Proof. We induct on \(m\). Write \(\mu(u)=\min_i u_i\) and \(\mathbf 1=(1,\ldots,1)\). Since the largest coordinate of \(X_h(u)\) is \(h^{\mu(u)}\), we first construct a family for each fixed value \(t\) of the smallest exponent. We then extend its guarantee to a neighborhood of \(t\) and combine finitely many such families by shifting their offsets. The fixed-minimum construction is a singleton when \(m=1\); the extension and combination therefore also prove the base case.

Throughout the proof, asymptotic comparisons along \(h_\nu\to0+\) and \(p^\nu\to p\) use fixed gaps between limiting exponents. In particular, if \(p_i>q\), then eventually \(p_i^\nu\geq q+\delta\) for some \(\delta>0\), so \(h_\nu^{p_i^\nu}=o(h_\nu^q)\); if \(p_i<q\), the ratio \(h_\nu^{p_i^\nu}/h_\nu^q\) tends to infinity. We shall use these comparisons only with strict limiting gaps.

A family for a fixed minimum. Fix \(t\in[a,b]\). We construct a finite family \(\Lambda_t\) containing \((t\mathbf 1,0)\), with \(\mu(r)=t\) for every slope, such that the conclusion of the lemma holds for activity within \(\Lambda_t\) whenever \(\mu(p)=t\).

If \(m=1\) or \(t=b\), the family \(\{(t\mathbf 1,0)\}\) suffices: a point \(p\in[a,b]^m\) with \(\mu(p)=t\) then equals \(t\mathbf 1\). Otherwise \(m>1\) and \(t<b\). Say a label has type \(k\) if exactly \(k\) entries of its slope exceed \(t\), the remaining entries being \(t\). Start with the type \(0\) label \((t\mathbf 1,0)\), and construct types \(k=1,\ldots,m-1\) successively. After type \(k-1\) has been constructed, choose \(\beta_k\) satisfying \[\begin{gather*} t<\beta_k<b,\qquad \beta_k<\beta_{k-1}\quad\text{if }k>1,\\ \beta_k-t<E(r) \quad\text{for every slope $r$ of type $k-1$.} \end{gather*}\] These conditions use only finitely many positive values of \(E\), so such a choice is possible. The new type \(k\) labels will force every limiting exponent on a coordinate left at \(t\) by a type \(k-1\) label that stays active to be at most \(\beta_k\). The last inequality will then bound the error by that label’s tolerance.

For each \(k\)-element coordinate set \(U\subset\{1,\ldots,m\}\) and \(s\in[\beta_k,b]^U\), let \(r(s)\) equal \(s\) on \(U\) and \(t\) off \(U\). Apply the induction hypothesis in dimension \(k\) to this cube with tolerance \(s\mapsto E(r(s))\), identifying \(U\) with its naturally ordered coordinates. For every resulting label \((s,d)\), add to \(\Lambda_t\) the type \(k\) label \[ r=r(s),\qquad c(h)=-h^{\beta_k}+d(h), \qquad d(h)=o(h^{\beta_k}). \tag{3}\] Here and below \(c\) denotes a local offset. The resulting family is finite, and all its offsets are \(o(h^t)\). The negative offset favors raising slope entries on coordinates where \(X_h(p)\) is small; the decreasing \(\beta_k\) make that preference stronger with each successive type.

