7.8 Density lower bounds and mixed products2044B813A798

Proof. Let $B$ be positive definite with $\det B=1$. The positive semidefinite matrix $B^{1/2}AB^{1/2}$ has trace $\operatorname {tr}(BA)$ and determinant $\det A$, so the arithmetic–geometric mean inequality for its eigenvalues gives $\operatorname {tr}(BA)/n\ge (\det A)^{1/n}$. If $A$ is positive definite, equality holds for $B=(\det A)^{1/n}A^{-1}$. If $A$ is singular, the matrices $B_\varepsilon =\det (A+\varepsilon I)^{1/n}(A+\varepsilon I)^{-1}$, $\varepsilon >0$, are admissible and

as $\varepsilon \downarrow 0$.

The formulas for $M_ n$ and $M_0$ follow from $\det (tA)=t^ n\det A$ and $\det (sC)=s^ n\det C$. Suppose first that $A$ and $C$ are positive definite, and let $\lambda _1,\ldots ,\lambda _ n>0$ be the eigenvalues of $A^{-1/2}CA^{-1/2}$. Since $\det (tA+sC)=\det A\prod _{i=1}^ n(t+s\lambda _ i)$, comparison of coefficients gives

the sum being over the subsets $J\subset \{ 1,\ldots ,n\} $ with $n-k$ elements. Each eigenvalue occurs in exactly $\binom {n-1}{n-k-1}$ of these $\binom n{n-k}$ products, so the geometric mean of the products is $(\lambda _1\cdots \lambda _ n)^{(n-k)/n}$. Since $\lambda _1\cdots \lambda _ n=\det C/\det A$, the arithmetic–geometric mean inequality gives

For semidefinite $A$ and $C$, apply this to $A+\varepsilon I$ and $C+\varepsilon I$ and let $\varepsilon \downarrow 0$; both sides depend continuously on the matrices. $\square$

Proof. We reduce to the Laplacian, treat the case $F=0$ by a maximum principle argument, and reduce the general case to it.

Step 1. Reduction to $L=\Delta $. Let $C=(c_{kl})$, let $P$ be the positive definite square root of $C^{-1}$, so that $PCP=I$, and put $\tilde U=P(U)$, $\tilde w=w\circ P^{-1}$ and $\tilde F=F\circ P^{-1}$. If $q$ is a $C^2$ function near $x\in U$ and $\tilde q=q\circ P^{-1}$, then $\partial _ k\partial _ lq(x)=\sum _{a,b}P_{ak}P_{bl}(\partial _ a\partial _ b\tilde q)(Px)$ by the chain rule, and hence

The map $q\mapsto \tilde q$ is a bijection from the upper test functions for $w$ at $x$ onto those for $\tilde w$ at $Px$, so $\tilde w$, $\Delta $ and $\tilde F$ satisfy the hypothesis of the lemma on $\tilde U$. Suppose that the lemma holds for them. Then $w=\tilde w\circ P$ is locally integrable, and for every nonnegative test function $\chi $ on $U$ the change of variables $y=Px$, with $\tilde\chi =\chi \circ P^{-1}$ and $L\chi =(\Delta \tilde\chi )\circ P$, gives

the first equality is the definition of distributional derivatives. We may therefore assume that $L=\Delta $.

Step 2. The case $F=0$. We show that $w$ satisfies assertion (1) of theorem 3.4.5 for every ball $D=B(y,r)$ with $\overline{B(y,r)}\subset U$; by the converse part of that theorem, $w$ is then subharmonic on $U$. Let $h$ be a real function which is harmonic on $B(y,r)$ and continuous on $\overline{B(y,r)}$, with $\varlimsup _{B(y,r)\ni x\to \zeta }w(x)\le h(\zeta )$ for every $\zeta \in \partial B(y,r)$, and suppose that $w(x_1)>h(x_1)$ for some $x_1\in B(y,r)$. Choose $\varepsilon >0$ so small that the upper semicontinuous function $g(x)=w(x)-h(x)-\varepsilon (r^2-|x-y|^2)$ on $B(y,r)$ satisfies $g(x_1)>0$.

Since $h$ is continuous on the closed ball, $\varlimsup _{B(y,r)\ni x\to \zeta }g(x)\le 0$ for every $\zeta \in \partial B(y,r)$. Hence every $\zeta \in \partial B(y,r)$ has an open neighborhood $N_\zeta $ with $g<g(x_1)/2$ on $N_\zeta \cap B(y,r)$; finitely many of these neighborhoods cover the sphere, and the complement $K$ of their union in $\overline{B(y,r)}$ is a compact subset of $B(y,r)$ containing $x_1$. The upper semicontinuous function $g$ attains its maximum over $K$ at a point $x_0$, and $c=g(x_0)\ge g(x_1)>0$; since $g<g(x_1)/2$ outside $K$, we have $g\le c$ on $B(y,r)$. The function $q=h+\varepsilon (r^2-|x-y|^2)+c$ is smooth on $B(y,r)$, satisfies $q-w=c-g\ge 0$ there and $q(x_0)=w(x_0)$, so it is an upper test function for $w$ at $x_0$. But $\Delta q=-2d\varepsilon <0$, which contradicts the hypothesis. Hence $w\le h$ on $B(y,r)$, and $w$ is subharmonic on $U$. Being real-valued, it is not identically $-\infty $ on any connected component of $U$, so theorem 3.4.12(1) shows that $w$ is locally integrable and $\Delta w\ge 0$.

