Proof. Suppose \(x_j\to b\in\partial\Omega\) and \(u(x_j)\le A\). Fix \(a\in\Omega\). Convexity, followed by continuity at the interior point \(a+q(b-a)\), gives \[u(a+q(b-a))\le (1-q)u(a)+qA,\qquad 0\le q<1.\] An affine supporting function at \(a\) gives a lower bound on the same segment. The resulting bounded convex function of \(q\) has finite total variation and a finite limit as \(q\uparrow1\). Its lifted curve has finite Euclidean length, since \[\int_0^1\sqrt{|b-a|^2+ \left|\frac{\mathop{}\!\mathrm d}{\mathop{}\!\mathrm dq}u(a+q(b-a))\right|^2}\,\mathop{}\!\mathrm dq \le |b-a|+\int_0^1 \left|\frac{\mathop{}\!\mathrm d}{\mathop{}\!\mathrm dq}u(a+q(b-a))\right|\,\mathop{}\!\mathrm dq<\infty.\] Its tail is therefore Cauchy in the graph metric, but its base coordinate tends to \(b\notin\Omega\). This contradicts completeness. The asserted boundary limit follows. It also shows that a bounded convergent sequence of points of \(D\) cannot have its base limit on \(\partial\Omega\). Thus \(D\) is closed, and its boundary is exactly the graph.

Choose \(r,c>0\) such that \(\overline B_r(a)\subset\Omega\) and \(D^2u\ge c\mathrm{Id}\) on this ball. Along every ray from \(a\), convexity of \(g_a\), together with \(g_a(a)=0\), gives \[g_a(x)\ge \frac{cr}{2}|x-a| \qquad (x\in\Omega,\ |x-a|\ge r).\] Indeed, \(g_a\ge cr^2/2\) at the point of the ray at distance \(r\), and its quotient by the distance is nondecreasing. Thus each stated sublevel is bounded. The boundary limit just proved makes it closed in \(\mathbb R^n\) and contained in \(\Omega\). On such a compact base, both \(l_a\) and the cap height are bounded, which proves the last claim. ◻

Proof. Choose \(q_j\in C_j\) with \(q_j\to q\), so \(\ell(q_j)<b\) eventually. If points \(p_j\in C_j\cap\{\ell\le b\}\) escape every bounded set, pass to a subsequence for which \[\frac{p_j-q_j}{|p_j-q_j|}\longrightarrow v,\qquad |v|=1.\] For each fixed \(t\ge0\), the points \(q_j+t(p_j-q_j)/|p_j-q_j|\) eventually lie on the segments \([q_j,p_j]\). Their limit \(q+tv\) therefore belongs to \(C\cap\{\ell\le b\}\), contradicting boundedness.

Uniform boundedness and local convergence show that every limit point of the approximating caps lies in the limit cap. Conversely, if \(x\) belongs to the limit cap, then \((1-\eta)x+\eta q\) lies strictly below the top for every \(0<\eta\le1\) and is approximated by points of the approximating caps. Letting \(\eta\downarrow0\) and using compactness gives the reverse Hausdorff inclusion. ◻

Proof. The recession assertion holds for any closed convex set containing the indicated orthant, without a full-dimensionality assumption. For \(X\in C\) and \(v_i\ge0\), take the limit as \(\eta\downarrow0\) of \[(1-\eta)X+\eta(v/\eta,0)\in C\] to obtain \(X+(v,0)\in C\). This proves fiber monotonicity. Since the ambient origin belongs to \(C\), contraction toward it proves the second inclusion in (2). Containment of zero and compactness follow from the definition.

Take an interior ball of \(C\) centered at \((s^0,y^0)\); all coordinates of \(s^0\) are positive. Given \(s_i>0\), choose \(0<\eta\le1\) with \(\eta s^0_i\le s_i\). Contract the transverse slice of this ball by \(\eta\) about the ambient origin, and increase the \(s\) coordinates to \(s\). The resulting transverse ball lies in \(K_s\), proving full dimension. When \(m=0\) this assertion uses the usual convention for \(\mathbb R^0\).

