[proofplan] The proof first fixes the Riemann data and the entropy pair associated to an arbitrary entropy level. For rarefactions, the self-similar profile is continuous and monotone, so bounded-variation calculus reduces the entropy identity to a one-dimensional distributional identity. For shocks, the entropy inequality is a single jump condition along the discontinuity, and strict convexity turns that jump condition into the usual admissibility inequality. The constant case is immediate. [/proofplan]

[step:Define the Riemann data and the Kruzhkov entropy flux associated to a level $k$] Let $u_L\in\mathbb{R}$ denote the left Riemann state and let $u_R\in\mathbb{R}$ denote the right Riemann state. Define the Riemann initial datum $u_0:\mathbb{R}\to\mathbb{R}$ by $u_0(x)=u_L$ for $x<0$ and $u_0(x)=u_R$ for $x>0$; the value at $x=0$ is irrelevant for local $L^1$ convergence. Let $u:\mathbb{R}\times(0,\infty)\to\mathbb{R}$ denote the shock, rarefaction, or constant wave selected by the complete convex scalar Riemann solver for these states. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let $\operatorname{sgn}:\mathbb{R}\to\{-1,0,1\}$ denote the sign function, defined by $\operatorname{sgn}(r)=1$ for $r>0$, $\operatorname{sgn}(r)=0$ for $r=0$, and $\operatorname{sgn}(r)=-1$ for $r<0$. Fix $k \in \mathbb{R}$. Define the Kruzhkov entropy $\eta_k: \mathbb{R} \to [0,\infty)$ and entropy flux $q_k: \mathbb{R} \to \mathbb{R}$ by \begin{align*} \eta_k(a) = |a-k| \end{align*} and \begin{align*} q_k(a) = \operatorname{sgn}(a-k)(f(a)-f(k)). \end{align*} The desired entropy inequality is the distributional inequality $\partial_t \eta_k(u) + \partial_x q_k(u) \leq 0$ on $\mathbb{R} \times (0,\infty)$ together with the initial trace $u(\cdot,t) \to u_0$ locally in $L^1(\mathbb{R})$ as $t \downarrow 0$. The initial trace follows directly from the explicit formulas. In the shock and constant cases it is pointwise away from $x=0$. In the rarefaction case, for every $R>0$ the set on which $u(\cdot,t)$ differs from $u_0$ inside $(-R,R)$ is contained in the interval between $f'(u_L)t$ and $f'(u_R)t$, whose length is $(f'(u_R)-f'(u_L))t$, while $u$ takes values in $[u_L,u_R]$ there. Hence $u(\cdot,t)\to u_0$ locally in $L^1(\mathbb{R})$ as $t\downarrow0$. In this proof, saying that the Riemann wave is a Kruzhkov [entropy solution](/page/Entropy%20Solution) means exactly that the interior distributional inequality above holds for every $k\in\mathbb R$ and that the displayed local $L^1$ initial trace holds. The remaining steps prove the interior distributional inequality in the rarefaction, shock, and constant cases. [/step]

