Proof. A sphere in \(T\) lifts to an embedded sphere in \(\mathbb R^3\). Its distinct lattice translates have disjoint interiors: equal-radius spheres with intersecting interiors would have intersecting boundaries. Thus it bounds an embedded ball on one side. For its radius \(R\), \[W=\frac{\pi R^3}{6},\qquad A=\frac{\pi R^2}{2} =(9\pi/2)^{1/3}W^{2/3},\] where \(W\) denotes the ball-side volume. Since \(W\ge v\), this area is at least the first expression in (3).

For a compact circular-cylinder quotient, the stabilizer of its lifted axis contains a nonzero axial lattice translation. Its primitive core therefore has length \(2k\), with \(k=|m|\ge1\) for a primitive nonzero \(m\in\mathbb Z^3\). Embeddedness gives a solid tube on one side, with \[W=\frac{k\pi R^2}{4},\qquad A=\frac{k\pi R}{2} =\sqrt{\pi kW}\ge\sqrt{\pi v}.\] This comparison concerns the area of the given tube and does not require a shortest tube of the possibly larger volume \(W\) to embed. If such a boundary minimizes, \(A=I(V)\le F(V)\), where \(I\) is the isoperimetric profile. Equality throughout forces \(k=1\), \(W=v\), and \(v\) in the closed cylindrical branch. The corresponding sphere comparison similarly forces \(W=v\) and the spherical branch.

A connected closed planar surface has a primitive integral normal \(m\) and physical area \(4|m|\). Disjoint planar surfaces must be parallel, and a single such torus does not separate \(T\). At least two are required to bound a region. Their total normalized area is therefore at least one, with equality only for two tori with \(|m|=1\). Their intervening regions are exactly coordinate slabs.

Finally the branches in (3) are realized by an embedded ball, a shortest tube, and a coordinate slab, respectively. At the last volume of its branch the ball radius is \(2/3\); the largest tube radius in its branch is \(2/\pi\). Both are less than the transverse injectivity radius one. The branch intersections are exactly \(V_1,V_2\). Complements give the remaining volumes. This proves both attainment and the asserted equality classification. ◻

Proof. Compactness of finite-perimeter sets gives lower semicontinuity of \(I\). A local volume-changing smooth variation of a minimizer gives upper semicontinuity at every interior volume. Since \(0\le I\le F\), continuity also holds at the endpoints. Complementation gives the symmetry.

Let \(K=k_1k_2\) be the intrinsic Gauss curvature of \(\Sigma\). Let \(a(\tau)\) and \(w(\tau)\) be normalized area and volume under outward normal parallel displacement by \(\tau\). At \(\tau=0\), \[w'=A,\qquad a'=hA,\qquad w''=hA,\qquad a''=\frac18\int_\Sigma 2K\,dS.\] Small parallels are embedded and \(w'>0\); expressing their area as a function \(C(w)\) provides the upper support. The chain rule gives \[C'=h,\qquad C^2C''=\frac18\int_\Sigma 2K\,dS-h^2A.\] For connected \(\Sigma\), Gauss–Bonnet gives (4). For an arbitrary boundary, \[\frac18\int_\Sigma(2K-h^2)\,dS =-\frac18\int_\Sigma(k_1^2+k_2^2)\,dS\le0,\] so \(C''\le0\). These are the parallel-variation formulas of [5], with the stated normalization.

For completeness, this contact property implies concavity without differentiability of \(I\). Suppose \(I(c)<L(c)\) for a chord \(L\) on \([a,b]\) and \(a<c<b\). Choose \(\varepsilon>0\) sufficiently small that \(I(c)<L(c)+\varepsilon(c-a)(c-b)\). The minimum of \[I(x)-L(x)-\varepsilon(x-a)(x-b)\] is negative and occurs at an interior point \(x_0\). Adding this minimum to the strictly convex quadratic produces a function \(q\le I\) with equality at \(x_0\). Any local upper support there satisfies \(C''(x_0)\ge q''=2\varepsilon\), a contradiction.

For a concave function, the slope of a smooth upper support lies between its one-sided derivatives: \(I'_+(V)\le h\le I'_-(V)\). The secant joining the equal values at \(V\) and \(1-V\) has slope zero. Concavity therefore gives the asserted signs on either side of \(1/2\). No curvature conclusion at exactly \(1/2\) is needed at this stage. ◻

Proof. The assumption, existence, and Lemma 3 give \(I=F\) at \(V_1,V_2\), and hence also at \(1-V_2,1-V_1\). On each open interval between consecutive points in \[0,\ V_1,\ V_2,\ 1-V_2,\ 1-V_1,\ 1,\] the function \(F\) is smooth and \[ F''+\frac{(F')^2}{F}\ge0. \tag{5}\] This follows directly for its power-law and constant branches.