Finally, for any \(a\in\Omega\), the coordinates \(s=z-l_a(x)\) and \(y=x-a\) make \(D\) a \(1\)-model by Lemma 2. ◻

Proof. Choose a maximum-volume inscribed ellipsoid and take the linear part of its normalization to a unit ball. John’s Theorem gives the two inclusions in (3). Since \(0\in AK\), the outer inclusion implies \(|c|\le d\). The centering assumption also gives \(\overline B_{1/C_*}(-c/C_*)\subset AK\). The convex combination of this ball and \(\overline B_1(c)\) with respective weights \(C_*/(C_*+1)\) and \(1/(C_*+1)\) is the centered ball in (4). ◻

Proof. Suppose that no uniform constant exists. Choose \(k\)-models and positive fibers for which the inclusion with constant \(j\) fails. Positive diagonal changes in \(s\) put those fibers at \(\mathbf1=(1,\ldots,1)\), and Lemma 6 provides linear changes in \(y\). Denote the resulting models by \(C_j\) and their fibers at \(\mathbf1\) by \(F_j\). Thus \[ \overline B_1(c_j)\subset F_j\subset\overline B_{2m}, \qquad |c_j|\le m, \qquad -F_j\not\subset jF_j. \tag{6}\] All these coordinate changes preserve the model properties and the failed inclusion.

For \(T\ge1\) and \(0\le s_i\le T\), Lemma 5 gives \[ K_s^{(j)}\subset K_{T\mathbf1}^{(j)} \subset T F_j\subset\overline B_{2mT}. \tag{7}\] Pass to a local limit \(C\) and a subsequence with \(c_j\to c\). The limit is contained in the nonnegative \(s\) halfspaces, contains the entire orthant at \(y=0\), and has compact caps by (7). Moreover, \[[1,2]^k\times\overline B_1(c_j)\subset C_j,\] so its limit has nonempty interior. The closure properties of \(\mathcal F(D)\) show that \(C\) is a \(k\)-model.

Its fiber \(F=K_{\mathbf1}\) is the Hausdorff limit of \(F_j\). Only one inclusion needs explanation. If \((s^{(j)},y_j)\in C_j\) tends to \((\mathbf1,y)\in C\), let \[a_j=\min\left\{1,\frac1{s^{(j)}_1},\ldots, \frac1{s^{(j)}_k}\right\}.\] The denominators are positive for large \(j\), and \(a_j\to1\). Contracting by \(a_j\) and then increasing the \(s\) coordinates gives \((\mathbf1,a_jy_j)\in C_j\). Thus every point of \(F\) is approximated from the exact fibers \(F_j\); the other inclusion follows from local convergence. Uniform boundedness makes this Hausdorff convergence. If zero were interior to \(F\), the \(F_j\) would eventually contain a fixed ball centered at zero. Their common outer bound would then give a fixed centering constant, contrary to (6). Hence \(0\in\partial F\).

A supporting linear functional of \(F\) at zero can be made the coordinate \(y_1\), with \(F\subset\{y_1\ge0\}\). This inequality holds on all of \(C\): contract any \((s,y)\in C\) until each \(s_i\le1\), then increase the \(s\) coordinates to \(\mathbf1\) and apply the inequality on \(F\). Since \(F\) has nonempty interior, an additional invertible linear choice of transverse coordinates makes an interior point of \(F\) equal to \((y_1,y')=(1,0)\), while retaining the global support \(y_1\ge0\).

