Local compactness and initial trace for BV finite-volume states

For every $t$, $W_m(\cdot ,t)$ is piecewise constant with values in $I$ and spatial variation at most $V$: on each time slab it is a convex combination of two states whose variations are at most $V$ by [F2]. Here the variation of a piecewise-constant section on an interval is the sum of the absolute jump sizes inside that interval. Fix an integer $R\ge 1$ and consider any sequence $v_m$ of such sections on $[-R,R]$. At each rational point in $[-R,R]$, the values lie in the compact interval $I$. The bounded-value subsequence at one rational point exists by [F6]; explicitly, partition $I$ into finitely many closed intervals of length at most $2^{-1}$, take the first interval containing infinitely many values, and repeat inside it with successive meshes $2^{-2},2^{-3},\dots$, always selecting the first subinterval with infinitely many retained indices. The nested intervals have diameters tending to zero, so the least increasing indices in the nested sets give a convergent subsequence. Use the enumeration fixed by [F5], and repeat this deterministic extraction successively on the current subsequence at each enumerated rational that lies in $[-R,R]$, skipping the others. The nested subsequences yield a diagonal subsequence converging at every rational point; each interval and index is selected by a least-index rule, so this uses no choice axiom. Given $\delta >0$, choose a finite rational partition of $[-R,R]$ with mesh at most $\delta$, possible by [F4]. For each subinterval $J=[r_i,r_{i+1}]$ and any two selected functions $v_p,v_q$, variation gives $\mathrm{\mid }{v}_p(x)-v_q(x)\mathrm{\mid }\le \mathrm{\mid }{v}_p(r_i)-v_q(r_i)\mathrm{\mid }+{\mathrm{Var}}_J(v_p)+{\mathrm{Var}}_J(v_q)$ for almost every $x\in J$. Integrating and summing yields $\mathrm{\parallel }{v}_p-v_q{\mathrm{\parallel }}_{L^1([-R,R])}\le 2\delta V+2R{\max }_i\mathrm{\mid }{v}_p(r_i)-v_q(r_i)\mathrm{\mid }\mathrm{.}$ For the fixed partition the maximum tends to zero as $p,q\to \mathrm{\infty }$; then taking $\delta \downarrow 0$ proves the selected sections are Cauchy in local $L^1$.

On each slab, ${\mathrm{\partial }}_t{W}_m=(U_j^{n+1}-U_j^n)\mathrm{/}{\tau }_m$ on cell $I_j$. By [F2], its $L^1(\mathbb{R})$ norm is at most $C:=(L_1+L_2)V$. Integrating over any time interval gives $\mathrm{\parallel }{W}_m(\cdot ,t)-W_m(\cdot ,s){\mathrm{\parallel }}_{L^1(\mathbb{R})}\le C\mathrm{\mid }t-s\mathrm{\mid }\mathrm{.}$ Use the enumeration fixed by [F5] and process its indices together with integers $R\ge 1$ in order of increasing sum. For each pair $(\mathrm{\ell },R)$, if $q_{\mathrm{\ell }}\in [0,T]$, apply step 1.1 at time $q_{\mathrm{\ell }}$ on $[-R,R]$; otherwise skip that pair. The resulting diagonal extraction covers every rational time in $[0,T]$ and every integer interval. The subsequence converges in $L^1([-R,R])$ at every rational time for every integer $R$; this recursive construction uses no choice axiom.