If \(I-F\) had a negative minimum on one of the closed intervals, it would occur in its interior. A minimizer there must be nonstandard by Lemma 3. For its upper support, \(C=I<F\), \(C'=F'\), and \(C''\ge F''\) at contact. Hence \[C''+\frac{(C')^2}{C} \ge F''+\frac{(F')^2}{F}\ge0,\] whereas (4) makes the left-hand side \(\pi(1-g)/C^2<0\). Thus \(I=F\) throughout.

Now consider any nonstandard minimizer at an interior point of one of the smooth intervals. Its support touches \(F=I\) from above, so \(C'=F'\) and \(C''\ge F''\), giving the same contradiction with \(C=F\). The partition points are covered by the hypothesis and complementation. In particular \(1/2\) belongs to the smooth constant interval, so it is included. Lemma 3 completes the equality classification. ◻

Proof. The complement of this pattern becomes the same pattern after reversing all three coordinates. We may therefore choose the smaller-volume side, so that \(V\in\{V_1,V_2\}\) and \(h\ge0\). Put \[a=\sqrt{\pi V},\qquad x=hA.\] The standard competitors and the curvature estimate (2) give \[ 0<A\le a\le1,\qquad 0\le x\le\sqrt{\pi A}\le\sqrt\pi. \tag{12}\] Since \(b_x+b_y+b_z=2\), choose a slicing direction, denoted by \(z\), for which \(b=b_z\ge2/3\), and write \(d=\sqrt{1-b}\). The strict positivity of \(N_z\) in the interior implies \(b<1\), so \[0<d\le\frac1{\sqrt3}.\] The top section is a corner region, separated from the two opposite edges of its square. The bottom section has a complement of the same kind at the opposite corner. Reflecting either corner region in its two adjacent edges gives a planar region of four times its area and four times its relative perimeter. The planar isoperimetric inequality therefore gives \[p(1)\ge\sqrt{\pi r},\qquad p(0)\ge\sqrt{\pi(1-s)}.\] Monotonicity gives \(r\le V\). Define \(w\in[0,1]\) by \(\sqrt{\pi r}=aw\). The endpoint flux identities (9) and the projection estimate (10) now imply \[ \begin{aligned} aw&\le Ab-\frac{ha^2}{\pi}(1-w^2),\\ \sqrt{\pi(1-s)} &\le Ab-\frac{ha^2}{\pi}(1-w^2)+xd,\\ s&\le\frac{a^2w^2}{\pi}+Ad. \end{aligned} \tag{13}\] Indeed, \(p(0)=p(1)+h(s-r)\), and \(s-r\le Ad\).

Dividing the first inequality in (13) by \(a\) and using \(A\le a\) and \(ha\ge x\) yields \[f(w):=w+\frac{x}{\pi}(1-w^2)\le b.\] Since \(f(1)=1>b\), continuity gives \(\tau\in[w,1)\) with \[ \tau+\frac{x}{\pi}(1-\tau^2)=1-d^2, \qquad x=\frac{\pi(1-\tau-d^2)}{1-\tau^2}. \tag{14}\] In particular \(\tau\le1-d^2\). No monotonicity of \(f\) is required. The function \(E(q)=Ab-ha^2(1-q^2)/\pi\) is nondecreasing for \(q\ge0\), and \[E(\tau) =A\tau+\frac{h(A^2-a^2)}{\pi}(1-\tau^2) \le a\tau.\] Consequently (13) gives \[\sqrt{\pi(1-s)}\le a\tau+xd, \qquad s\le\frac{a^2\tau^2}{\pi}+Ad.\] Both sides of the first inequality are nonnegative. Squaring it, combining the two inequalities, and using \(A,a\le1\), we obtain the necessary condition \[ 1\le R(\tau,d):= d+\frac{\tau^2+(\tau+xd)^2}{\pi}. \tag{15}\] The scalar inequality in Lemma 21 contradicts (15); its hypotheses follow from (12) and (14), together with \(0<d\le1/\sqrt3\). This excludes the four-vertex pattern at both transition volumes. ◻