[step:Verify the entropy identity for the centered rarefaction]Assume $u_L<u_R$. Strict convexity of $f$ implies that $f':\mathbb{R}\to\mathbb{R}$ is strictly increasing, because for $a<b<c$ the slopes of the two secants satisfy \begin{align*} \frac{f(b)-f(a)}{b-a}\leq\frac{f(c)-f(a)}{c-a}\leq\frac{f(c)-f(b)}{c-b}. \end{align*} Since $f\in C^2(\mathbb{R})$, the map $f'$ is continuous. Hence the restriction of $f'$ to $[u_L,u_R]$ has a continuous inverse from $[f'(u_L),f'(u_R)]$ to $[u_L,u_R]$. Define the self-similar profile $U:\mathbb{R}\to\mathbb{R}$ by setting $U(\xi)=u_L$ for $\xi\leq f'(u_L)$, $U(\xi)=(f'|_{[u_L,u_R]})^{-1}(\xi)$ for $f'(u_L)\leq\xi\leq f'(u_R)$, and $U(\xi)=u_R$ for $\xi\geq f'(u_R)$. Then the rarefaction is the continuous map $u:\mathbb{R}\times(0,\infty)\to\mathbb{R}$ given by $u(x,t)=U(x/t)$. Let $A:\mathbb{R}\to[0,\infty)$ and $B:\mathbb{R}\to\mathbb{R}$ be defined by $A(\xi)=\eta_k(U(\xi))$ and $B(\xi)=q_k(U(\xi))$. It suffices to prove, in distributions on $\mathbb{R}$, \begin{align*} \frac{d}{d\xi}(B(\xi)-\xi A(\xi))=-A(\xi). \end{align*} Indeed, after the change of variables $x=t\xi$, the identity above is exactly the weak form of $\partial_t\eta_k(u)+\partial_xq_k(u)=0$ on $\mathbb{R}\times(0,\infty)$. The function $U$ is monotone. The maps $\eta_k$ and $q_k$ are Lipschitz on the compact interval with endpoints $u_L$ and $u_R$, because $f$ is $C^1$ there and the two one-sided formulas for $q_k$ agree at $k$. Hence $A$ and $B$ have bounded variation on compact intervals. Split the real line at the fan endpoints and, if $k\in(u_L,u_R)$, at the single point $\xi=f'(k)$. On each resulting open interval the sign of $U-k$ is fixed, so $\eta_k$ and $q_k$ are $C^1$ functions of $U$ there. The Stieltjes chain rule for $C^1$ functions composed with the monotone function $U$ therefore applies on each such interval, including the absolutely continuous and singular continuous parts of the measure $dU$. If $U<k$, then \begin{align*} B(\xi)-\xi A(\xi)=-f(U(\xi))+f(k)-\xi(k-U(\xi)), \end{align*} and using $\xi=f'(U(\xi))$ on the fan gives \begin{align*} d(B-\xi A)=(U(\xi)-k)\,d\mathcal{L}^1(\xi)=-A(\xi)\,d\mathcal{L}^1(\xi). \end{align*} If $U>k$, then \begin{align*} B(\xi)-\xi A(\xi)=f(U(\xi))-f(k)-\xi(U(\xi)-k), \end{align*} and again $\xi=f'(U(\xi))$ yields \begin{align*} d(B-\xi A)=(k-U(\xi))\,d\mathcal{L}^1(\xi)=-A(\xi)\,d\mathcal{L}^1(\xi). \end{align*} On the constant regions outside the fan, $B-\xi A$ is affine with derivative $-A$. The preceding open intervals account for all non-atomic parts of the [distributional derivative](/page/Distributional%20Derivative). At any interface where the formula for $U$ changes, $U$ is continuous, so $B-\xi A$ has no jump atom. If $k\in(u_L,u_R)$, then at the point $\xi=f'(k)$ the two displayed formulas agree and both give $B-\xi A=0$, so there is no atom there either. Therefore $d(B-\xi A)=-A\,d\mathcal{L}^1$ on all of $\mathbb{R}$, which proves $\partial_t\eta_k(u)+\partial_xq_k(u)=0$ in distributions.[/step]