Fix $R$ and $\epsilon >0$, and put $\delta :=\epsilon \mathrm{/}(4(C+1))$. Choose a finite set of rational times in $[0,T]$ that is a $\delta$-net: include $0$, choose $q\in \mathbb{Q}\cap [0,T]$ with $T-q<\delta$ (using $q=0$ if $T<\delta$), and subdivide $[0,q]$ by rational points with gaps below $\delta$, possible by [F4]. At those finitely many times the subsequence in step 2.1 is Cauchy in $L^1([-R,R])$. For any $t\in [0,T]$, choose a net time $s$ within the gap bound; the triangle inequality and the uniform time estimate in step 2.1 give $\mathrm{\parallel }{W}_p(\cdot ,t)-W_q(\cdot ,t){\mathrm{\parallel }}_{L^1([-R,R])}\le 2C\mathrm{\mid }t-s\mathrm{\mid }+\mathrm{\parallel }{W}_p(\cdot ,s)-W_q(\cdot ,s){\mathrm{\parallel }}_{L^1([-R,R])}\mathrm{.}$ Thus the subsequence is uniformly Cauchy in time in local $L^1$. For fixed $t,R$, its bounded measurable sections have an $L^1([-R,R])$ limit: for each $j$, let $N_j$ be the least integer such that all $p,q\ge N_j$ have $L^1([-R,R])$ difference below $2^{-j}$, which exists by Cauchyness. Put ${\widehat{N}}_j:={\max }_{1\le r\le j}{N}_r$ and recursively choose the least increasing indices $n_j\ge {\widehat{N}}_j$. Then both $n_j,n_{j+1}\ge N_j$, so successive differences have norms below $2^{-j}$. By [F8] their pointwise absolute sum is finite almost everywhere, and dominated convergence [F7] gives the $L^1$ limit. Uniqueness makes these limits consistent as $R$ increases, and the uniform Cauchy estimate makes $t\mapsto u(t)$ continuous in each local $L^1$ norm. The subsequence is also Cauchy in $L^1([-R,R]\times (0,T))$ for every integer $R$, by integrating its uniform-in-time Cauchy bound. For each $j$, let $M_j$ be the least integer such that all $p,q\ge M_j$ have difference below $2^{-j}$ on every cylinder with integer $R\le j$; this exists because there are finitely many such cylinders and the sequence is Cauchy on each. Put ${\widehat{M}}_j:={\max }_{1\le r\le j}{M}_r$ and recursively choose the least increasing indices $m_j\ge {\widehat{M}}_j$. For every fixed integer $R$ and every $j\ge R$, both $m_j,m_{j+1}\ge M_j$, so their successive differences have norm below $2^{-j}$ on that cylinder. [F8] then gives a finite pointwise sum of differences almost everywhere on each cylinder, and [F7] yields a measurable space-time limit $v$ with strong local $L^1$ convergence. For each integer $R$, the sum of the $L^1([-R,R]\times (0,T))$ norms of these successive differences is finite. By [F9], for almost every $t$ their spatial absolute sum is integrable on $[-R,R]$; hence $W_{m_j}(\cdot ,t)$ converges almost everywhere in $x$ there to $v(\cdot ,t)$. Since all these sections and their limit take values in the bounded interval $I$, dominated convergence [F7] gives convergence to $v(\cdot ,t)$ in $L^1([-R,R])$. But this is a subsequence of the sequence that converges in $L^1([-R,R])$ to $u(t)$ for every $t$, so uniqueness of the $L^1$ limit gives $v(\cdot ,t)=u(t)$ for almost every $t$. Intersecting these full-measure time sets over integer $R$ shows that the space-time limit and the continuous local-$L^1$ map are the same limit on all bounded intervals.

Every $W_m$ takes values in $I$, so its local $L^1$ limit also takes values in $I$ almost everywhere: if, for example, $u(t)>a_{+}$ on a set of positive measure, then some set where $u(t)\ge a_{+}+1\mathrm{/}r$ has positive measure, contradicting $L^1$ convergence from functions bounded by $a_{+}$. The same argument applies to the lower bound.

For $r<0$, translation invariance of Lebesgue measure gives the same bound as for $-r$, so $\mathrm{\parallel }{u}_0(\cdot +r)-u_0{\mathrm{\parallel }}_{L^1(\mathbb{R})}\le \mathrm{\mid }r\mathrm{\mid }V$ for every $r\in \mathbb{R}$. On each cell, the initial cell-average error is controlled by translations: ${\sum }_j{\int }_{I_j}\mathrm{\mid }{u}_0(x)-U_j^0\mathrm{\mid }dx\le \frac{1}{h}{\int }_{-h}^h{\int }_{\mathbb{R}}\mathrm{\mid }{u}_0(x+r)-u_0(x)\mathrm{\mid }dxdr\le hV\mathrm{.}$ The first inequality follows by writing each cell average as an integral over that same cell, applying the triangle inequality, and changing variables to $r=y-x$; the restriction that $x$ and $x+r$ lie in the same cell can only decrease the integral. Since $W_m(\cdot ,0)=U^0$, step 3.1 gives $u(0)=u_0$ in local $L^1$.

On each slab, $\mathrm{\mid }{W}_m-U_j^n\mathrm{\mid }=\theta \mathrm{\mid }{U}_j^{n+1}-U_j^n\mathrm{\mid }$. Integrating in time contributes the factor ${\tau }_m\mathrm{/}2$; the time-increment bound in [F2] gives, on $[0,T]$, $\mathrm{\parallel }{W}_m-{\mathcal{U}}_m{\mathrm{\parallel }}_{L^1(\mathbb{R}\times (0,T))}\le \frac{(T+{\tau }_m){\tau }_m}{2}(L_1+L_2)V⟶0.$ Consequently ${\mathcal{U}}_m$ converges strongly in local space-time $L^1$ to the same limit as $W_m$.