Proof. Set \(a=\sqrt{\pi V_1}=2\pi/9\) and \(x=hA\). As before, \(A\le a\) and \(0\le x\le M:=\sqrt{\pi a}=\pi\sqrt2/3<\pi\). Since \(V_1=a^2/\pi\ge A^2/\pi\), (16) gives \[s^2\le A^2-xV_1\le A^2(1-x/\pi), \qquad s\le A\sqrt{1-x/\pi}.\] Combining this with \(\pi(1-s)\le h^2s^2\) yields \[ 1\le a\sqrt{1-x/\pi} +\frac{x^2}{\pi}(1-x/\pi). \tag{17}\] The first summand decreases in \(x\). The second, denoted by \(g(x)\), has derivative \(g'(x)=x(2\pi-3x)/\pi^2\ge0\) on \([0,M]\), since \(M<2\pi/3\). For \(0\le x\le1\), the right side of (17) is at most \[a+g(1) <\frac{44}{63}+\frac29=\frac{58}{63}<1.\] Here \((\pi-1)/\pi^2<2/9\) because \((p-1)/p^2\) decreases for \(p>2\) and \(\pi>3\). For \(1\le x\le M\), it is at most \[\begin{align*} a\sqrt{1-1/\pi}+g(M) &=a\left(\sqrt{1-1/\pi}+1-\frac{\sqrt2}{3}\right)\\ &<\frac{44}{63}\left(\frac56+\frac8{15}\right) =\frac{902}{945}<1. \end{align*}\] The strict bounds \(\sqrt{1-1/\pi}<5/6\) and \(\sqrt2>7/5\) follow by squaring, using \(\pi<22/7\). Both intervals contradict (17). ◻

Proof. The slab competitor gives \(A\le1\). The lower bound on \(h\) in (18) follows from the bottom perimeter inequality in (16). Its projection inequality gives \[h\le\pi\left(A-\frac{s^2}{A}\right) \le\pi(1-s^2),\] since \(A-s^2/A\) is increasing for \(A>0\). Solving \(h^2s^2+\pi s-\pi\ge0\) for \(s>0\) gives \(s\ge t(h)\).

Combining the two bounds on \(h\), and dividing by \(1-s>0\), gives \[\Phi(s):=s^2(1-s)(1+s)^2\ge\frac1\pi.\] Its derivative is \(\Phi'(s)=s(1+s)(2+s-5s^2)\), so \(\Phi\) increases up to \(s_*=(1+\sqrt{41})/10\) and decreases thereafter. The points \(107/200\) and \(89/100\) lie on the respective sides of \(s_*\). Exact arithmetic gives \[\frac{22}{7}\Phi(107/200) =\frac{1103875107423}{1120000000000}<1, \qquad \frac{22}{7}\Phi(89/100) =\frac{4890924423}{5000000000}<1.\] Since \(\pi<22/7\), unimodality implies \(107/200<s<89/100\). Finally, \(\sqrt{\pi(1-s)}/s\) decreases with \(s\), and hence \[h>\sqrt{\frac{3300}{7921}}>\frac{16}{25}, \qquad h<\frac{22}{7}\left(1-\left(\frac{107}{200}\right)^2\right) =\frac{314061}{140000}<\frac94.\] For the first strict comparison, \(\pi>3\) was used; its last step follows from \(3300/7921-(16/25)^2=34724/4950625>0\). These strict bounds imply the closed curvature interval stated in (18). ◻

Proof. For fixed \(J\in[0,1]\), let \[f_J(X)=\frac1X-\lambda X-J\frac{\sqrt{1-X^2}}X, \qquad 0<X\leq1.\] On \((0,1)\), \[X^3f_J''(X)=2-J\frac{2-3X^2}{(1-X^2)^{3/2}}\geq0.\] Indeed, the function \((2-3r)/(1-r)^{3/2}\) decreases from \(2\) to \(0\) on \([0,2/3]\), its derivative being \(-3r/[2(1-r)^{5/2}]\); it is nonpositive on \([2/3,1)\). Convexity extends to the continuous endpoint value at \(X=1\).

Use normalized arclength on the slice curve as a probability measure. The flux identity says that the mean of \(\alpha\) is \(m=hv/p\in(0,1]\). Take \(J=G(m)\). For \(m<1\), \[f_J(m)=\frac{\sqrt{1-m^2}}m\bigl(g(m)-G(m)\bigr)\geq0;\] for \(m=1\), \(f_J(1)=1-\lambda>0\). Jensen’s inequality now yields \(\int f_J(\alpha)\,d\ell\geq0\), which is (22).

Down-set position gives \(N_z\geq0\), hence \(\sqrt{1-\alpha^2}=N_z\). The coarea identities therefore turn (22) into \[A-\lambda hV \geq\int_0^1G(hv(z)/p(z))\,(-v'(z))\,dz \geq\int_0^1\gamma(v(z))\,(-v'(z))\,dz.\] The integrands are set to zero at empty or exceptional levels. Lemma 9, with \(r=0\), identifies the last integral with \(\int_0^s\gamma(v)\,dv\). Since \(V=1/\pi\), this proves (23). ◻

Proof. The far vertex \((1,1)\) is absent even from the bottom slice and hence from every slice. A nonempty regular slice has, by monotonicity, one, two adjacent, or three occupied square vertices. In the first case the slice is confined to the occupied corner. Reflection across its two adjacent sides gives a bounded planar set of area \(4v\) and perimeter \(4p\); the Euclidean isoperimetric inequality gives \(p^2\geq\pi v\). In the three-vertex case the same argument applies to the opposite-corner complement and gives \(p^2\geq\pi(1-v)\).

