7.8.5 Several independent factors7E603FDE8D9A

Proof. For $n=1$ both sides equal $A_1$. Let $n\geq 2$, and assume first that all $A_ i$ are positive definite. By lemma 5.3.3(3), $D$ is positive on positive definite matrices.

We first derive the reverse Cauchy–Schwarz inequality. Fix positive definite matrices $A_3,\ldots ,A_ n$, and consider the symmetric bilinear form

on Hermitian matrices. Let $A,B$ be positive definite, put $c=q(A,B)/q(B,B)$ and $M=A-cB$. Then $q(M,B)=0$, so $q(M,M)\leq 0$ by lemma 5.3.4. Since $q(M,M)=q(A,A)-q(A,B)^2/q(B,B)$, we obtain

Put $\widetilde A_ i=(\det A_ i)^{-1/n}A_ i$, so that $\det \widetilde A_ i=1$. By multilinearity it suffices to prove that $D(\widetilde A_1,\ldots ,\widetilde A_ n)\geq 1$. Consider the finitely many ordered $n$-tuples whose entries belong to the set $\{ \widetilde A_1,\ldots ,\widetilde A_ n\} $, and let $F>0$ be the smallest mixed discriminant of such a tuple. A tuple whose entries all equal $\widetilde A_ i$ has discriminant $\det \widetilde A_ i=1$, so $F\leq 1$. The multiplicity of a matrix in a tuple is the number of entries equal to it. Among the tuples with discriminant $F$, choose one for which the sum of the squares of the multiplicities is largest.

Suppose that this tuple contains two different matrices $A$ and $B$. By symmetry of $D$ we may place them in the first two slots, so that $F=q(A,B)$, where $q$ is formed with the remaining entries of the tuple. Replacing the pair $(A,B)$ by $(A,A)$ or by $(B,B)$ gives tuples with discriminants $q(A,A)\geq F$ and $q(B,B)\geq F$, whereas (7.8.7) gives $q(A,A)q(B,B)\leq F^2$. Hence $q(A,A)=q(B,B)=F$, and both new tuples are minimizing. If $A$ and $B$ have multiplicities $r$ and $s$, the two replacements change the sum of the squared multiplicities by $2(r-s)+2$ and $2(s-r)+2$, respectively. These two numbers add up to $4$, so one of them is positive, which contradicts the choice of the tuple. Therefore all entries of the chosen tuple are equal, and $F=1$. In particular $D(\widetilde A_1,\ldots ,\widetilde A_ n)\geq 1$.

Finally, for positive semidefinite $A_ i$, apply the inequality to $A_ i+\varepsilon I$ and let $\varepsilon \downarrow 0$. Both sides are continuous: $D$ is a polynomial in the matrix entries, and $(\det A)^{1/n}$ is continuous on positive semidefinite matrices. $\square$

Proof. For $n=1$ this is the hypothesis, so let $n\geq 2$. Put $u_{i,\varepsilon }=u_ i*\rho _\varepsilon $, with the same radial mollifiers for all $i$. By the last assertion of proposition 7.8.3, applied to each $u_ i$ with the density $f_ i$,

By (7.2.4), applied to the smooth function $\sum _ it_ iu_{i,\varepsilon }$, the exterior powers of smooth forms satisfy

Both sides are polynomials in $t_1,\ldots ,t_ n$. Since $2$-forms commute, the coefficient of $t_1\cdots t_ n$ on the left is $n!\, \mathrm{dd}^{\mathrm{c}}u_{1,\varepsilon }\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}u_{n,\varepsilon }$, and on the right it is $n!\, D(\mathcal H(u_{1,\varepsilon }),\ldots ,\mathcal H(u_{n,\varepsilon }))\eta ^ n$. Since the Bedford–Taylor product of smooth functions is the exterior product of the forms $\mathrm{dd}^{\mathrm{c}}u_{i,\varepsilon }$ (proposition 7.2.2(1)), we obtain

where the inequality is lemma 7.8.6, applied to the complex Hessians of the smooth plurisubharmonic functions $u_{i,\varepsilon }$, which are positive semidefinite (proposition 5.9.2). As $\varepsilon \downarrow 0$, the density on the right converges locally uniformly to $\prod _ if_ i^{1/n}$, and the left side converges weakly to $\mathrm{dd}^{\mathrm{c}}u_1\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}u_ n$ by theorem 7.3.7, since $u_{i,\varepsilon }\downarrow u_ i$ for each $i$ (proposition 5.9.2). $\square$