To verify the guarantee, let \(p^\nu\to p\in[a,b]^m\) with \(\mu(p)=t\), and suppose a type \(k\) label \((r,c)\) stays active within \(\Lambda_t\) along \(h_\nu\to0+\). Let \(U\) be its raised coordinate set, and abbreviate \(h=h_\nu\). For \(k>0\), comparison with \((t\mathbf 1,0)\) gives \[\sum_{i\in U}(r_i-t)h^{p_i^\nu} \leq h^{\beta_k}-d(h)=(1+o(1))h^{\beta_k}.\] Every coefficient on the left is at least \(\beta_k-t>0\). A limiting exponent \(p_i<\beta_k\) would make the corresponding term divided by \(h^{\beta_k}\) tend to infinity. Therefore \[ p_i\geq\beta_k\qquad(i\in U). \tag{4}\] Among labels made from this same \(U\), all contributions off \(U\) and the term \(-h^{\beta_k}\) are common. Thus \((s,d)\) is active in the family furnished by induction. Its limiting exponent vector lies in \([\beta_k,b]^U\), so \[\max_{i\in U}\left|p_i-r_i\right|<E(r).\] The approximating vectors \(p_U^\nu\) may approach a face of that cube from outside. This is why the induction statement allows them outside the cube, including when \(p_i=\beta_k\) or \(p_i=b\).

It remains to bound the coordinates off \(U\). This argument also includes type \(0\), for which \(U=\varnothing\). If \(k<m-1\), we claim that \[p_i\leq\beta_{k+1}\qquad(i\notin U).\] Otherwise choose \(j\notin U\) with \(p_j>\beta_{k+1}\) and any type \(k+1\) label raised on \(U\cup\{j\}\). Every limiting exponent on this set exceeds \(\beta_{k+1}\), by (4) and \(\beta_k>\beta_{k+1}\) when \(k>0\). The competitor’s affine value minus that of \((t\mathbf 1,0)\) is therefore \[-h^{\beta_{k+1}}+o(h^{\beta_{k+1}}).\] For the supposed active label, the same difference is zero if \(k=0\), and is bounded below by \(-h^{\beta_k}+o(h^{\beta_k})\) if \(k>0\). Since \(h^{\beta_k}=o(h^{\beta_{k+1}})\), the competitor is strictly smaller for all sufficiently small \(h\) along the sequence, a contradiction. Consequently, off \(U\), \[\left|p_i-r_i\right|\leq\beta_{k+1}-t<E(r).\] If \(k=m-1\), there is just one coordinate off \(U\). Its limiting exponent is \(t\), since \(\mu(p)=t\) and all coordinates on \(U\) have limiting exponents greater than \(t\). Its error is zero. This proves the required strict bound for every active label at the fixed minimum.

Extension to nearby minima. For a fixed label \(\lambda=(r,c)\in\Lambda_t\), let \(A_\lambda\subset[a,b]^m\) consist of the limits of all sequences \(p^\nu\in\mathbb R^m\) for which \(h_\nu\to0+\) and \(\lambda\) is active within \(\Lambda_t\) at \(X_{h_\nu}(p^\nu)\). This set is closed. Indeed, if \(z_n\in A_\lambda\) and \(z_n\to z\), choose from a sequence realizing \(z_n\) a parameter \(0<h_n<1/n\) and a vector \(u_n\) with \(\left\|u_n-z_n\right\|_\infty<1/n\) such that \(\lambda\) is active at \(X_{h_n}(u_n)\). Then \(u_n\to z\), so this selected sequence realizes \(z\in A_\lambda\). This argument uses actual active points and requires no continuity of the offsets in \(h\).

The set of limits violating the desired estimate, \[B_t=\bigcup_{\lambda=(r,c)\in\Lambda_t} \{p\in A_\lambda:\left\|p-r\right\|_\infty\geq E(r)\},\] is compact: the union is finite and each \(E(r)\) here is a fixed number. The preceding argument shows that \(B_t\) contains no point with \(\mu(p)=t\). Continuity of \(\mu\) and compactness therefore give \(\rho_t>0\) such that \(B_t\) contains no point with \[\mu(p)\in I_t:=(t-\rho_t,t+\rho_t).\] If \(B_t\) is empty any positive radius initially suffices; otherwise take \(\rho_t<\min_{p\in B_t}\left|\mu(p)-t\right|\). Thus \(\Lambda_t\) has the asserted limiting activity property whenever \(\mu(p)\in I_t\).