In the two-vertex case, after interchanging the square coordinates if necessary, the occupied vertices form the bottom edge. This entire edge is occupied and the opposite edge is vacant. Each vertical segment crosses the separating curve exactly once: monotonicity gives uniqueness, and the strict normal inequality gives a transverse crossing. The implicit function theorem, also in the reflected endpoint neighborhoods, writes the curve as a single decreasing graph \(y=f(x)\) for \(0\leq x\leq1\). Its length is at least \(1\). These arguments establish (24), including both strip orientations.

Since \(G\) is non-increasing, (24) implies \(G(hv/p)\geq G(hv/p_*(v))\). Apply Lemma 14 and use \(s\geq t\) and \(G\geq\ell\) on the remaining interval \([t,s]\) to obtain (26). ◻

Proof. Use coordinates \((\xi,\eta)\) measured from the complement’s corner, so that its monotone separating curve runs from \((0,L_y)\) to \((L_x,0)\) with \(d\xi\geq0\), \(d\eta\leq0\). For \(L=L_y\), the vector \[\left(\frac\eta L,-\sqrt{1-\frac{\eta^2}{L^2}}\right)\] has unit length. Pairing it with the tangent differential and integrating gives \[\begin{split} hs&\geq \frac1L\int\eta\,d\xi -\int\sqrt{1-\eta^2/L^2}\,d\eta\\ &=\frac{1-s}{L}+\int_0^L\sqrt{1-\eta^2/L^2}\,d\eta =\frac{1-s}{L}+\frac{\pi L}4. \end{split}\] The first integral is the complement area. This calculation uses only monotonicity of the curve, and so requires no convexity. Exchanging \(\xi\) and \(\eta\) proves the same estimate for \(L_x\). Both extents are positive, because the complement has positive area.

Solving the quadratic inequality (27) gives, for each extent, \[L\geq\frac2\pi \left(hs-\sqrt{h^2s^2-\pi(1-s)}\right).\] The discriminant is nonnegative by \(hs\geq\sqrt{\pi(1-s)}\). A strip \(y=f(x)\) is contained in the bottom slice, so its smaller endpoint satisfies \(f(1)\leq1-L_y\leq e\). For a strip \(x=f(y)\) the same argument uses \(L_x\). Thus one common bound controls both orientations without imposing a symmetry on the two extents.

Finally, write \(s=t+Y^2\) and use \(h^2t^2=\pi(1-t)\) to obtain \[h^2s^2-\pi(1-s) =(\pi+2h^2t)Y^2+h^2Y^4.\] Its square root is at most \(\sqrt{\pi+2h^2t}\,Y+hY^2\). Substituting this upper bound in \(e\) cancels the term \(hY^2\) coming from \(hs\), and proves (28). ◻

Proof. The definition of \(u\) gives \(u\geq0.96/1.62=16/27\), \(u\leq b_0\), and \(hu\leq0.96\). Hence the first assertion follows from \(16/27-0.21=1033/2700\), using \(\pi>3\). Moreover, \[h^2u^2\leq0.96^2<1\leq\pi(1-u).\] The function \(x\mapsto h^2x^2+\pi x-\pi\) is strictly increasing for \(x\geq0\) and vanishes at \(t\), so \(u<t\). This proves (30); in particular \([u-\delta,u]\subset[a_0,b_0]\cap[0,s]\), where \(p_*=1\). For the remaining assertion assume \(\delta>0\). The definition of \(\delta\) and (28) then give \[ e\leq u-P+QY\leq u-\delta. \tag{33}\]

We verify (32) separately for all three slice types of Lemma 15.

Occupied corner. Here \(p^2\geq\pi v\). Since \(v>0.38\) and \(\pi>3\), \[p^2-1>0.14>c d_m^2=0.11907\geq c\delta^2H(q).\] In the last step we used \(H(q)\leq q^2\leq1\).

Opposite-corner complement. Here \(p^2\geq\pi(1-v)\), and \(v=u-\delta+\delta q\) gives \[\begin{split} p^2-1&\geq\pi(b_0-v) \geq\pi\delta(1-q)\\ &\geq c\delta^2\frac{\pi(1-q)}{c d_m} \geq c\delta^2H(q). \end{split}\] This includes the endpoint \(q=1\), \(u=b_0\), where \(H(1)=0\).