We now construct a model with \(k+1\) directions. Put \(d=m-1\). For \(0<\epsilon<1/2\), let \[F_\epsilon= \{y'\in\mathbb R^d:(\mathbf1,\epsilon,y')\in C\}.\] This compact slice contains zero in its interior when \(d>0\). Indeed, if \(\overline B_r((1,0))\subset F\), contracting by \(\epsilon\) about the ambient origin and then adding \((1-\epsilon)\mathbf1\) in the \(s\) coordinates shows that \[\overline B_{\epsilon r}\subset F_\epsilon\subset\mathbb R^d.\] When \(d>0\), apply Lemma 6 to choose an invertible linear map \(A_\epsilon\) on \(\mathbb R^d\) so that \[\overline B_1(c_\epsilon)\subset A_\epsilon F_\epsilon \subset\overline B_{2d},\qquad |c_\epsilon|\le d.\] If \(d=0\), take \(A_\epsilon\) to be the unique linear automorphism of \(\mathbb R^0\) and omit all statements about its balls. Use the affine images \[C_\epsilon= \{(s,r,z):(s,\epsilon r,A_\epsilon^{-1}z)\in C\}.\] They are supported in \(s_i,r\ge0\), contain the origin, and their slice at \((s,r)=(\mathbf1,1)\) has the stated normalization.

Figure 1 illustrates the supporting fiber and the slice used in this rescaling. The remaining step is to turn control of that one slice into a uniform bound for every fixed cap.

Here is a uniform bound for their caps. Suppose \((s,r,z)\in C_\epsilon\) with \(s_i,r\le T\), where \(T\ge1\). Increase \(s\) to \(T\mathbf1\) and contract by \(1/T\). This gives \[(\mathbf1,q,z/T)\in C_\epsilon, \qquad q=r/T\in[0,1].\] The point \((\mathbf1,2,0)\) also belongs to \(C_\epsilon\), since \(2\epsilon<1\) and it is obtained by contracting \((\mathbf1,1,0)\in C\) and then increasing \(s\). Interpolate these two points with weight \[a=\frac1{2-q}\in[1/2,1]\] on \((\mathbf1,q,z/T)\). The resulting point is \((\mathbf1,1,az/T)\), so the normalized slice gives \[ |z|\le 4dT\qquad(d>0). \tag{8}\] Equivalently, a slice outer radius \(R\) gives the bound \(2TR\). When \(d=0\), the bounds on \(s\) and \(r\) themselves suffice. Thus caps in the new nonnegative coordinates are uniformly bounded.

Extract a local limit \(C_0\) as \(\epsilon\downarrow0\), also taking \(c_\epsilon\to c_0\) if \(d>0\). For each fixed \(s_i>0\) and \(r\ge0\), the point \((s,r,0)\) belongs to \(C_\epsilon\) as soon as \[\epsilon r\le\min\{1,s_1,\ldots,s_k\}.\] For \(r>0\), this follows by contracting \((\mathbf1,1,0)\in C\) by \(\epsilon r\) and then increasing \(s\); for \(r=0\) it follows from the original orthant. Hence \(C_0\) contains these points. Closedness adds all faces with some \(s_i=0\), so it contains the entire nonnegative \((s,r)\) orthant at \(z=0\). The support inequalities pass to the limit, and (8) makes its caps compact.

For \(d>0\), the limit also contains \(\{(\mathbf1,1)\}\times\overline B_1(c_0)\). The recession argument in Lemma 5, which does not require full dimension, permits addition of every nonnegative \((s,r)\) direction. Thus \[[1,2]^{k+1}\times\overline B_1(c_0)\subset C_0,\] which proves full dimension. For \(d=0\), the contained orthant already has full dimension. Finally, affine invariance and diagonal closure of \(\mathcal F(D)\) give \(C_0\in\mathcal F(D)\). It is a \((k+1)\)-model, contradicting maximality.

This contradiction allowed the initial models, fibers, and admissible coordinates to vary arbitrarily. It therefore proves the uniform assertion (5). Its last conclusion follows, for example, from (4). ◻

Proof. The normalization follows from Proposition 7 and Lemma 6. Monotonicity and contraction toward zero give \(aK_{\mathbf1}\subset K_s\subset bK_{\mathbf1}\) on \([a,b]^k\), proving (9). If \(\ell(s)=\sum_i a_i s_i\) with \(a_i>0\) and \(B>0\), put \(T=\max\{1,B/\min_i a_i\}\). Then \[C\cap\{\ell\le B\} \subset [0,T]^k\times\overline B_{TR_0}.\] Also \(K_s\supset\overline B_{r_0/2}\) whenever each \(s_i\ge1/2\), which supplies the fixed interior ball about \((\mathbf1,0)\).