[guided]The rarefaction proof must not assume $f''>0$ pointwise. Strict convexity together with $f\in C^2(\mathbb{R})$ guarantees that $f'$ is continuous and strictly increasing, but $f''$ may vanish at isolated or even more complicated sets. Therefore we prove the entropy identity in the self-similar variable using monotonicity and bounded-variation calculus rather than differentiating the inverse of $f'$. First define the rarefaction profile carefully. Strict convexity implies the secant slopes are monotone: for $a<b<c$, \begin{align*} \frac{f(b)-f(a)}{b-a}\leq\frac{f(c)-f(a)}{c-a}\leq\frac{f(c)-f(b)}{c-b}. \end{align*} Letting the endpoints approach shows that $f'$ is increasing, and strict convexity makes it strictly increasing. Since $f'$ is continuous, its restriction to $[u_L,u_R]$ has a continuous inverse. Define $U:\mathbb{R}\to\mathbb{R}$ by setting $U(\xi)=u_L$ for $\xi\leq f'(u_L)$, $U(\xi)=(f'|_{[u_L,u_R]})^{-1}(\xi)$ for $f'(u_L)\leq\xi\leq f'(u_R)$, and $U(\xi)=u_R$ for $\xi\geq f'(u_R)$. Then $u(x,t)=U(x/t)$ is continuous, and $U$ is monotone. For the entropy level $k$, define $A:\mathbb{R}\to[0,\infty)$ and $B:\mathbb{R}\to\mathbb{R}$ by $A(\xi)=\eta_k(U(\xi))$ and $B(\xi)=q_k(U(\xi))$. The self-similar weak computation reduces the entropy identity to \begin{align*} \frac{d}{d\xi}(B(\xi)-\xi A(\xi))=-A(\xi) \end{align*} in distributions on the $\xi$-line. This reduction follows by testing against a smooth compactly supported function and changing variables $x=t\xi$; the derivative of the factor $t$ from $d\mathcal{L}^1(x)=t\,d\mathcal{L}^1(\xi)$ is exactly the extra $-A$ term in the displayed one-dimensional identity. Now compute $B-\xi A$ on the regions where the sign of $U-k$ is fixed. If $U<k$, then \begin{align*} B(\xi)-\xi A(\xi)=-f(U(\xi))+f(k)-\xi(k-U(\xi)). \end{align*} Using the Stieltjes chain rule for $C^1$ functions of the monotone function $U$, including any singular continuous part of $dU$, and the identity $\xi=f'(U(\xi))$ on the fan, the terms involving $dU$ cancel, leaving \begin{align*} d(B-\xi A)=(U(\xi)-k)\,d\mathcal{L}^1(\xi)=-A(\xi)\,d\mathcal{L}^1(\xi). \end{align*} If $U>k$, then \begin{align*} B(\xi)-\xi A(\xi)=f(U(\xi))-f(k)-\xi(U(\xi)-k), \end{align*} and the same cancellation gives \begin{align*} d(B-\xi A)=(k-U(\xi))\,d\mathcal{L}^1(\xi)=-A(\xi)\,d\mathcal{L}^1(\xi). \end{align*} On the constant regions outside the fan, the formula is affine in $\xi$ and has derivative $-A$. At the fan boundaries, $U$ is continuous, so no jump measure appears. If $k$ lies between $u_L$ and $u_R$, the point where $U=k$ also creates no atom because $A=0$ and $B=0$ from both sides there. Thus the one-dimensional distributional identity holds globally. Consequently $\partial_t\eta_k(u)+\partial_xq_k(u)=0$ in $\mathbb{R}\times(0,\infty)$, which is stronger than the required entropy inequality.[/guided]

[step:Reduce the shock entropy inequality to a scalar jump condition] Assume $u_L>u_R$. Define the shock speed $s\in\mathbb{R}$ by \begin{align*} s = \frac{f(u_L)-f(u_R)}{u_L-u_R}. \end{align*} For the piecewise constant shock, $\eta_k(u)$ and $q_k(u)$ are constant on the two sides of the line $x=st$. The distributional entropy inequality $\partial_t \eta_k(u)+\partial_x q_k(u)\leq 0$ is therefore equivalent to the jump condition \begin{align*} s(\eta_k(u_L)-\eta_k(u_R))-(q_k(u_L)-q_k(u_R))\leq 0. \end{align*} This condition follows by direct testing. Let $\varphi:\mathbb{R}\times(0,\infty)\to[0,\infty)$ be any function in $C_c^\infty(\mathbb{R}\times(0,\infty))$. Integrating the two constant states on the regions $x<st$ and $x>st$ and applying one-dimensional [integration by parts](/theorems/210) in the normal direction to the line $x=st$ gives a boundary contribution equal to the displayed coefficient multiplied by the non-negative trace of $\varphi$ on the shock curve. Since this trace is non-negative for every such [test function](/page/Test%20Function), the distributional inequality is equivalent to the displayed coefficient being non-positive. If $k\notin[u_R,u_L]$, then $\operatorname{sgn}(u_L-k)=\operatorname{sgn}(u_R-k)$. The defining formula for the shock speed gives \begin{align*} s(u_L-u_R)=f(u_L)-f(u_R), \end{align*} so the jump coefficient is $0$. Hence only $k\in[u_R,u_L]$ must be considered. [/step]