Strip. Write the strip as \(y=f(x)\), interchanging the two square coordinates if required. Its smooth decreasing graph has \(f(1)\leq e\) by Lemma 16. Integration by parts and (33) give \[\int_0^1x|df|=\int_0^1f(x)\,dx-f(1) =v-f(1)\geq v-u+\delta=:j=\delta q.\] Set \(k=2.5j\leq2.5d_m=0.525\). Pair the graph tangent \((dx,|df|)\) with the unit vector \((\sqrt{1-k^2x^2},kx)\) to obtain \[ p\geq kj+\int_0^1\sqrt{1-k^2x^2}\,dx. \tag{34}\] For \(0\leq w\leq0.525^2\), \[\sqrt{1-w}\geq1-\frac w2-\frac{w^2}5.\] Indeed the right side is positive and its square minus \(1-w\) is \(w^2(-3/20+w/5+w^2/25)\leq0\); the expression in parentheses is increasing and is negative at \(w=0.525^2\). It follows from (34) that \[\begin{split} p&\geq1+\left(\frac{35}{24}-\frac{25j^2}{16}\right)j^2\\ &\geq1+\frac{26677}{19200}j^2 \geq1+1.35j^2. \end{split}\] Squaring gives \(p^2\geq1+2.7j^2\geq1+c\delta^2H(q)\). This establishes the same estimate for every slice type, even if the type or strip orientation changes within the area interval. Exceptional levels are not needed in this argument and have zero measure after the substitution in Lemma 14. ◻

Proof. The assertion is immediate at \(X=0\). For \(X>0\), divide the left side by \(X\) and write the result as \[\frac{N(X)}{D_1(X)},\qquad N(X)=0.49-\frac{0.01}{1-X^2},\qquad D_1(X)=X\sqrt{1-X^2}.\] The numerator decreases; the denominator is at most \(1/2\) everywhere and decreases for \(X\geq1/\sqrt2\). The following rational bounds therefore suffice: \[\begin{array}{c|cc} \text{interval for }X&N(X)\text{ lower bound}&D_1(X)\text{ upper bound}\\ \hline {[0,0.84]}&0.456&0.50\\ {[0.84,0.89]}&0.44&0.46\\ {[0.89,0.94]}&0.40&0.41\\ {[0.94,0.96]}&0.36&0.33 \end{array}\] For clarity, each entry is certified by arithmetic on rational numbers. If \(b\) is the right endpoint and \(n\) the numerator bound, the four values of \((0.49-n)(1-b^2)-0.01\) are respectively \(0.0000096\), \(0.000395\), \(0.000476\), \(0.000192\), all positive. For the last three rows the values of \(D_1(a)^2\) at the left endpoints are \(0.20772864\), \(0.16467759\), \(0.10285104\), smaller than \(0.46^2\), \(0.41^2\), \(0.33^2\) respectively. Finally each numerator lower bound exceeds \(0.91\) times the denominator upper bound; the respective differences are \(0.001\), \(0.0214\), \(0.0269\), \(0.0597\). ◻

Proof. The two branches of \(H\) meet once, at \[\theta=\frac{2}{1+\sqrt{1+4cd_m/\pi}}, \qquad cd_m\theta^2=\pi(1-\theta).\] Thus \(H(q)=q^2\) up to \(\theta\) and \(H(q)=\pi(1-q)/(cd_m)\) thereafter, with maximum \(\theta^2\). Integration, followed by the identity \(\pi/(cd_m)=\theta^2/(1-\theta)\), gives \[ \begin{split} M&=\frac{\theta^3}3+ \frac{\pi(1-\theta)^2}{2cd_m} =\frac{\theta^2(3-\theta)}6,\\ \int_0^1qH(q)\,dq &=\frac{\theta^4}4+ \frac{\pi(1-\theta)^2(1+2\theta)}{6cd_m},\\ q_m&=\frac{2+2\theta-\theta^2}{2(3-\theta)},\qquad p_2=1+cd_m^2\theta^2. \end{split} \tag{40}\] Only \(3.14<\pi<22/7\) is needed to bound these expressions. The function \(0.567x^2-\pi(1-x)\) increases on \([0,1]\), and \[\begin{split} 0.567(32/37)^2-3.14(5/37)&=-\frac{73}{342250}<0,\\ 0.567(13/15)^2-(22/7)(2/15)&=\frac{3587}{525000}>0. \end{split}\] It follows that \(32/37<\theta<13/15\). Both \(\theta^2(3-\theta)/6\) and \((2+2\theta-\theta^2)/(2(3-\theta))\) increase on \((0,1)\); for the latter, the derivative is \((\theta-2)(\theta-4)/(2(3-\theta)^2)>0\). Consequently \[M>\frac{40448}{151959}>0.2659, \qquad q_m>\frac{2041}{2923}>0.697, \qquad p_2<\frac{2723587}{2500000}<1.09.\] This proves (37).