Shrink \(\rho_t\) further, if necessary, so that every local offset obeys \[ c(h)=o(h^{t+\rho_t}). \tag{5}\] For a nonsingleton family, choose \(t+\rho_t<\min_k\beta_k\) and use (3); for a singleton the offset is zero. We have obtained a family valid near each possible minimum exponent. The remaining task is to ensure that a globally active label comes from a family valid at its own limiting point.

Combining the families. Take a finite subcover of \([a,b]\) from the intervals \(I_t\). Discard intervals contained in others, keeping just one if two coincide, and order the remaining centers as \(t_1<\cdots<t_N\). Write \(I_{t_j}=(L_j,R_j)\). Noncontainment forces both endpoints to have the same strict order; since \(t_j=(L_j+R_j)/2\), it is the order of the centers. Hence \[L_1<\cdots<L_N,\qquad R_1<\cdots<R_N.\] Consecutive intervals overlap. If \(R_j\leq L_{j+1}\), a point of \([R_j,L_{j+1}]\subset(t_j,t_{j+1})\subset[a,b]\) would be uncovered. The first interval contains \(a\) and the last contains \(b\).

For \(1\leq j<N\), choose a cut \[q_j\in \bigl(\max\{t_j,L_{j+1}\},\min\{t_{j+1},R_j\}\bigr).\] The interval displayed is nonempty because consecutive intervals overlap and each contains its own center. With \(q_0=a\) and \(q_N=b\), we have \[ t_j<q_j<t_{j+1}\quad(1\leq j<N), \qquad [q_{j-1},q_j]\subset I_{t_j}\quad(1\leq j\leq N). \tag{6}\] For \(N=1\), the sole interval contains \([a,b]\) and no interior cuts are needed. The closed intervals in (6) are the ranges of minimum exponents assigned to the respective families; see Figure 1.

Combine the selected families, replacing every local offset \(c(h)\) in \(\Lambda_{t_j}\) by \[ \widetilde c(h)=c(h)-\sum_{\ell=1}^{j-1}h^{q_\ell}. \tag{7}\] These offsets are \(o(h^a)\): (5) has \(t_j+\rho_{t_j}>a\), and each shift exponent satisfies \(q_\ell>t_\ell\geq a\). A common shift preserves activity within each family. Between consecutive families, the new term \(-h^{q_j}\) will force any limiting minimum exponent to the correct side of the cut.

Suppose a fixed label from \(\Lambda_{t_j}\) is active in the combined family along \(h_\nu\to0+\) and \(p^\nu\to p\in[a,b]^m\). Write \(c(h)\) for its original local offset and \(h=h_\nu\). We claim \[q_{j-1}\leq\mu(p)\leq q_j.\] If \(j<N\) and \(\mu(p)>q_j\), each coordinate \(h^{p_i^\nu}\) is \(o(h^{q_j})\). Also \(c(h)=o(h^{q_j})\), by (5) and \(q_j<R_j=t_j+\rho_{t_j}\). Subtracting the value of the shifted label \((t_{j+1}\mathbf 1,0)\) from the active label’s value gives \[(r-t_{j+1}\mathbf 1)\cdot X_h(p^\nu)+c(h)+h^{q_j} =h^{q_j}+o(h^{q_j})>0,\] a contradiction.

If \(j>1\) and \(\mu(p)<q_{j-1}\), compare instead with the shifted label \((t_{j-1}\mathbf 1,0)\). The difference is \[(r-t_{j-1}\mathbf 1)\cdot X_h(p^\nu)+c(h)-h^{q_{j-1}}.\] Every coefficient in the scalar product is at least \(t_j-t_{j-1}>0\). Some coordinate has \(p_i<q_{j-1}\), so its exponential divided by \(h^{q_{j-1}}\) tends to infinity. Meanwhile, \(c(h)=o(h^{q_{j-1}})\) because \(q_{j-1}<t_j<t_j+\rho_{t_j}\). Thus the displayed difference divided by \(h^{q_{j-1}}\) tends to \(+\infty\), again contradicting activity. The outer bounds for \(j=1\) or \(j=N\) follow from \(\mu(p)\in[a,b]\).