Step 3. The general case. Let $V\Subset U$ be open. Every compact subset of $U$ lies in such a set $V$, so it suffices to show that $w$ is locally integrable on $V$ and that $\Delta w\ge F$ on $V$. Choose an open set $V'$ with $\overline V\subset V'\Subset U$. By lemma 2.5.1(1), the mollifications of $F$ converge to $F$ uniformly on $\overline{V'}$; subtracting from each of them its maximal deviation from $F$ on $\overline{V'}$, we obtain smooth functions $F_ j$ on a neighborhood of $\overline{V'}$ such that $F_ j\le F$ on $\overline{V'}$ and $F_ j\to F$ uniformly there. Let $\vartheta \in C^\infty _ c(V')$ be equal to one on an open set $W\supset \overline V$ (lemma 2.5.1(4)). By proposition 3.1.13, the convolution $h_ j=E_ d*(\vartheta F_ j)$, where $\vartheta F_ j$ is extended by zero, is a smooth function on $\mathbb R^ d$ with $\Delta h_ j=\vartheta F_ j$; in particular $\Delta h_ j=F_ j$ on $W$.

If $q$ is an upper test function for $w-h_ j$ at $x\in V$, then $q+h_ j$ is an upper test function for $w$ at $x$, so that $\Delta q(x)+F_ j(x)\ge F(x)$, that is, $\Delta q(x)\ge F(x)-F_ j(x)\ge 0$. Step 2, applied to the real upper semicontinuous function $w-h_ j$ on $V$, shows that $w-h_ j$ is locally integrable on $V$ and $\Delta (w-h_ j)\ge 0$ there. Hence $w$ is locally integrable on $V$ and $\Delta w\ge F_ j$ on $V$. Testing against a nonnegative test function on $V$ and letting $j\to \infty $, we obtain $\Delta w\ge F$ on $V$. $\square$

Proof. We first observe that the complex Hessian $\mathcal H(q)(x)$ of an upper test function $q$ for $u$ at $x$ is positive semidefinite. Indeed, let $\xi \in \mathbb C^ n\setminus \{ 0\} $, and let $r>0$ be so small that the closed disc $\{ x+\zeta \xi :|\zeta |\le r\} $ lies in the neighborhood of $x$ on which $q\ge u$. The function $\zeta \mapsto u(x+\zeta \xi )$ is subharmonic near this disc and finite at $\zeta =0$, so theorem 3.4.5(2) gives

Write $q_ j=\partial _{z_ j}q$ and $q_{jk}=\partial _{z_ j}\partial _{z_ k}q$. Since $q$ is real and of class $C^2$, Taylor's formula gives

For $h=r\mathrm e^{\mathrm i\theta }\xi $ the terms containing $\mathrm e^{\mathrm i\theta }$ and $\mathrm e^{2\mathrm i\theta }$ have mean zero over $\theta \in [0,2\pi ]$, so the right-hand side of the preceding display equals $q(x)+r^2\sum _{j,k}q_{j\bar k}(x)\xi _ j\bar\xi _ k+o(r^2)$. Dividing by $r^2$ and letting $r\downarrow 0$, we obtain $\sum _{j,k}q_{j\bar k}(x)\xi _ j\bar\xi _ k\ge 0$.

(1) implies (2). Assume (1), let $q$ be an upper test function for $u$ at $x$, and suppose that $\det \mathcal H(q)(x)<f(x)$. Since $\mathcal H(q)(x)$ is positive semidefinite, we can choose $\delta >0$ so small that $q_\delta =q+\delta |z-x|^2$ satisfies $\mathcal H(q_\delta )(x)=\mathcal H(q)(x)+\delta I>0$ and still $\det \mathcal H(q_\delta )(x)<f(x)$. Since $q$ is of class $C^2$ and $f$ is continuous, there is $r>0$ such that $\overline{B(x,r)}\subset \Omega $, $q\ge u$ on $\overline{B(x,r)}$, the complex Hessian $\mathcal H(q_\delta )$ is positive definite on $\overline{B(x,r)}$, and $\det \mathcal H(q_\delta )<f$ on $\overline{B(x,r)}$; in particular, $q_\delta $ is plurisubharmonic on $B(x,r)$ by proposition 5.9.2. Put $b=q_\delta -\delta r^2/2$. Then $b=q+\delta r^2/2>u$ on the sphere $\partial B(x,r)$, while $b(x)=u(x)-\delta r^2/2<u(x)$.