Lemma 3 transfers each cap bound to the approximants; the strict point below its top is the origin. For a fixed box \([a,b]^k\), a smaller product \([a,b]^k\times\overline B_{ar_0/2}\) is a compact subset of the interior of the model, as is seen by first enlarging the positive \(s\) box slightly. Interior convergence therefore gives the lower fiber bound on the entire box, and a containing cap gives the upper bound. These constants depend only on the displayed fixed parameters. Taking the maximum of the finitely many necessary indices proves the final statement. ◻

Proof. The Hessian cofactor matrix satisfies \(\partial_iU^{ij}=\partial_jU^{ij}=0\). This follows by differentiating its alternating determinant formula: terms cancel in pairs by symmetry of the third derivatives. Since \[(\det D^2u)^\delta(D^2u)^{-1}=wU,\] the first variation in a compactly supported graph direction \(\eta\) is \[\delta\int_\Omega \mathop{\mathrm{tr}}\bigl((D^2u)^{-1}D^2\eta\bigr)\,\mathop{}\!\mathrm dA =\delta\int_\Omega \eta U^{ij}w_{ij}\,\mathop{}\!\mathrm dx=0.\] Here and below repeated coordinate indices are summed. For a compactly supported parametrized variation, its normal part can be represented locally as a graph variation, while its tangential part is a reparametrization. A partition of unity therefore gives stationarity for parametrized variations. Formula (11) transfers this statement to every affine image. ◻

Proof. The Gauss map is injective: two distinct contact points with the same supporting normal would force a boundary segment in their supporting hyperplane, contrary to positive curvature. Thus \(\int K\,\mathop{}\!\mathrm dS\) over the part in the ball is at most the area of \(S^n\).

For the Euclidean area bound, cover this part by the \(2(n+1)\) sets on which some signed component of the unit normal is at least \((n+1)^{-1/2}\). On each set, convexity makes projection to the corresponding coordinate hyperplane injective: points with a given projection lie on a line meeting the body in an interval, and the chosen sign selects one endpoint. The projection Jacobian is at least \((n+1)^{-1/2}\), and its image lies in the radius-\(R\) ball. This bounds \(\int\mathop{}\!\mathrm dS\). Finally, Hölder’s inequality gives \[\int K^\delta\,\mathop{}\!\mathrm dS \le \left(\int K\,\mathop{}\!\mathrm dS\right)^\delta \left(\int\mathop{}\!\mathrm dS\right)^{1-\delta}.\] ◻

Proof. The identity is affine invariant up to the common area factor in (11), so it suffices to work on the original graph. Put \(H=D^2u\), write \(o=(o',o_{n+1})\), and use \(\nu=(-Du,1)\). Set \[q(x)=\ell(x,u(x)),\qquad Z(x)=\nu(o-X)=o_{n+1}-u+(x-o')\cdot Du.\] If \(\ell_z\) denotes the ambient vertical coefficient of \(\ell\), then \[DZ=H(x-o'),\qquad \mathop{\mathrm{tr}}(H^{-1}D^2Z)=n+(x-o')\cdot D\log\det H,\qquad D^2q=\ell_z H.\] Testing Lemma 9 with \(Z\psi(q)\) gives \[\begin{align*} 0=\int_M\bigl[ &Z|\mathop{}\!\mathrm dq|_{H^{-1}}^2\psi''(q) +nZ\ell_z\psi'(q) +2(x-o')\cdot Dq\,\psi'(q)\\ &+n\psi(q) +\psi(q)(x-o')\cdot D\log\det H \bigr]\,\mathop{}\!\mathrm dA . \end{align*}\] Because \(\mathop{}\!\mathrm dA=(\det H)^\delta\,\mathop{}\!\mathrm dx\), integration by parts in the last term replaces it by \[-\delta^{-1} \left[n\psi(q)+(x-o')\cdot Dq\,\psi'(q)\right].\] The coefficient of \(\psi\) is consequently \(-n(n+1)\). The coefficient of \(\psi'\) is \[n\bigl[Z\ell_z-(x-o')\cdot Dq\bigr] =n\bigl[\ell(o)-q\bigr].\] This is (12). Compact support justifies both integrations by parts. ◻