Here are explicit rational margins for the remaining coefficient. Put \(\beta_0=0.303d_m=0.06363\). The bounds just proved imply \[\frac{0.91cM}{p_2^{3/2}} >\frac{0.91\cdot2.7\cdot0.2659}{1.09^{3/2}}>0.574.\] The last inequality follows by squaring positive quantities, since \[(0.91\cdot2.7\cdot0.2659)^2 -0.574^2\,1.09^3 =\frac{14121304169}{10^{14}}>0.\] Also \((1-q_m)d_m<\beta_0\). If \(u=b_0\), then \(u-\beta_0>1-1/3.14-0.06363>0.617\), and \[0.574(0.617)^3=0.134824054862>0.134.\] Hence the left side of (38) exceeds \(0.134h^3\). If \(u=0.96/h\), then \[h(u-\beta_0)\geq0.96-1.62\beta_0 =0.8569194>0.856, \qquad 0.574(0.856)^3=0.360025437184>0.36.\] Both branches therefore dominate \(C_0(h)\), including their common endpoint and the switch in the definition of \(C_0\). ◻

Proof. The first assertion follows from (26). For the second, first suppose \(\delta>0\) and restrict to the area interval \([u-\delta,u]\) of Lemma 17. Put \(K=c\delta^2H(q)\). Since \(p\geq\sqrt{1+K}\) and \(G\) is non-increasing, \[G(hv/p)\geq G\!\left(\frac{hv}{\sqrt{1+K}}\right).\] On this interval the baseline uses \(p_*=1\), and \(hv\leq hu\leq0.96<D\). Thus the whole squared-argument interval \([(hv)^2/(1+K),(hv)^2]\) lies in the differentiable branch of \(G\). By (21) and Lemma 18, \[ \begin{split} G\!\left(\frac{hv}{\sqrt{1+K}}\right)-G(hv) &\geq \int_{(hv)^2/(1+K)}^{(hv)^2}0.91\sqrt r\,dr\\ &\geq\frac{0.91(hv)^3K}{(1+K)^{3/2}} \geq\frac{0.91(hv)^3c\delta^2H(q)}{p_2^{3/2}}. \end{split} \tag{43}\] The second inequality bounds \(\sqrt r\) below by its value at the left endpoint, and the last uses \(1+K\leq p_2\).

Changing variables \(v=u-\delta+\delta q\) supplies a further factor \(\delta\). Jensen’s inequality for the convex function \(v^3\) on \(v\geq0\), with probability density \(H/M\), gives \[\begin{split} \int_0^1 (u-\delta+\delta q)^3H(q)\,dq &\geq M\bigl(u-(1-q_m)\delta\bigr)^3\\ &\geq M\bigl(u-(1-q_m)d_m\bigr)^3. \end{split}\] All bases are positive by (30). Integrating (43) and applying Lemma 19 therefore gives an additional area contribution at least \(C_0\delta^3\) beyond the baseline. If \(\delta=0\), that conclusion holds trivially with zero gain; no variable \(q\) is then introduced.

Together with the baseline contribution from \([t,s]\), this proves \[A-1\geq B(h)+\ell(s-t)+ C_0\min\{d_m,(P-Q\sqrt{s-t})_+\}^{\,3}.\] Taking the infimum over all \(y\geq0\) weakens this inequality and yields (42). ◻

Proof. The reciprocal of the planar profile gives the increasing function \[q_h(v):=\frac{hv}{p_*(v)}= \max\left\{h\sqrt{v/\pi},\ hv, \frac{hv}{\sqrt{\pi(1-v)}}\right\}.\] For \(0\le X\le D<1\), its sublevel set in \([0,t]\) has length \[\min\{\pi(X/h)^2,X/h,\rho(X/h)\}.\] Indeed these are the three separate sublevel restrictions; the additional restriction \(v\le t\) is redundant since \(\rho(X/h)<\rho(1/h)=t\). As a function of \(k=X/h\), the selected branches are \(\pi k^2\) for \(k\le a_0\), \(k\) for \(a_0\le k\le b_0\), and \(\rho(k)\) for \(k\ge b_0\). This is also obtained by inverting the three successive pieces of \(p_*\). The identity \[G(q)=\ell+\int_0^D W(X)\mathbf 1_{\{q\le X\}}\,dX \qquad(q\ge0)\] and Tonelli’s theorem now prove (48).

