Quadratic forms, volumes and separation

This companion retains AN03-P001 Sections 4–5 and supplies the connecting finite-dimensional facts used by the selected metric and multiplier proofs.

This is a separate modified selection from the earlier AN-03 programme. Original principal author and publisher: AN-03 course-writing task / AN-03 local course project, 2026. Earlier modification: AN-03 course-writing task and OpenAI Codex. Selection and the identified connecting proofs: GPT-6 Astra (OpenAI), Ultra, 4 October 2026; publisher: AN-04 local course project.

Original text: CC0.

0. Exact earlier inputs

The measure companion, M0–M8 supplies the declared choice principle, rational enumeration, completed measure and convergence theorems. The Fourier companion, L0–L3 supplies the full $L^2$ extension and distributional compatibility. The exact U001 Schwartz and spectral proofs, Q3–Q5 supply all finite-dimensional Fourier and spectral steps. The finite-dimensional compactness, algebra and differential inputs are the exact earlier proofs named in that companion's F0 contract. Smooth cutoffs are U001 P14.3. Compact and affine substitution, with every determinant, is U001 P21.3–P21.4. M8 proves compatibility of these integrals with the Lebesgue integrals.

1. Positive forms and coordinate maps

Let $Q(X)=X^{\mathsf T}GX$, where $G$ is real symmetric positive definite. U001 Q5 gives $G=U\operatorname{diag}(\lambda_j)U^{\mathsf T}$, $\lambda_j>0$. Put $L=U\operatorname{diag}(\lambda_j^{-1/2})$. Then $L^{\mathsf T}GL=I$, $G^{-1}=LL^{\mathsf T}$, and $J=|\det L|=(\det G)^{-1/2}>0$, by direct multiplication and the proved determinant rules. Thus $Q(Ly)=|y|^2$ and $Q^{-1}(\Xi)=\Xi^{\mathsf T}G^{-1}\Xi=|L^{\mathsf T}\Xi|^2$. The Euclidean Cauchy–Schwarz inequality follows by expanding $|v-tv'|^2\geq0$ and minimizing in the real scalar $t$; the zero-vector case is immediate. Transfer this inequality by $L$ to obtain the triangle inequality for $Q^{1/2}$.

If $B$ is another real symmetric matrix, its expression under these maps is $L^{-1}BL^{-\mathsf T}$, which is again symmetric and hence has an orthonormal eigenbasis by Q5. This proves the simultaneous normalization used in the multiplier companion, without a new spectral or Gram-matrix theorem. The chain rule gives the corresponding directional derivative identities. In an orthonormal basis each coordinate of a unit direction has absolute value at most one; expansion of a $k$-linear form then compares its directional norm and its maximum coordinate coefficient with factors at most $d^k$.

2. Lebesgue volume of each ellipsoid

Let $K=\{Q(X)<r^2\}$, $r>0$. Choose a smooth function $b:\mathbb R\to[0,1]$, zero for $t\leq0$ and one for $t\geq1$, using the supplied flat cutoff, and set $f_k(X)=b(k(r^2-Q(X)))$. Each $f_k$ is compact smooth, is supported in the closed ellipsoid, and converges pointwise to $\mathbf1_K$. All supports lie in a fixed cube: positivity of the eigenvalues bounds $|X|$ on them. Compact substitution, followed by M8 and dominated convergence on that cube and its inverse image, therefore gives

The second equality uses the same argument for $y=rz$. The unit ball has positive finite measure, since it contains a positive-volume cube and lies in a bounded cube. Translations have Jacobian one. Consequently every ellipsoid volume comparison used in the covering and counting proofs follows from (F1), inclusions, and countable additivity. No unproved general measure substitution theorem is being invoked for indicator functions.

3. Dual ellipsoids and phase distances

Suppose $B:E^*\to E$ is symmetric and $A(\Xi)=\langle B\Xi,\Xi\rangle$. Use Section 1 to make $Q$ Euclidean and $B=\operatorname{diag}(\beta_j)$. On the range of $B$, write

This is exactly the supremum defining the phase-dual form: put $w_j=\beta_j\Xi_j$, apply Cauchy–Schwarz to $\sum Z_jw_j/\beta_j$, and approach the unit vector in that direction. A nonzero coordinate in $\ker B$ makes the supremum infinite by scalar dilation. This proves both directions, including $B=0$.