The Borel set $S=\{ z\in B(x,r):b(z)<u(z)\} $ has positive Lebesgue measure. Otherwise $\max \{ u,b\} =b$ almost everywhere on $B(x,r)$; since both functions are plurisubharmonic, the uniqueness assertion of theorem 5.9.1 gives $\max \{ u,b\} =b$ everywhere, contradicting $b(x)<u(x)$. The function $b-u$ is lower semicontinuous and positive on $\partial B(x,r)$, so $\varliminf _{z\to \zeta }(b(z)-u(z))>0$ for every $\zeta \in \partial B(x,r)$. The comparison principle theorem 7.3.11, applied on $B(x,r)$ with the bounded plurisubharmonic functions $b$ and $u$ in place of $u$ and $v$, therefore gives

Here $(\mathrm{dd}^{\mathrm{c}}b)^ n=\det \mathcal H(q_\delta )\, \eta ^ n$ on $B(x,r)$ because $b$ is a plurisubharmonic $C^2$ function there (see the beginning of this section), and the integrals are finite because $(\mathrm{dd}^{\mathrm{c}}u)^ n$ is a Radon measure on $\Omega $ and $\overline{B(x,r)}$ is a compact subset of $\Omega $. Since $\det \mathcal H(q_\delta )<f$ on $S$ and $S$ has positive measure, the right side is strictly smaller than the left side, a contradiction. Thus (2) holds.

(2) implies (3). Let $q$ be an upper test function for $u$ at $x$, and let $B$ be positive definite with $\det B=1$. Since $\mathcal H(q)(x)$ is positive semidefinite, the infimum formula of lemma 7.8.1 and (2) give

Since $u$ is real-valued and upper semicontinuous, lemma 7.8.2, applied to $L_ B$ and the continuous function $nf^{1/n}$, gives (3).

(3) implies (1). The mollifications $u_\varepsilon =u*\rho _\varepsilon $ of section 2.5 are smooth plurisubharmonic functions on $\Omega _\varepsilon $ which decrease to $u$ as $\varepsilon \downarrow 0$ (proposition 5.9.2). Fix $B$ as in (3). The distribution $L_ Bu-nf^{1/n}$ is positive, so its mollification is a nonnegative function on $\Omega _\varepsilon $ (proposition 2.5.3(3)). Since $L_ B$ has constant coefficients, proposition 2.5.3(1) identifies the mollification of $L_ Bu$ with $L_ Bu_\varepsilon $, and we obtain the pointwise inequality

Since $\mathcal H(u_\varepsilon )$ is positive semidefinite (proposition 5.9.2), taking the infimum over $B$ and using lemma 7.8.1, we obtain $\det \mathcal H(u_\varepsilon )\ge (f^{1/n}*\rho _\varepsilon )^ n$, which is the last assertion of the proposition. By the formula recalled at the beginning of this section, it reads

As $\varepsilon \downarrow 0$, the density on the right converges locally uniformly to $f$ by lemma 2.5.1(1), and $(\mathrm{dd}^{\mathrm{c}}u_\varepsilon )^ n\to (\mathrm{dd}^{\mathrm{c}}u)^ n$ weakly by theorem 7.3.7, applied along any sequence $\varepsilon _ j\downarrow 0$. Testing against nonnegative functions in $C_ c(\Omega )$ gives (1). $\square$

Proof. Let $u_\varepsilon =u*\rho _\varepsilon $ and $v_\varepsilon =v*\rho _\varepsilon $, with the same mollifiers for both functions. By the last assertion of proposition 7.8.3, applied to $u$ with the density $f$ and to $v$ with the density $g$,

For real $t$ and $s$, (7.2.4), applied to the smooth function $tu_\varepsilon +sv_\varepsilon $, gives $(t\mathrm{dd}^{\mathrm{c}}u_\varepsilon +s\mathrm{dd}^{\mathrm{c}}v_\varepsilon )^ n=\det \bigl (t\mathcal H(u_\varepsilon )+s\mathcal H(v_\varepsilon )\bigr )\eta ^ n$ for the exterior powers of smooth forms. Since $2$-forms commute, the left side can be expanded by the binomial formula. Comparing the coefficients of $\binom nkt^ ks^{n-k}$, and recalling that Bedford–Taylor products of smooth functions are exterior products of smooth forms (proposition 7.2.2(1)), we find

where the inequality is the second assertion of lemma 7.8.1, applied to the matrices $\mathcal H(u_\varepsilon )$ and $\mathcal H(v_\varepsilon )$, which are positive semidefinite by proposition 5.9.2. As $\varepsilon \downarrow 0$, the density on the right converges locally uniformly to $f^{k/n}g^{(n-k)/n}$, while the left side converges weakly to $(\mathrm{dd}^{\mathrm{c}}u)^ k\wedge (\mathrm{dd}^{\mathrm{c}}v)^{n-k}$ by theorem 7.3.7, since $u_\varepsilon $ and $v_\varepsilon $ are smooth plurisubharmonic functions decreasing to $u$ and $v$ (proposition 5.9.2). $\square$

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). $\square$