8.2 The admissible coneF677D35E0BD7

Proof. The polynomial

is a positive multiple of the $(n-m)$th derivative of $\prod _ i(x+\lambda _ i)$. Hence all its roots are real, by Rolle's theorem. Its coefficients are positive, so all its roots are strictly negative. Write $P(x)=\prod _{i=1}^ m(x+r_ i)$ with $r_ i>0$. Logarithmic differentiation at zero gives

Thus $\min _ i r_ i\geq e_ m$, with strict inequality if $m\geq 2$. For $x> -\min _ i r_ i$, every derivative of $P$ is positive: differentiating its product expansion produces sums of products of positive factors. Since

all the required symmetric polynomials are positive at $x=-t$. When $m=1$, the sole root is $-e_1$. $\square$

Proof. Write $\sigma _ k=\alpha ^ k\wedge \omega ^{n-k}/\omega ^ n$, and $M_ k=\alpha ^{k-1}\wedge \gamma \wedge \omega ^{n-k}/\omega ^ n$. For $1\leq k<m$, the cone shift with index $k+1$ and Gårding positivity give

Let $t$ increase to the displayed endpoint. Iterating, starting with $M_1=\gamma \wedge \omega ^{n-1}/\omega ^ n$, yields

All denominators are positive. Approximating a form in the closed cone by $\gamma +\delta \omega $, $\delta \downarrow 0$, gives the last assertion. $\square$

Proof. In the wedge notation of the mixed-trace estimate, differentiation gives $DF_\Lambda (M)=fM_ m/\sigma _ m$. The estimate $M_ m\geq \sigma _ m M_1/(m!\sigma _1)$, with $M_1=\operatorname {tr}M/n$ and $\sigma _1=S/n$, proves the first inequality. The positive semidefinite rank-one coordinate projections belong to the closed $m$-cone; taking $M$ to be each projection proves the second inequality. The closed $m$-cone is contained in the closed $2$-cone. Its eigenvalues $\mu $ satisfy

Thus the operator norm of $M$ is at most its trace, which proves the last inequality. The wedge pairing computes the derivative in the diagonal frame of $\Lambda $, regardless of the eigenvectors of $M$. $\square$

Proof. The derivative of the normalized root is $F_ i=fS_{m-1}(\lambda |i)/(mS_ m)$. The splitting identity gives

For $m=n$, set $S_{n+1}=0$. If $S_{m+1}\leq 0$, the second term is nonnegative. Otherwise $\lambda \in \Gamma _{m+1}$ and Maclaurin gives $S_{m+1}\leq \binom n{m+1}f^{m+1}$. Since $S_ m=\binom nm f^ m$, the subtracted term is at most $(n-m)f^2/m$. This proves the first assertion. Summing the derivatives gives $\mathcal F=f\sigma _{m-1}/\sigma _ m\geq 1$, again by Maclaurin. $\square$

Proof. First take a single form $\alpha $, diagonal in an $\omega $-unitary frame, with ordered eigenvalues $\lambda \in \Gamma _ m$. For $1\leq r<m$, repeated splitting and deleted-coordinate positivity give $S_ r(\lambda )\geq \lambda _1\cdots \lambda _ r$. The sum of the last $n-r$ eigenvalues is positive, so each negative eigenvalue has modulus at most $(n-r)\lambda _{r+1}$. Thus every product of $r$ distinct eigenvalues has modulus at most

Indeed, replace its negative factors by this bound and its positive factors by the corresponding largest eigenvalues. In the diagonal frame only complementary diagonal coefficients of $\Xi $ contribute to $\alpha ^ r\wedge \Xi $. This proves $|\alpha ^ r\wedge \Xi |\leq C_{n,m,\Xi }\alpha ^ r\wedge \omega ^{n-r}$.

Apply this bound to the polynomial $(\alpha _1+t_2\alpha _2+\cdots +t_ r\alpha _ r)^ r\wedge \Xi $, with $0\leq t_ i\leq 1$. Its absolute value is bounded by the stated right-hand side, by polarized positivity and monotonicity on the cone. A fixed invertible interpolation matrix on this finite-dimensional polynomial space bounds each coefficient by its supremum on the cube. The coefficient of $t_2\cdots t_ r$ is a fixed positive multiple of $\alpha _1\wedge \cdots \wedge \alpha _ r\wedge \Xi $. This proves the mixed bound. The closed-cone case follows by adding $\varepsilon \omega $ to each form and taking the limit. $\square$

Proof. First prove the unnormalized Newton inequality for an arbitrary real list $\mu $ of length $N$. For $1\leq q<N$, put $R(x)=\prod _{a=1}^ N(x+\mu _ a)$ and $G=R^{(N-q-1)}$. Repeated Rolle's theorem makes $G$ a real-rooted polynomial of degree $d=q+1$. Away from its roots $r_ a$, logarithmic differentiation and Cauchy–Schwarz give

Continuity extends this polynomial inequality to every real $x$. At zero,

Consequently $S_{q-1}(\mu )S_{q+1}(\mu )\leq c_{N,q}S_ q(\mu )^2$. No sign restriction on $\mu $ or on these coefficients was used. For $q=N$, its left side is zero and the same assertion holds with $c_{N,N}=0$.

Apply this to the list with $i,j$ deleted. Put $A=S_{q-1;ij}$, $B=S_{q;ij}$, $C=S_{q+1;ij}$ and $D_ i=S_{q;i}$, $D_ j=S_{q;j}$. Deleted-coordinate positivity gives $A,D_ i,D_ j\geq 0$: the required degrees are $q+1\leq m$. Splitting in the two omitted coordinates gives the exact identity

Since $S_{q+1}(\lambda )\geq 0$ and $AC\leq c_{N,q}B^2$, this is at least $(1-c_{N,q})B^2$. There is no division by $A$ or $D_ i$; all degenerate endpoints are therefore included. Equivalently one may first add $\epsilon \mathbf1$ and pass to the closed cone. $\square$

Proof. Choose a nonnegative even smooth probability density $\zeta $ supported in $[-1,1]$, and set

The function $r_\epsilon $ is smooth and convex, $|r_\epsilon '|\leq 1$, and $r_\epsilon (t)=|t|$ for $|t|\geq \epsilon $. Jensen's inequality and the triangle inequality give

Its partial derivatives in $x,y$ are nonnegative. For fixed $t$, $r_\epsilon (t)$ is an even convex function of $\epsilon $, since $\zeta $ is even. It is therefore nondecreasing for $\epsilon \geq 0$. Consequently $w_ j=M_{\epsilon _ j}(u_ j,v_ j)$ decreases pointwise to $\max \{ u,v\} $ and has the asserted equalities and bounds.

Put $h_ j=u_ j-v_ j$ and $a_ j=(1+r_{\epsilon _ j}'(h_ j))/2$. The chain rule gives

Here $0\leq a_ j\leq 1$, and the last form is semipositive. Convexity of the cone and its stability under semipositive addition prove admissibility. For smooth local functions satisfying the unshifted cone condition, omit $\omega $ from this chain rule; the same argument proves preservation of that condition. The pointwise maximum is itself admissible: an upper test for it at a point is an upper test for a potential attaining the maximum there. Its canonical representative is thus the indicated pointwise maximum. $\square$

Proof. By 8.2.9, $w=\max \{ u,v\} $ is bounded admissible with its indicated canonical representative. Since $w=v$ almost everywhere, they represent the same distribution. The canonical subharmonic representative used in (8.2.2) is determined by that distribution. Thus $w=v$ pointwise, which is the assertion. $\square$