By (6), the claimed bounds place \(\mu(p)\) inside \(I_{t_j}\), including when it equals a cut. The label is active within its original local family, so the guarantee on \(I_{t_j}\) gives \(\left\|p-r\right\|_\infty<E(r)\). This completes the induction. ◻

Proof. Otherwise there are \(h_\nu\to0+\), points \(p^\nu\in[a,b]^m\), and active labels \(\lambda_\nu\) such that \(\left\|p^\nu-r_{\lambda_\nu}\right\|_\infty\geq E(r_{\lambda_\nu})\). By finiteness and compactness, pass to a subsequence on which \(\lambda_\nu=\lambda\) is fixed and \(p^\nu\to p\in[a,b]^m\). The inequality passes to the limit and contradicts Lemma 2. No continuity of \(E\) is needed because its argument \(r_\lambda\) is fixed on this subsequence. ◻

Proof. Write \(p_c=\log c/\log h\). If a label active at \(c\) had a nonzero coordinate \(r_{\lambda,i}\), (12) with \(v=e_i\) would give \(\left|p_{c,i}\right|\geq H\), contrary to (13). Thus every label active at \(c\) has zero slope. At least one label is active there, so one of the affine functions is the constant \(M_h\). It follows that \(F_h\leq M_h\) everywhere, and that \(M_h\) is the least value of any constant label. Hence \[K_h=\bigcap_{\lambda\in\Lambda} \{x\in\mathbb R^d:L_{\lambda,h}(x)\geq M_h\}\] is closed and convex, and contains \(c\).

If \(x=X_h(p)\in K_h\cap B_h\), the constant label with value \(M_h\) is active at \(x\). Applying (10) to its zero slope gives \[\left\|p\right\|_\infty<E(0)=\frac12, \qquad\text{and thus}\qquad x\in\operatorname{int}B_{\sqrt h}.\] In particular, \(K_h\) misses the boundary of \(B_h\). If some point of \(K_h\) lay outside \(B_h\), the segment joining it to \(c\in\operatorname{int}B_h\) would meet that boundary at a point of \(K_h\), by convexity. This is impossible. Thus \(K_h\subset B_h\), and the preceding argument proves (14) for every point of \(K_h\). Closedness and boundedness give compactness. ◻

Proof. First, let \(x\in P\setminus K_h\). If \(r_\lambda\in S^\perp\), then \(x-c\in S\) implies \[L_{\lambda,h}(x)=L_{\lambda,h}(c)\geq F_h(c)=M_h.\] But \(F_h(x)<M_h\) by Proposition 4. Such a label cannot be active at \(x\).

If \(R\) is empty, then \(f=0\) and \(S=\{0\}\), so (15) holds and (16) is vacuous. For the flux comparison, assume \(R\ne\varnothing\) and fix \(x=X_h(p)\in B_h\) and an active slope \(r\). Whenever \(y,z\in C\) satisfy \(r\cdot y<r\cdot z\), (12) applied to \(z-y\) gives \[ \frac{x^z}{x^y}=h^{p\cdot(z-y)}\leq h^H. \tag{17}\] List the distinct values of \(r\cdot y\), for \(y\in C\), as \(a_1<\cdots<a_m\). For \(1\leq j<m\), define the net upward flux across the cut at \(a_j\) by \[\begin{align*} Q_j(x)={}& \sum_{\substack{u\to u'\in R\,:\\r\cdot u\leq a_j<r\cdot u'}} k_{u\to u'}x^u -\sum_{\substack{u\to u'\in R\,:\\r\cdot u'\leq a_j<r\cdot u}} k_{u\to u'}x^u. \end{align*}\] An edge crossing the cut in the first sum is called upward, and an edge in the second sum is called downward.