Proof. Lemma 3 gives a uniform bound for these caps. Choose a closed ball \(\overline B_r(o)\) in \(\operatorname{int}C\cap\{\ell<b\}\). After decreasing \(r\), the same ball lies in \(D_j\) for all sufficiently large \(j\). Fix \(\varepsilon_0>0\) with \(\ell(o)\le b-\varepsilon_0\).

Let \(e_j\) be the unit vector in the image of the original upward vertical direction under the affine map defining \(D_j\). It is a recession direction of \(D_j\), and \(\partial D_j\) is graphical along it. The component \(\ell(e_j)\) is bounded below by a positive constant. Indeed, the ray \(o+t e_j\) lies in \(D_j\); if \(\ell(e_j)\le0\), the whole ray lies in the cap, and otherwise its initial segment of length \((b-\ell(o))/\ell(e_j)\) lies there. Uniform boundedness of the caps excludes both a nonpositive component and components tending to zero. Consequently \[V_j=\frac{e_j}{\ell(e_j)} \quad\text{satisfies}\quad \ell(V_j)=1,\qquad |V_j|\le C.\] Use the coordinates \[X=\zeta+\tau V_j,\qquad \zeta\in\ker\ell.\] These coordinate maps and their inverses are uniformly bounded. With an orthonormal basis of \(\ker\ell\), their determinants have absolute value \(1/|\ell|\), where \(|\ell|\) is the Euclidean norm of the covector. In these coordinates \(\partial D_j\) is the graph of a smooth strictly convex function \(g_j\), and the cap corresponds to \(\tau\le b\).

The sets \[E_j=\{\zeta:g_j(\zeta)<b\}\] are uniformly bounded. Every closed sublevel of \(g_j\) at a level at most \(b\) is compact within its domain: its lifted graph lies in a compact cap, and the closed hypersurface \(\partial D_j\) is precisely the graph in this direction. Writing \(\zeta_j=o-\ell(o)V_j\), the common interior ball gives \[g_j(\zeta)\le b-\varepsilon_0 \qquad\bigl(\zeta\in B_{\ker\ell}(\zeta_j,r)\bigr).\] Here \(B_{\ker\ell}(\zeta_j,r)\) is the radius-\(r\) ball in \(\ker\ell\) centered at \(\zeta_j\). The centers \(\zeta_j\) and the sets \(E_j\) are uniformly bounded. We may therefore choose one \(\alpha>0\) so small that the quadratics \[q_j(\zeta)=b-\frac{2\varepsilon_0}{3} +\alpha|\zeta-\zeta_j|^2\] satisfy \[b-\frac{2\varepsilon_0}{3}\le q_j \le b-\frac{\varepsilon_0}{3} \qquad\text{on }E_j.\]

Let \(\chi\) be a smooth convex regularization of the positive part, with \(0\le\chi'\le1\), equal to \(0\) for arguments at most \(-\eta\) and equal to the argument for arguments at least \(\eta\), where \(0<\eta<\varepsilon_0/6\). On \(E_j\) set \[\widetilde g_j=g_j+\chi(q_j-g_j).\] This equals \(q_j\) on the indicated base ball and equals \(g_j\) near the level \(b\). Extending by \(g_j\) outside \(E_j\) gives a smooth perturbation whose difference from \(g_j\) has compact support in \(E_j\). Its Hessian is positive definite, since it is a convex combination of \(D^2g_j\) and \(D^2q_j\), plus a positive semidefinite matrix.

