Combining Theorem 2.3 and Corollary 2.5 of this paper gives that, for a strictly convex conservation law
$$u_t + f(u)_x = 0,$$ satisfying the entropy condition
$$\eta(u)_t + q(u)_x \le 0$$ in the sense of distribution for *one* entropy-entropy flux pair $(\eta,q)$ is equivalent to satisfying it for *all* entropy-entropy flux pair $(\eta,q)$.

On a high level, the key point of the paper is based on the fact that one entropy is enough to guarantee that the solution of the conservation law corresponds to the unique viscosity solution Hamilton-Jacobi equation $$h_t + f(h_x) = 0$$ (which is, indeed, the content of Theorem 2.3).

**Question:** Can you give me a high-level overview of this passage of the proof? Specifically, what goes on in Section 4.2 (that is, pages 7--8) of the paper? I feel like I'm loosing the intuition behind their argument and the motivation behind their estimates.

**Let me ask more precisely what are some points of pag. 7--8 of the linked paper that I'd like to clarify.**

- We want to prove that, if $\zeta$ is a smooth function such that $h-\zeta$ has a minimum at some $(t, x) \in \Omega$, then $\left[\zeta_{t}+f\left(\zeta_{x}\right)\right](t, x) \geq 0$. For simplicity we assume that $(t, x)=(0,0)$ and $[h-\zeta](0,0)=0$. Moreover, we assume that the minimum is strict. Indeed, if we choose $\varepsilon>0$ and consider $\zeta^{\varepsilon}(t, x):=\zeta(t, x)+\varepsilon\left(t^{2}+x^{2}\right)$, then $\left[h-\zeta^{\varepsilon}\right]$ has a strict minimum at $(0,0)$ and $\left[\zeta_{t}^{\varepsilon}+f\left(\zeta_{x}^{\varepsilon}\right)\right](0,0)=\left[\zeta_{t}+f\left(\zeta_{x}\right)\right](0,0)$

For any $\delta>0$ consider \begin{align*} \Omega_{\delta}:=\text { connected component of }\{(t, x):[h-\zeta](t, x)<\delta\} \text { containing }(0,0) \text {. } \end{align*}

Since $h$ is continuous and the origin is a strict minimum, $\Omega_{\delta}$ is an open set and$\operatorname{diam}\left(\Omega_{\delta}\right) \downarrow 0$ as $\delta \downarrow 0$. We introduce the notation \begin{align*} \langle g\rangle_{\delta}:=\frac{1}{\left|\Omega_{\delta}\right|} \int_{\Omega_{\delta}} g(t, x) d t d x \end{align*}

Why, heuristically, are we considering this average?

- By definition of (h),

\begin{align}\label{eq:19} \left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)\right\rangle_{\delta}=-\left\langle\left(\begin{array}{c} (h-\zeta)_{t} \\ (h-\zeta)_{x} \end{array}\right)\right\rangle_{\delta}+\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right)\right\rangle_{\delta} \end{align}

For $\delta$ sufficiently small we have $\Omega_{\delta} \subset \subset B_{1}$. Thus \begin{align*} \left\langle(h-\zeta)_{t}\right\rangle_{\delta}=\int_{\Omega_{\delta}}(h-\zeta)_{t}=\frac{1}{\left|\Omega_{\delta}\right|} \int_{\Omega_{\delta}}(h-\zeta)_{t}=\frac{1}{\left|\Omega_{\delta}\right|} \int_{B_{1}}(\min \{h-\zeta-\delta, 0\})_{t} \end{align*}

Since the function $\min \{h-\zeta-\delta, 0\}$ is continuous and identically zero on a neighborhood of $\partial B_{1}$, the right hand side above vanishes. The same argument applies to $\left\langle(h-\zeta)_{x}\right\rangle_{\delta}$. Hence, from \eqref{eq:19} we get \begin{align*} \left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)\right\rangle_{\delta}=\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right)\right\rangle_{\delta} \end{align*}

Why is it true that For $\delta$ sufficiently small we have $\Omega_{\delta} \subset \subset B_{1}$? Why exactly, by continuity, $\min \{h-\zeta-\delta, 0\}$ is identically zero on a neighborhood of $\partial B_{1}$?

On the other hand we have \begin{align*} \left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right) \cdot\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta} \geq\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right) \cdot\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta} \end{align*}

It seems that, up to this point, we stated an entropy inequality for the "averaged" equation. Is this a good intuition of what happens? Why does it help?

- Using the Proposition 3.2 (already discussed in the answer below), we obtain \begin{align*} \begin{aligned} &\frac{c^{2}}{3}\left\langle\left(u-\langle u\rangle_{\delta}\right)^{4}\right\rangle_{\delta} \\ &\quad \leq\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right) \cdot\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta}-\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right)\right\rangle_{\delta} \cdot\left\langle\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta} \\ &\leq \quad C \sup _{\Omega_{\delta}}\left|\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)-\left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)\right\rangle_{\delta}\right| \end{aligned} \end{align*}

And, since $\zeta$ is smooth, we deduce

\begin{align*} \lim _{\delta \downarrow 0}\left\langle\left(u-\langle u\rangle_{\delta}\right)^{4}\right\rangle_{\delta}=0 \end{align*} and, assuming $f$ Lipschitz, \begin{align*} \left|\langle f(u)\rangle_{\delta}-f\left(\langle u\rangle_{\delta}\right)\right| \leq C\left\langle\left|u-\langle u\rangle_{\delta}\right|\right\rangle_{\delta} \leq C\left\langle\left(u-\langle u\rangle_{\delta}\right)^{4}\right\rangle_{\delta}^{1 / 4} \longrightarrow 0 \text{ as \delta \to 0} \end{align*}

Are we here only applying Lebesgue's dominated convergence theorem? The final conclusion seems to be that the average of the flux is equal to the flux of the average of $u$; where does the intuition for this come from and lead to ?

- Plugging this back into the equations above gives \begin{align*} \lim _{\delta \downarrow 0}\left|-\left\langle\zeta_{t}\right\rangle_{\delta}-f\left(\left\langle\zeta_{x}\right\rangle_{\delta}\right)\right|=0 \end{align*} and, since $\zeta$ is smooth, \begin{align*} -\zeta_{t}(0,0)-f\left(\zeta_{x}(0,0)\right)=0 \end{align*} which is the conclusion.

What is the final message of this proof? Can it be streamlined/simplified a bit?