If there is a downward edge \(u\to u'\), weak reversibility supplies a directed path from \(u'\) back to \(u\). This path starts below the cut and ends above it, so it contains an upward edge, say \(w\to w'\). Every source \(u\) of a downward edge has \(r\cdot u>r\cdot w\). By (17) and (13), \[\sum_{\substack{u\to u'\in R\,:\\r\cdot u'\leq a_j<r\cdot u}} k_{u\to u'}x^u \leq h^H x^w\sum_{e\in R}k_e < k_{w\to w'}x^w.\] Thus one upward edge already contributes more than the entire downward sum, and \(Q_j(x)>0\). If there is no downward edge, then \(Q_j(x)\geq0\), with strict inequality whenever there is an upward edge. In all cases \(Q_j(x)\) is nonnegative, and it is positive if any reaction crosses the cut.

The contribution of an edge to \(r\cdot f(x)\) is its flux multiplied by the difference between the levels of its target and source. Writing this difference as the signed sum of the intervening gaps gives \[ r\cdot f(x)=\sum_{j=1}^{m-1}(a_{j+1}-a_j)Q_j(x). \tag{18}\] Every term is nonnegative. If \(r\notin S^\perp\), some reaction \(y\to y'\) satisfies \(r\cdot(y'-y)\ne0\) and therefore crosses at least one cut; the corresponding term is positive. This proves (15). When \(m=1\), the sum is empty and every reaction has zero scalar product with \(r\), as required. Self-loops never cross a cut and contribute zero to the vector field.

For the uniform bound, the active set of a fixed label, \[T_{\lambda,h}= \{x\in B_h:L_{\lambda,h}(x)=F_h(x)\},\] is compact. If \(r_\lambda\notin S^\perp\) and \(T_{\lambda,h}\) is nonempty, the continuous function \(r_\lambda\cdot f\) is strictly positive on this set and has a positive minimum there. Take the minimum of these positive numbers over the finitely many eligible labels. If there are none, choose any \(\delta_h>0\). This proves (16), including the case \(R=\varnothing\). ◻

Proof. For a solution \(x(t)\) and each fixed time \(t\) before its maximal existence time, the right derivative of its affine minimum is \[ \frac{d^+}{dt}F_h(x(t)) =\min_{\lambda\in A_h(x(t))}r_\lambda\cdot f(x(t)). \tag{20}\] Indeed, all inactive labels have a positive gap above the minimum at time \(t\). There are finitely many of them, so continuity excludes them from the minimum for a sufficiently short time to the right. For each active label, differentiability of the solution gives \[L_{\lambda,h}(x(t+s)) =F_h(x(t))+s\,r_\lambda\cdot f(x(t))+o(s) \qquad(s\downarrow0).\] Taking the minimum of these finitely many expansions proves (20).

We use the elementary fact that a continuous function \(g\) on \([a,b]\) with right derivative at least \(q\) at every point of \([a,b)\) satisfies \(g(b)-g(a)\geq q(b-a)\). For completeness, for any \(\eta>0\) the function \(g(u)-qu+\eta u\) has positive right derivative before \(b\). Its maximum on the compact interval therefore cannot occur before \(b\). Comparing its values at \(a\) and \(b\) and then letting \(\eta\downarrow0\) proves the assertion. Formula (20) and Proposition 5 now give monotonicity in \(B_h\), and give (19) when the segment lies in \(P\setminus K_h\).