The function \(F(H)=(\det H)^\delta\) is concave on positive definite matrices: the determinant \(n\)-th root is concave, and \(n\delta=n/(n+2)<1\). Concavity and Lemma 9 give \[\begin{align*} \int_{E_j}F(D^2\widetilde g_j)\,\mathop{}\!\mathrm d\zeta &\le \int_{E_j}F(D^2g_j)\,\mathop{}\!\mathrm d\zeta\\ &\quad+\delta\int_{E_j} F(D^2g_j)(D^2g_j)^{-1}:D^2(\widetilde g_j-g_j)\,\mathop{}\!\mathrm d\zeta\\ &=\int_{E_j}F(D^2g_j)\,\mathop{}\!\mathrm d\zeta. \end{align*}\] The first integral is bounded below by \(\lvert B_{\ker\ell}(0,r)\rvert(2\alpha)^{n\delta}>0\). Affine covariance of area, with the uniformly controlled coordinate determinants above, proves the assertion. ◻

Proof. For a fixed \(a\), use the ambient affine coordinates \[s=z-l_a(x),\qquad y=x-a.\] The epigraph lies in \(s\ge0\), contains the ray \(\{s\ge0,\ y=0\}\), and has compact caps by Lemma 2. It is therefore a model with one direction, whose fiber at height \(t\) is \(K_a(t)\). Proposition 14 says that the maximal direction number is one. The uniformity over models and coordinates in Proposition 7 consequently gives \(-K_a(t)\subset C K_a(t)\) with one finite constant \(C\). Taking \(\rho=\min\{1,C^{-1}\}\) proves the claim. ◻

Proof. Initially let \(I=\{r:a+re\in\Omega\}\), an open interval containing zero, and choose \(R\in I\) with \(R>0\). For \(0<r_0<R\), the tangent section based at \(a+r_0e\) and having height \[b=r_0g'(r_0)-g(r_0)>0\] contains \(a\) on its boundary. Applying (44) to the displacement \(-r_0e\) puts the displacement \(\rho r_0e\) in the same translated section. In particular \((1+\rho)r_0\in I\), and the section inequality gives \[g((1+\rho)r_0)\le(1+\rho)r_0g'(r_0).\] Convexity and \(g\ge0\) give \[g'(r_0)\le \frac{g(R)-g(r_0)}{R-r_0} \le\frac{g(R)}{R-r_0}.\] Set \[\theta=\frac{2+\rho}{2(1+\rho)},\qquad r_0=\theta R.\] Then \((1+\rho)\theta=\beta\), with \(\beta\) as in (46), and \[\beta R\in I,\qquad g(\beta R)\le\frac{\beta}{1-\theta}g(R)=A g(R).\] Iterating the domain inclusion shows that the positive ray from \(a\) in direction \(e\) is contained in \(\Omega\). Every oriented direction admits an initial positive interval by openness, so all rays from \(a\) lie in \(\Omega\). This proves \(\Omega=\mathbb R^n\), as well as (45) on every ray. ◻

Proof. Write \(v=v_t\). Since \(v(0)=Dv(0)=0\) and \(v\le1\) on \(\overline{B_1}\), convexity gives \(0\le v(x)\le1/4\) on \(B_{1/4}\). Fix such an \(x\). For every unoriented line through \(x\), choose its unit direction \(e\) so that \(Dv(x)\cdot e\le0\). The ray in this direction meets \(\partial S_v(0,1)\) at a distance at most \(2R\). At that intersection, its tangent-subtracted gap \[g_{x,e}(r)=v(x+re)-v(x)-r\,Dv(x)\cdot e\] is at least \(1-v(x)\ge3/4\). Thus the positive-ray radius of the closed tangent section of height \(1/4\), based at \(x\), is at most \(2R\). Balance of that section bounds the opposite-ray radius by \(2R/\rho\): a point on the opposite ray, reflected and contracted by \(\rho\), must still fit on the first ray. Consequently every unit direction \(e\) satisfies \[ g_{x,e}(R_*)\ge\frac14,\qquad R_*=\frac{2R}{\rho}. \tag{50}\]