For $\eta\in E^*$, the support value of $\{Z:Q^A(Z)<a^2\}$ is

Again Cauchy–Schwarz gives the upper bound. When the square root is positive, take $Z_j$ proportional to $\beta_j^2\eta_j$ and approach the boundary from inside; when it is zero, every pairing vanishes and both sides are zero. The same calculation with the ordinary dual form gives support value $rQ^{-1}(\eta)^{1/2}$ for $\{Q<r^2\}$. For $Q_1\leq M Q_2$, inclusion of the defining ellipsoids and scaling gives $Q_1^A\geq M^{-1}Q_2^A$ on the common finite domain. Off it both sides are interpreted as infinite. Sections 4–5 below prove the separation and sum rules needed to use these support values.

4. Strict separation of an open convex set

Theorem 4.1 (Strict separation). Let $V$ be a finite-dimensional real vector space, let $C\subset V$ be nonempty, open and convex, and let $x\notin C$. There is a nonzero real covector $\eta$ such that

The second inequality includes $x$ on the boundary. The supremum need not be attained. The proof uses only finite-dimensional compactness and the inner product.

Choose an auxiliary Euclidean norm, put $K=\overline C$, and fix $c\in C$ with $B(c,r)\subset C$, $r>0$. First, if $z\in K$ and $0\leq t<1$, then

For $t=0$ this is immediate. For $0<t<1$, choose $z_j\in C$ tending to $z$. Convexity gives a ball of radius $(1-t)r$ centered at $(1-t)c+tz_j$ inside $C$. For sufficiently large $j$, this ball contains $(1-t)c+tz$, proving the assertion.

For $\varepsilon>0$, put $z_\varepsilon=x+\varepsilon(x-c)$. This point lies outside $K$: if it lay in $K$, then

would lie in $C$ by the preceding assertion. The nonempty closed set $K$ has a point $q_\varepsilon$ nearest to $z_\varepsilon$. To see existence without a compactness assumption on $K$, take a minimizing sequence. Its distance to $z_\varepsilon$ is bounded, so it has a convergent subsequence in finite dimensions; closedness places its limit in $K$.

Write $v_\varepsilon=z_\varepsilon-q_\varepsilon\ne0$. For $y\in K$, the segment $q_\varepsilon+t(y-q_\varepsilon)$, $0\leq t\leq1$, lies in $K$. Differentiating the squared distance at its minimum $t=0$ gives

Consequently, for $u_\varepsilon=v_\varepsilon/|v_\varepsilon|$,

Choose $\varepsilon=1/k$ and a convergent subsequence of the unit vectors, with limit $u$, $|u|=1$. Since $z_\varepsilon\to x$, the last inequality implies $\langle u,y\rangle\leq\langle u,x\rangle$ for every $y\in K$. The same subsequence works for every $y$: it was selected using only the unit vectors, and the inequality holds for every $y$ at every index.

Set $\eta(y)=\langle u,y\rangle$. It is nonzero. For $y\in C$, openness gives $y+\delta u\in C$ for some $\delta>0$, whence

This proves strict pointwise separation. The supremum statement follows because $\eta(C)$ is nonempty and bounded above. In dimension zero the hypotheses cannot occur, since the only nonempty open set is all of $V$.

5. Support functions of Minkowski sums

Proposition 5.1. For nonempty $A,B\subset V$, put

Then $h_{A+B}(\eta)=h_A(\eta)+h_B(\eta)$, where $A+B=\{a+b:a\in A,b\in B\}$. Nonemptiness excludes $-\infty$, so the right side is unambiguous.

Linearity first gives the inequality $\leq$. If both right-hand suprema are finite, choose $a,b$ within $\varepsilon/2$ of the two suprema. Their sum gives the reverse inequality up to $\varepsilon$; let $\varepsilon\downarrow0$. If $h_A(\eta)=+\infty$, fix $b_0\in B$. The values $\eta(a+b_0)$ are unbounded above, so $h_{A+B}(\eta)=+\infty$. Interchange $A,B$ for the other case. No boundedness, compactness or convexity is needed for this identity.