[step:Use strict convexity to prove the shock jump condition]Let $k\in[u_R,u_L]$. If $u_R<k<u_L$, convexity applied to the representation \begin{align*} k=\frac{u_L-k}{u_L-u_R}u_R+\frac{k-u_R}{u_L-u_R}u_L \end{align*} gives \begin{align*} f(k)\leq\frac{u_L-k}{u_L-u_R}f(u_R)+\frac{k-u_R}{u_L-u_R}f(u_L). \end{align*} Rearranging this single inequality gives both \begin{align*} f(k)-f(u_R)\leq s(k-u_R) \end{align*} and \begin{align*} f(u_L)-f(k)\geq s(u_L-k). \end{align*} When $k=u_R$ or $k=u_L$, the corresponding inequality is an equality and the other follows from the same convexity inequality by continuity. For $k\in[u_R,u_L]$, we have \begin{align*} \eta_k(u_L)=u_L-k \end{align*} and \begin{align*} \eta_k(u_R)=k-u_R. \end{align*} Also \begin{align*} q_k(u_L)=f(u_L)-f(k) \end{align*} and \begin{align*} q_k(u_R)=f(k)-f(u_R). \end{align*} Using the two convexity inequalities, \begin{align*} f(u_L)-f(k)\geq s(u_L-k) \end{align*} and \begin{align*} f(k)-f(u_R)\leq s(k-u_R), \end{align*} we obtain \begin{align*} q_k(u_L)-q_k(u_R)=(f(u_L)-f(k))-(f(k)-f(u_R))\geq s((u_L-k)-(k-u_R)). \end{align*} Since \begin{align*} \eta_k(u_L)-\eta_k(u_R)=(u_L-k)-(k-u_R), \end{align*} it follows that \begin{align*} s(\eta_k(u_L)-\eta_k(u_R))-(q_k(u_L)-q_k(u_R))\leq 0. \end{align*} The shock satisfies the Kruzhkov entropy inequality for every $k\in\mathbb{R}$.[/step]

[guided]The delicate point is the sign of the jump coefficient. The entropy inequality requires the coefficient of the measure supported on $x=st$ to be non-positive, so we must prove that the entropy flux jump is at least the speed $s$ times the entropy jump. Fix $k\in[u_R,u_L]$. We derive the needed convexity estimates directly. If $u_R<k<u_L$, then \begin{align*} k=\frac{u_L-k}{u_L-u_R}u_R+\frac{k-u_R}{u_L-u_R}u_L. \end{align*} Convexity of $f$ therefore gives \begin{align*} f(k)\leq\frac{u_L-k}{u_L-u_R}f(u_R)+\frac{k-u_R}{u_L-u_R}f(u_L). \end{align*} Rearranging this inequality once yields \begin{align*} f(k)-f(u_R)\leq s(k-u_R), \end{align*} and rearranging it the other way yields \begin{align*} f(u_L)-f(k)\geq s(u_L-k). \end{align*} If $k=u_R$ or $k=u_L$, the same conclusions follow by the displayed inequalities with the vanishing endpoint term interpreted as $0$. For this range of $k$, the signs in the Kruzhkov pair are fixed: $u_L-k\geq0$ and $u_R-k\leq0$. Hence \begin{align*} \eta_k(u_L)=u_L-k \end{align*} and \begin{align*} \eta_k(u_R)=k-u_R, \end{align*} while \begin{align*} q_k(u_L)=f(u_L)-f(k) \end{align*} and \begin{align*} q_k(u_R)=f(k)-f(u_R). \end{align*} Subtracting the two convexity estimates in the correct direction gives \begin{align*} q_k(u_L)-q_k(u_R)=(f(u_L)-f(k))-(f(k)-f(u_R))\geq s(u_L-k)-s(k-u_R). \end{align*} The right-hand side is $s$ times the entropy jump, because \begin{align*} \eta_k(u_L)-\eta_k(u_R)=(u_L-k)-(k-u_R). \end{align*} Therefore \begin{align*} q_k(u_L)-q_k(u_R)\geq s(\eta_k(u_L)-\eta_k(u_R)). \end{align*} Rearranging yields the required non-positive jump coefficient: \begin{align*} s(\eta_k(u_L)-\eta_k(u_R))-(q_k(u_L)-q_k(u_R))\leq 0. \end{align*} This proves the Kruzhkov entropy inequality for shocks.[/guided]

[step:Handle the constant state and conclude] If $u_L=u_R$, then $u(x,t)=u_L$ is constant. Hence, for every $k\in\mathbb{R}$, $\partial_t \eta_k(u)+\partial_x q_k(u)=0$ as a distribution on $\mathbb{R}\times(0,\infty)$, and the initial trace is immediate. The rarefaction case gives entropy equality, the shock case gives entropy inequality by the convex jump computation, and the constant case gives entropy equality. Combining these conclusions with the initial trace verification proves that every wave selected by the convex scalar Riemann solver is a Kruzhkov entropy solution with Riemann initial datum $u_0$. [/step]