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Apply the entropy solution inequality for $u(t,x)$ with level $v(s,y)$ and for $v(s,y)$ with level $u(t,x)$, integrate both in the other pair of variables, and add them. Symmetry of the Kruzhkov entropy flux produces derivatives ${\partial }_t+{\partial }_s$ and ${\partial }_x+{\partial }_y$. A nonnegative mollifier concentrated near the diagonal then yields the Kato inequality for scalar conservation laws. Boundary-time terms require an initial trace argument, not merely interior translation continuity.

If $|f^{\prime }|\le M$ on the solutions' state range, the Kato inequality for scalar conservation laws implies the displayed estimate for almost every $s\in [0,t]$. Approximate the shrinking interval by smooth cutoffs $W$ satisfying $W_s+M|W_x|\le 0$ and multiply by a temporal cutoff ending at $s$. Since $|Q|\le M|u-v|$, the lateral flux cannot increase the integral. The estimate gives uniqueness and a finite propagation speed without global $L^1$ assumptions on the data.

Subtract the weak equation for $u-v$ from its Kato inequality for scalar conservation laws and divide by two. The result is an inequality for $(v-u{)}_{+}$ with a flux bounded in magnitude by $M(v-u{)}_{+}$. The same shrinking-interval argument proves $u_0\ge v_0\Rightarrow u\ge v$ almost everywhere. Choosing the constant solution $v=0$ proves preservation of nonnegativity even when $f(0)\ne 0$. Two-sided absolute-value contraction alone should not be mistaken for this one-sided comparison proof.

For the stated mollifier, $({\partial }_t+{\partial }_s){\chi }_{\epsilon }=({\partial }_x+{\partial }_y){\chi }_{\epsilon }=0$. Thus the interior doubled integral isOn a bounded state range, $H$ is Lipschitz in each argument, and so is $Q$: away from the diagonal each partial derivative has modulus at most $\sup |f^{\prime }|$, and the function is continuous across the diagonal. Local $L^1$ translation continuity therefore makes this converge to the corresponding single-time, single-space integral. The compact time support is chosen below $T$, with a margin for $s\le t+2\epsilon$.

The initial terms need a separate argument. Since the time kernel is supported at $t-s\in [-2\epsilon ,-\epsilon ]$, $B_v$ is identically zero for $t\ge 0$, whereas $B_u$ samples $v(s,\cdot )$ at $\epsilon \le s\le 2\epsilon$.

The averaged initial trace of an entropy solution follows directly from its inequalities. Test the constant-level entropy for $v$ with $(1-s/\delta {)}_{+}\rho (y)$, using smooth approximations of that temporal cutoff. Boundedness of $v$ and its entropy flux givesApproximate $v_0$ locally in $L^1$ by finitely many constants with a smooth nonnegative partition of unity. Applying this estimate on each partition member and using the triangle inequality makes the limsup of ${\delta }^{-1}{\int }_0^{\delta }\Vert v(s)-v_0{\Vert }_{L^1(K)}ds$ arbitrarily small. This proves its convergence to zero for each compact $K$.

The time kernel has size $O({\epsilon }^{-1})$ and the spatial kernel is a normalized approximate identity. The averaged trace, followed by spatial $L^1$ translation continuity of $v_0$, therefore gives $B_u\to \int |u_0-v_0|\phi (0)$. Its time mass is one, not one-half, because its whole support lies on the positive $s$ side. ConsequentlyThis is the Kato inequality for scalar conservation laws with its initial contribution. Assuming an initial trace solely from interior translation continuity would leave a gap; the entropy argument above supplies it.

Put $h=|u-v|$, $q=Q(u,v)$ and $h_0=|u_0-v_0|$. The mean value theorem gives $|q|\le Mh$ on the bounded state range. Fix a final time $t$ and let $L(\tau )=a-M(t-\tau )$ and $R(\tau )=b+M(t-\tau )$. Choose a smooth increasing function $\rho$ equal to zero on $(-\infty ,0]$ and one on $[1,\infty )$. For small $\delta >0$, defineIts two nonnegative derivative contributions imply $(W_{\delta }{)}_{\tau }+M|(W_{\delta }{)}_x|\le 0$. Therefore $h(W_{\delta }{)}_{\tau }+q(W_{\delta }{)}_x\le 0$.

Use the nonnegative test $\phi (\tau ,x)={\eta }_{ϵ}(\tau )W_{\delta }(\tau ,x)$, where ${\eta }_{ϵ}(0)=1$, it is one up to just before $s$, decreases smoothly to zero near $s$, and remains zero afterward. The Kato inequality for scalar conservation laws givesbecause the other interior contribution is nonpositive. At Lebesgue times $s$, the temporal approximate identity tends to $\int h(s)W_{\delta }(s)$; take a countable sequence $\delta \downarrow 0$ and use dominated convergence on the bounded cone. This yieldsfor almost every $s\in [0,t]$. This is local L1 contraction for scalar conservation laws. It uses only local integrability, so the bounded data need not have finite global $L^1$ norm. If $M=0$, the flux difference is zero and the same proof uses a stationary interval.

If $u_0=v_0$, the right-hand side of the local L1 contraction for scalar conservation laws estimate is zero. Applying it on a countable collection of rational intervals and final times proves $u=v$ almost everywhere in space-time. Thus

To deduce nonnegativity one needs a one-sided comparison, rather than merely putting $v=0$ in the absolute-value estimate. The difference $w=u-v$ satisfies the weak equation with flux $f(u)-f(v)$. Subtract its weak integral identity from the Kato inequality for scalar conservation laws and divide by two. The result is the same inequality forWhen $w\ge 0$ both expressions vanish; when $w<0$, $|q_{-}|\le M{w}_{-}$. The shrinking-interval proof therefore bounds the negative part by its initial negative part. This is order preservation for scalar entropy solutions.

The constant $v=0$ is an entropy solution for every $f$, even if $f(0)\ne 0$, because the constant flux has zero spatial derivative. If $u_0\ge 0$, its initial negative part is zero, and comparison givesThis explicitly establishes the additional sign conclusion without incorrectly identifying uniqueness alone with order preservation.