The gap \(g_{x,e}\) is nondecreasing on the positive ray. Define \[ N(r)=\max\left\{0,\left\lceil\log_\beta\frac{R_*}{r}\right\rceil\right\}, \qquad b(r)=\frac{1}{8A^{N(r)}}. \tag{51}\] Iterating (45) and using (50) gives \[A^{N(r)}g_{x,e}(r) \ge g_{x,e}\bigl(\beta^{N(r)}r\bigr) \ge\frac14.\] Thus the tangent gap is at least \(2b(r)\) at every point of distance \(r\) from \(x\), and at every point farther away. This proves (48).

If \(x\in\gamma S_v(0,1)\), then \(|x|<1/4\). Since \(B_1\subset S_v(0,1)\), we obtain \[S_v(x,b(r))\subset x+B_r\subset x+rS_v(0,1).\] Applying \(A_t\) gives (49). All constants and the function \(b\) are independent of the height \(t\), the base \(x\) in the stated range, and the line direction. ◻

Proof of Theorem 1. Proposition 14 applies for every \(3\le n\le9\). Lemmas 17 and 18 therefore give \(\Omega=\mathbb R^n\) and uniform section balance. Fix any base point, translate it to zero, subtract its tangent, and use the normalized functions \(v_t\) from (47).

These functions solve the same affine maximal equation. Indeed, for \(v(y)=t^{-1}u(Ay)\), set \[c_t=t^{-n}(\det A)^2,\qquad a=\frac{n+1}{n+2}.\] Then \[w_v=c_t^{-a}w_u\circ A,\qquad U_v=t^{1-n}(\det A)^2 A^{-1}(U_u\circ A)A^{-\mathsf T},\] so the contraction \(U_v^{ij}(w_v)_{ij}\) is a nonzero constant multiple of \((U_u^{ij}(w_u)_{ij})\circ A\), and vanishes. Subtracting an affine function leaves the equation unchanged as well.

Apply Theorem 20 to \(f_t=v_t-1\) on \(O=B_{1/4}\), with \(O'=B_{1/8}\). There \(-1\le f_t\le-3/4\), and Lemma 19 supplies the same lower modulus at every point of \(O\). Restricting a section to \(O\) only decreases it. We obtain constants independent of \(t\) such that \[ D^2v_t(0)\ge c_0\mathrm{Id},\qquad |D^3v_t(0)|\le C_0. \tag{52}\]

Put \(H=D^2u(0)>0\). The identity \[D^2v_t(0)=t^{-1}A_t^{\mathsf T}HA_t\] and the first bound in (52) imply \[t c_0|z|^2 \le (A_tz)^{\mathsf T}H(A_tz) \le\lambda_{\max}(H)|A_tz|^2,\] and hence \[ \|A_t^{-1}\| \le\sqrt{\lambda_{\max}(H)/c_0}\,t^{-1/2}. \tag{53}\] For any three vectors \(z_1,z_2,z_3\), differentiation of \(u(x)=t\,v_t(A_t^{-1}x)\) gives \[D^3u(0)[z_1,z_2,z_3] =t\,D^3v_t(0) [A_t^{-1}z_1,A_t^{-1}z_2,A_t^{-1}z_3].\] Equations (52) and (53) therefore give \[|D^3u(0)|\le Ct^{-1/2}\longrightarrow0 \qquad\text{as }t\longrightarrow\infty.\] The base point was arbitrary, so \(D^3u\equiv0\). Thus \[u(x)=\tfrac12 x^{\mathsf T}Qx+b\cdot x+c\] on \(\mathbb R^n\), with \(Q>0\). Its graph is an invertible affine image of \(\{(z,|z|^2):z\in\mathbb R^n\}\), as required. ◻