The function inverted to obtain \(\rho\) has positive first and second derivatives. Hence \(\rho\) is increasing and concave, with \(\rho(0)=0\), and \(k\rho'(k)\le\rho(k)\le1\). Consequently \[\frac{d^2}{dh^2}\rho(X/h) =\frac{k^2\rho''(k)+2k\rho'(k)}{h^2}\le\frac2{h^2}.\] The same bound holds for \(t=\rho(1/h)\). On the selected quadratic piece, \(k\le a_0\), so \[\frac{d^2}{dh^2}(\pi k^2)=\frac{6\pi k^2}{h^2} \le\frac6{\pi h^2}<\frac2{h^2}.\] On the selected linear piece the second derivative is \(2k/h^2<2/h^2\), since \(k\le b_0<1\).

These bounds concern only the pieces that are actually selected. For each fixed \(X>0\), as \(h\) increases they occur in the order \(\rho(X/h)\), \(X/h\), \(\pi(X/h)^2\). At the first change, \(\rho'(b_0)=2/(\pi+1)<1\), so the first derivative decreases. At the second it changes from \(-a_0/h\) to \(-2a_0/h\), again decreasing. There is thus no positive atom in the distributional second derivative. The case \(X=0\) is the constant zero function. Each integrand, and also \(t\), becomes concave on \([a,b]\) after subtracting \(h^2/a^2\). Their nonnegative weights in (48) sum to \(\ell+.72=1\). The remaining term is affine, proving the assertion. ◻

Proof. Fix \(y\ge0\), and write \[F_y(h)=\ell y^2+C_0(h)f(w(h)),\qquad f(w)=\min\{d_m,w_+\}^3, \quad w=u-1+R-Qy,\quad R=2ht/\pi.\] Then \(S=\inf_{y\ge0}F_y\). Differentiation of \(h^2t^2=\pi(1-t)\) gives \[ \begin{gathered} R'=\frac{2t^2}{\pi(2-t)},\quad R''\le0, \qquad Q=\frac2{\sqrt\pi}(1+4h^2/\pi)^{1/4},\\ \frac{Q'}Q=\frac{2h}{\pi+4h^2},\qquad \frac{Q''}Q=\frac{2\pi-4h^2}{(\pi+4h^2)^2}. \end{gathered} \tag{49}\] In particular, \(0<Q'/Q\le1/(2h)\) and \(Q''/Q\ge-1/(4h^2)\).

Here are rational bounds sufficient for the argument. The inequalities \(3.14<\pi<22/7\) and the equation for \(t\) imply \[.64<t<.8,\qquad t(1.62)<.65.\] For example these follow by substituting the proposed endpoints into \(h^2t^2-\pi(1-t)\), which is strictly increasing in \(t\). Since \(t\) decreases and \(R\) increases, they yield \[.19<\frac{1792}{9350}<R' <\frac{160}{471}<.34, \qquad R\le R(1.62)<\frac{1053}{1570}<.671.\] On a piece where \(0<w<d_m\), it follows that \[0\le Qy<P\le b_0-1+R<.36.\] The switch in \(u=\min\{b_0,.96/h\}\) occurs at \(h_c=.96/b_0>1.4\). On its constant piece, \[w'>.19-\frac{.36}{2(1.02)}>0, \qquad w''\le\frac{.36}{4(1.02)^2}<.8.\] On its reciprocal piece, \[w'>-\frac{.96}{1.4^2}+.19-\frac{.36}{2(1.4)}>-.50, \qquad w''\le\frac{1.92}{1.4^3}+\frac{.36}{4(1.4)^2}<.8.\] On both pieces \(w'\le R'<.34\). Also, on every smooth piece, \[0<C_0\le.36,\qquad 0\le C_0'\le1.08, \qquad 0\le C_0''\le2.16;\] these follow directly from \(C_0=\min\{.134h^3,.36\}\) and \(h\le1.62\). Thus on an active cubic piece the product rule gives \[\begin{align*} (C_0w^3)'' &=C_0''w^3+6C_0'w^2w'+6C_0w(w')^2+3C_0w^2w''\\ &\le2.16d_m^3+6(1.08)d_m^2(.34) +6(.36)d_m(.50)^2+3(.36)d_m^2(.8)\\ &=.26866728<.4. \end{align*}\] On \(w<0\) the added term is zero; on \(w>d_m\) its second derivative is at most \(2.16d_m^3<.4\).

