18.34 The determinant functional3AC0CF30AE00

Proof. Choose a basis $s_1,\ldots ,s_{N_ m}$ that is orthonormal for the weight $v$, so that $G_ m(v)=I$ is the identity matrix of size $N_ m$; the indices $i,j$ below range from $1$ to $N_ m$. Since the weights are bounded, we may differentiate under the integral sign. With pointwise inner products taken in $h^ m$, put

The first and second derivatives of $t\mapsto G_ m(v+tf)$ at $t=0$ are $-mA$ and $m^2C$. The identity $(\log \det G)'=\operatorname {tr}(G^{-1}G')$ gives the first derivative of $\mathcal F_ m^{(r)}$, and differentiating it once more yields

Let $P_ v$ be the orthogonal projection of the weighted $L^2$ space of measurable sections of $L^ m$ onto $V_ m$. Since $\int _ Xf^2B_{m,v,\varphi ,\nu }\, \mathrm{d}\nu =\sum _ j\| fs_ j\| _{m,v,\nu }^2$ and $P_ v(fs_ j)=\sum _ iA_{ji}s_ i$, Pythagoras' theorem shows that the quantity in parentheses equals

This proves concavity. Multiplying every Gram entry by $\mathrm{e}^{-mc}$ shows that $\det G_ m(v+c)=\mathrm{e}^{-mcN_ m}\det G_ m(v)$, which is the translation rule. By the first-variation formula and the total mass identity, the derivative of $\mathcal F_ m^{(r)}$ along the segment from $w$ to $v$ is bounded by $(r!N_ m/m^ r)\| v-w\| _\infty $, which gives the Lipschitz estimate.

Suppose finally that the second derivative vanishes. Then $fs_ j\in V_ m$ in $L^2$ for every $j$, so there are $0\ne s\in V_ m$ and $\widetilde s\in V_ m$ with $fs=\widetilde s$ almost everywhere. Both sides are continuous and $\nu $ has a positive density, so $fs=\widetilde s$ everywhere. On a ball where $s\ne 0$, the quotient $\widetilde s/s=f$ is a real-valued holomorphic function, hence a constant $c$. The identity theorem gives $\widetilde s=cs$ on the connected manifold $X$, so $f=c$ wherever $s\neq 0$, and by continuity everywhere. Conversely, for constant $f$ the translation rule shows that the second derivative vanishes. $\square$

Proof. The measures $\beta _ m^{(n)}$ are positive with total masses $n!N_ m/m^ n\to 0$. The second assertion follows from the Lipschitz estimate, since $\mathcal F_ m^{(n)}(0)=0$. $\square$

Proof. If $V_ m=\{ 0\} $ there is nothing to prove. Fix a real weight $v_0$. For $f\in C(X,\mathbb C)$ define an operator on $V_ m$ by

Since the exponential series converges in the uniform norm, $T$ is an entire function of $f$ with values in the endomorphisms of $V_ m$. Since $P_{v_0}$ is a contraction,

Hence, for $m\| f\| _\infty <\log (3/2)$, we have $\| T(f)-I\| <1/2$, so the Neumann series for $T(f)^{-1}$ converges in operator norm and $\| T(f)^{-1}\| \leq 2$. For real $f$, the operator $T(f)$ is selfadjoint and represents the new inner product:

In particular it is positive definite, bounded below by $\mathrm{e}^{-m\| f\| _\infty }I$. The Riesz identity therefore gives

For complex $f$ the right-hand side is still a holomorphic function of $f$. Consequently

is a holomorphic extension of $f\mapsto B_{v_0+f}(x)$. Evaluations on the finite-dimensional space $V_ m$ are uniformly bounded over the compact manifold $X$, so the linear map $A\mapsto [x\mapsto \ell _ x(Ak_{v_0,x})]$ from the endomorphisms of $V_ m$ to $C(X)$ is bounded; it is well defined because the phases of $\ell _ x$ and $k_{v_0,x}$ cancel. Hence the extension is holomorphic as a map with values in $C(X)$. Finally, $B\mapsto (r!/m^ r)B\nu $ is a bounded linear map into finite measures with the total variation norm, which proves the assertion about $\beta _ v$. At points of $Z_ m$ the Riesz vector is zero, so the argument includes the common zero set. $\square$

Proof. The derivative formula. By proposition 18.34.5, $k_{v_ t,x}$ depends smoothly on $t$, for any fixed norm on $V_ m$. Fix a real parameter $t$ and set

Then $H(0)=I$, $H^{(j)}(0)=-\mathcal T_{t,j}$ by Faà di Bruno's formula, and $k_{v_{t+a},x}=H(a)^{-1}k_{v_ t,x}$ as in the proof of proposition 18.34.5. Since $H(a)-I=O(|a|)$, truncating the Neumann series of $H(a)^{-1}$ after $k$ powers leaves a remainder $O(|a|^{k+1})$. Expanding the truncated series to order $k$ in $a$ gives (18.34.7), with its ordered products and coefficients $k!/\alpha !$. Only a finite Taylor jet of the smooth curve is used.