Suppose next that \(x(0)\in K_h\), and consider the maximal solution of the polynomial vector field \(f\). Since \(K_h\subset\operatorname{int}B_h\), the solution initially stays in \(B_h\). On any interval on which it does so, monotonicity and the global upper bound \(F_h\leq M_h\) force \(F_h(x(t))=M_h\). Hence it remains in \(K_h\). It cannot have a first exit from \(\operatorname{int}B_h\) within its interval of existence: continuity would place the exit point in both \(K_h\) and the boundary of \(B_h\), contradicting Proposition 4. Thus the solution stays in the compact set \(K_h\) throughout its maximal interval. A solution of a smooth vector field that remains in a compact subset of its domain extends past any finite endpoint of its interval of existence. To see the relevant point here, \(f\) is bounded on a neighborhood of \(K_h\), so a finite endpoint would give a limit of \(x(t)\) in \(K_h\), at which local existence continues the solution. The solution is therefore global. It remains positive by (14); moreover, \(f(x)\in S\) implies \(x(t)-x(0)\in S\), so a solution starting in \(P\) stays in \(P\).

Finally, let a global solution in \(P\) remain in \(B_h\) for all \(t\geq t_0\). If it never enters \(K_h\) after \(t_0\), then (19) gives \[F_h(x(t))\geq F_h(x(t_0))+\delta_h(t-t_0) \qquad(t\geq t_0),\] contradicting \(F_h\leq M_h\) for sufficiently large \(t\). It must enter \(K_h\) at a finite time, and the forward invariance just proved keeps it there. ◻

Proof of Theorem 1. Let \(x^0\in P\). Since \(x^0\) is positive, we may choose \(0<h_0\leq h_P\) so small that \[\left\|\frac{\log x^0}{\log h_0}\right\|_\infty<H,\] where the logarithm is componentwise and \(H>0\) is the fixed separation constant from Section 3. In particular, \(x^0\in B_{h_0}\). The coordinate sign comparisons force every label active at \(x^0\) to have slope zero, just as they did at \(c\). An active constant label must have the least value among all constant labels, namely \(M_{h_0}=F_{h_0}(c)\). Hence \[x^0\in K_{h_0}.\] Lemma 6 implies that the solution through \(x^0\) is global, stays positive, and remains in \(K_{h_0}\). Since \(f(x)\in S\), it also remains in \(P\).

Starting at \(h_0\), define \[ h_{j+1}=\min\{\sqrt{h_j},h_P\}. \tag{22}\] We claim that for every \(j\) there is a finite time \(T_j\) such that \(x(t)\in K_{h_j}\) for all \(t\geq T_j\). The case \(j=0\) holds with \(T_0=0\). Suppose it holds at \(j\). Because \(h_{j+1}\leq\sqrt{h_j}\), the boxes satisfy \(B_{\sqrt{h_j}}\subset B_{h_{j+1}}\). Thus (21) yields \[x(t)\in B_{h_{j+1}}\quad(t\geq T_j).\] The finite-entry assertion of Lemma 6, followed by its invariance assertion, supplies a finite \(T_{j+1}\geq T_j\) with \(x(t)\in K_{h_{j+1}}\) for all \(t\geq T_{j+1}\). This proves the claim. The argument uses containment in successive boxes; no nesting of the sets \(K_h\) is assumed.

Repeated square roots of \(h_0\in(0,1)\) tend to \(1\). Since \(h_P<1\), the capped sequence (22) therefore reaches \(h_P\) after finitely many steps. We conclude that every solution through a point of \(P\) enters \(K_{h_P}\) in finite time and stays there.

Finally set \[K_P=K_{h_P}\cap(c+S).\] This is nonempty because it contains \(c\), and it is compact and convex as the intersection of the compact convex set \(K_{h_P}\) with a closed affine subspace. Since \(K_{h_P}\) lies in the positive orthant, \(K_P\subset P\). Both \(K_{h_P}\) and the affine subspace \(c+S\) are forward invariant, so \(K_P\) is forward invariant. The preceding argument proves its absorbing property. Moreover, \(K_P\subset B_{h_P}\), and therefore \(\varepsilon_P=h_P\in(0,1)\) gives \[\varepsilon_P\leq x_i(t)\leq\varepsilon_P^{-1} \quad(1\leq i\leq d)\] for all sufficiently large \(t\), with the entry time allowed to depend on \(x^0\). The scale \(h_P\), and hence these bounds, was fixed before choosing the initial point. ◻