For completeness, all possible simultaneous changes of formula preserve this bound. The only jumps in \(w'\) and \(C_0'\) are nonpositive, coming respectively from the minima defining \(u\) and \(C_0\). At a point with \(0<w<d_m\) the jump in \(F_y'\) is \[w^3\Delta C_0'+3C_0w^2\Delta w'\le0.\] At \(w=0\), both \(f\) and its derivative vanish, so the derivatives match even at a simultaneous switch. At \(w=d_m\), the one-sided derivatives of \(f(w(h))\) are \[3d_m^2\max\{w'_-,0\}\quad\hbox{from the left},\qquad 3d_m^2\min\{w'_+,0\}\quad\hbox{from the right}.\] Their difference is nonpositive; a simultaneous switch of \(C_0\) adds \(d_m^3\Delta C_0'\le0\). These formulas include tangencies. On a capped piece only the nonpositive jump of \(C_0'\) remains. For fixed \(y\) these functions are algebraic on each branch, so the threshold points are finite unless a threshold holds identically on a piece; an identical threshold has the matching derivatives just described. It follows that \(F_y-.2h^2\) is concave on the full interval. Finally, \[S(h)-.2h^2=\inf_{y\ge0}\bigl(F_y(h)-.2h^2\bigr)\] is concave: the pointwise infimum of concave functions is concave, as is seen either from its defining inequality or its hypograph. The infimum is finite, since \(F_y\ge0\) and \(F_0\le C_0d_m^3\). This also handles changes of the minimizing value of \(y\). ◻

Proof. Differentiating (46) gives \[\rho'''(k)=\frac{3\sqrt\pi\,\sigma}{(1+\sigma k^2)^{5/2}}, \qquad \rho''''(k)=-\frac{15\sqrt\pi\,\sigma^2 k} {(1+\sigma k^2)^{7/2}}.\] The magnitude of the latter expression decreases when \(6\sigma k^2\ge1\), in particular for \(k\ge b_0\). Since \(.68<b_0<.69\) and \(3.14<\pi<22/7\), \[|\rho''''(b_0)| <\frac{15\sqrt{22/7}(11/14)^2(.69)} {(1+(3.14/4)(.68)^2)^{7/2}}<4.\] The final inequality follows by squaring positive quantities and cross-multiplying rational numbers. The midpoint Taylor remainder is therefore at most \(4(L/2)^4/24\) in absolute value. Multiplying by the total weight \(.72\) bounds its contribution to the integral by \(.0075L^4\le.008L^4\). ◻

Proof. Positivity of \(Q,C_0\) is immediate. To see that \(P>0\) uniformly, note that \(t(h)\) decreases, so \(ht(h)=\sqrt{\pi(1-t(h))}\) increases. At \(h=1\) the equation for \(t\) gives \(t(1)>3/4\), since \(9/16-\pi/4<0\). Using \(h\geq1\), \(u\geq16/27\), and \(\pi<22/7\), we get \[P>\frac{16}{27}-1+\frac{21}{44} =\frac{83}{1188}>0.\]

For a fixed \(h\), write \(w=P-Qy\), so \(w\leq P\). Define the uncapped polynomial objective on \([0,P]\) by \[F(w)=\frac{\ell(P-w)^2}{Q^2}+C_0w^3.\] It agrees with the original objective on the feasible uncapped interval \([0,\min\{P,d_m\}]\). The polynomial is strictly convex on \([0,P]\), with \(F'(0)<0<F'(P)\), and its unique critical point lies in \((0,P)\). Solving \(3C_0Q^2w^2=2\ell(P-w)\) gives exactly \(w=q_*\) in (54).

The branch \(w\leq0\) has minimum \(\ell P^2/Q^2=F(0)\), which is greater than \(F(q_*)\). If \(P\geq d_m\), the capped branch \(d_m\leq w\leq P\) has minimum \(C_0d_m^3\), attained at \(w=P\). When \(q_*\leq d_m\) both candidates in (55) are consequently sufficient. When \(q_*>d_m\), the minimum of \(F\) on \([0,d_m]\) is \(F(d_m)\), which is at least \(C_0d_m^3\); moreover \(F(q_*)\geq C_0q_*^3>C_0d_m^3\). Thus the capped value still wins in the displayed formula even though \(q_*\) is infeasible. Finally, if \(P<d_m\), the cap is never reached, but its displayed value is greater than \(C_0P^3=F(P)\geq F(q_*)\) and hence cannot alter the minimum. The equalities \(P=d_m\) and \(q_*=d_m\) are included in these comparisons. This proves the formula in all cases. ◻

Proof of Proposition 22. Apply (59) with the corrected knot lower bounds in (60). Table 2 gives a strictly positive lower bound on every interval, and those intervals cover the stated ranges, including all endpoints. The smallest certified uniform margin is \(9/100000\), on \([1.42,1.46]\) and \([1.46,1.50]\). In particular, the corrected lower bound at \(1.46\) is exactly \(.00057\) and the interior interpolation loss is \(.00048\). ◻