by $\phi$ and integrating over spacetime gives
\begin{align*}
0 = \int_0^{\,T} \int_U G(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Expanding $G$ gives
\begin{align*}
0 = \int_0^{\,T} \int_U \partial_t u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t) + \sum_{i=1}^n \int_0^{\,T} \int_U f_i'(u(t,x))\partial_{x_i}u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
We now move the derivatives from the smooth solution onto the test function. For the time derivative, choose $a,b \in \mathbb{R}$ with $0<a<b<T$ such that $\phi(t,x)=0$ whenever $t\notin(a,b)$. Integration by parts is performed on the compact interval $[a,b]$, where $u$ is $C^1$ and $\phi$ is smooth. The boundary term is
\begin{align*}
\int_U u(b,x)\phi(b,x)\,d\mathcal{L}^n(x)-\int_U u(a,x)\phi(a,x)\,d\mathcal{L}^n(x),
\end{align*}
and it is zero because $\phi(a,x)=\phi(b,x)=0$ for every $x\in U$. Therefore
\begin{align*}
\int_0^{\,T} \int_U \partial_t u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
=
-\int_0^{\,T} \int_U u(t,x)\partial_t\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
For the spatial terms, fix $i\in\{1,\dots,n\}$. The chain rule applies because $f_i\in C^1(\mathbb{R};\mathbb{R})$ and $u\in C^1(Q;\mathbb{R})$, giving
\begin{align*}
\partial_{x_i}\bigl(f_i(u(t,x))\bigr)=f_i'(u(t,x))\partial_{x_i}u(t,x).
\end{align*}
Thus the $i$th spatial integral is
\begin{align*}
\int_0^{\,T} \int_U \partial_{x_i}\bigl(f_i(u(t,x))\bigr)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Integration by parts in $x_i$ moves $\partial_{x_i}$ onto $\phi$. The spatial boundary term vanishes because $\phi(t,\cdot)$ has compact support in $U$ for each $t$. Hence
\begin{align*}
\int_0^{\,T} \int_U f_i'(u(t,x))\partial_{x_i}u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
=
-\int_0^{\,T} \int_U f_i(u(t,x))\partial_{x_i}\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Summing this identity over $i=1,\dots,n$ converts the spatial terms into
\begin{align*}
-\int_0^{\,T} \int_U f(u(t,x))\cdot\nabla_x\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Combining the time and spatial integrations by parts gives
\begin{align*}
0 =
-\int_0^{\,T} \int_U \left(u(t,x)\partial_t\phi(t,x)+f(u(t,x))\cdot\nabla_x\phi(t,x)\right)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Multiplying by $-1$ proves the weak formulation.
[/guided]
[/step]
[step:Integrate the weak identity back onto the smooth residual]
Assume that $u$ satisfies the weak identity. Let $\phi \in C_c^\infty(Q;\mathbb{R})$ be arbitrary. The same integrations by parts as above, now read in the reverse direction, give
\begin{align*}
\int_0^{\,T} \int_U u(t,x)\partial_t\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
=
-\int_0^{\,T} \int_U \partial_tu(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
\end{align*}
and, for each $i\in\{1,\dots,n\}$,
\begin{align*}
\int_0^{\,T} \int_U f_i(u(t,x))\partial_{x_i}\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
=
-\int_0^{\,T} \int_U f_i'(u(t,x))\partial_{x_i}u(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
Substituting these identities into the weak formulation yields
\begin{align*}
\int_0^{\,T} \int_U G(t,x)\phi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)=0
\end{align*}
for every $\phi \in C_c^\infty(Q;\mathbb{R})$.
[/step]
[step:Use compactly supported bumps to force the residual to vanish pointwise]
We prove that $G=0$ on $Q$. Suppose instead that there exists $(t_0,x_0)\in Q$ with $G(t_0,x_0)\ne 0$. Replacing $G$ by $-G$ in the following argument if necessary, assume $G(t_0,x_0)>0$. Since $G$ is continuous, there exist $r>0$ and an open Euclidean ball $B((t_0,x_0),r)\subset Q$ such that
\begin{align*}
G(t,x) \geq \frac{G(t_0,x_0)}{2}
\end{align*}
for every $(t,x)\in B((t_0,x_0),r)$.
Define the smooth bump function $\psi: Q \to \mathbb{R}$ by setting
\begin{align*}
\psi(t,x) = \exp\left(-\frac{1}{1-\frac{|(t,x)-(t_0,x_0)|^2}{r^2}}\right)
\end{align*}
when $|(t,x)-(t_0,x_0)|<r$, and by setting $\psi(t,x)=0$ when $|(t,x)-(t_0,x_0)|\geq r$.
Then $\psi\in C_c^\infty(Q;\mathbb{R})$, $\psi\geq 0$, and $\psi(t_0,x_0)>0$. Applying the identity from the previous step with $\phi=\psi$ gives
\begin{align*}
0=\int_0^{\,T} \int_U G(t,x)\psi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t).
\end{align*}
But the support of $\psi$ is contained in $B((t_0,x_0),r)$, and on this ball $G\geq G(t_0,x_0)/2$, so
\begin{align*}
\int_0^{\,T} \int_U G(t,x)\psi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
\geq
\frac{G(t_0,x_0)}{2}
\int_0^{\,T} \int_U \psi(t,x)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(t)
>0.
\end{align*}
This contradicts the integral identity. Therefore $G(t,x)=0$ for every $(t,x)\in Q$, which is precisely the pointwise conservation law
\begin{align*}
\partial_t u(t,x)+\operatorname{div}_x f(u)(t,x)=0.
\end{align*}
[/step]