Gevrey bounds. For nonnegative integers $n_\ell $ with $\sum \ell n_\ell =j$,

The sum $\sum \prod _\ell (n_\ell !)^{-1}$ over these tuples is the coefficient of $z^ j$ in $\exp (z/(1-z))$, and evaluation at $z=1/2$ shows that it is at most $\mathrm{e}\, 2^ j$. Together with the assumed bounds this gives

for some $L_0>0$. Since $P_{v_ t}$ is a contraction, each ordered product in (18.34.7) has operator norm at most $L_0^ k(\alpha !)^ s$. There are $2^{k-1}$ compositions of $k$, and $\alpha !\leq k!$. Since $v_ t$ is uniformly bounded for $t\in J$, the norms $\| \cdot \| _{m,v_ t,\nu }$ are uniformly equivalent, and the evaluations and Riesz vectors are bounded uniformly on $J\times X$. Hence the $t$-derivatives of the unweighted diagonal $x\mapsto \ell _ x(k_{v_ t,x})$ are bounded in $C(X)$ by $C(2L_0)^ k(k!)^ s$. The factor $\mathrm{e}^{-mv_ t}$ satisfies bounds of the same Gevrey order, by the estimate for $\mathcal B_ j(-mv_ t)$. Leibniz's formula and $j!(k-j)!\leq k!$ then give (18.34.8) for $B_{v_ t}$, and the bounded linear map $B\mapsto (r!/m^ r)B\nu $ gives the bound in total variation. $\square$

Proof. Since $\mathcal T_{t,1}u=mP_{v_ t}(qu)$, the case $k=1$ of (18.34.7) gives $\partial _ tk_{v_ t,x}=mP_{v_ t}(qk_{v_ t,x})$. Applying the fixed evaluation $\ell _ x$ and the Riesz identity gives

Dividing by the positive number $\ell _ x(k_{v_ t,x})=\| k_{v_ t,x}\| _{m,v_ t,\nu }^2$ and differentiating the factor $\mathrm{e}^{-mv_ t(x)}$ in (18.34.4) proves (18.34.11); the density in (18.34.10) has integral one by the definition of the norm. At points of $Z_ m$ the evaluations and Riesz vectors vanish for every weight. Finally, differentiating the identity $\int _ XB_{v_ t}\, \mathrm{d}\nu =N_ m$ in $C(X)$ proves the last assertion. For constant $q$ the right-hand side of (18.34.11) vanishes, in accordance with the invariance of the kernel under adding a constant to the weight. $\square$

Proof. The density $\rho _ t$ is radial and lies between one and two. By radial symmetry, the constant functions are orthogonal to the holomorphic functions vanishing at the origin, so the reproducing value at the center is the reciprocal of $\int _ D\rho _ t\, \mathrm{d}A$. Polar coordinates and the substitution $a=1-r^2$ give, for $t\ne 0$,

For $t=0$ the integral is $\pi $. This proves (18.34.13) and the expansion. For every $t\ne 0$ we have $\| \log \rho _ t\| _{L^\infty (D)}=\log 2$, whereas $\rho _0=1$. Thus the curve $t\mapsto \log \rho _ t$ is not continuous at $t=0$ for the uniform norm, and the example does not contradict proposition 18.34.5. $\square$

Proof. Orthonormality gives $\int _ X|C_ v(x,y)|^2\, \mathrm{d}\nu (y)=B_ v(x)$, so the first marginal of $J_ v$ is $\mu _ v$. By symmetry so is the second, and the total mass identity shows that $J_ v$ has mass one. Off $Z_ m$, dividing $|C_ v(x,\cdot )|^2\, \mathrm{d}\nu $ by $B_ v(x)$ gives the squared Riesz-vector density in (18.34.10). Thus the probabilities $\sigma _{v,x}$ are the conditional probabilities of $J_ v$ given the first coordinate, and $\mathcal Q_ v$ is the associated Markov operator. Since these conditional probabilities have mass one and the marginal of $J_ v$ is $\mu _ v$, the operator $\mathcal Q_ v$ fixes constants and preserves integrals; Jensen's inequality shows that it is a contraction of $L^2(\mu _ v)$, and the symmetry of $J_ v$ shows that it is selfadjoint. For $f\in L^2(\mu _ v)$, expanding the finite sum gives

Each of these integrals is independent of the phases, and it exists because $|F_ i\overline{F_ j}|\leq B_ v$ and $f\in L^1(\mu _ v)$. This proves positivity. $\square$

Proof. The formula for $D\mu _ v[q]$ follows from corollary 18.34.9, since $N_ m$ does not depend on the weight. For the bound, put $g=q-\mathcal Q_ vq$, $A=\{ g>0\} $ and $p=\mu _ v(A)$. Since $\mathcal Q_ v$ preserves integrals, $g$ has mean zero, and selfadjointness gives

The function $(I-\mathcal Q_ v)\mathbf1_ A$ is nonnegative on $A$ and nonpositive on its complement, almost everywhere, and its integral is zero. Subtracting $\inf _ Xq$ from $q$ does not change the pairing, which is therefore at most

Positivity of $\mathcal Q_ v$, applied to $\mathbf1_ A-p$, gives $\langle \mathbf1_ A,\mathcal Q_ v\mathbf1_ A\rangle \geq p^2$: since $\mathcal Q_ v$ fixes constants and preserves integrals, $\langle \mathbf1_ A-p,\mathcal Q_ v(\mathbf1_ A-p)\rangle =\langle \mathbf1_ A,\mathcal Q_ v\mathbf1_ A\rangle -p^2$. As $g$ has mean zero, it follows that

The total variation norm of a signed measure is the integral of the absolute value of its density, so $\| D\mu _ v[q]\| _{\mathrm{TV}}=m\| g\| _{L^1(\mu _ v)}$, and (18.34.19) follows.

For the global estimate, take $q=w-v$, $v_ t=v+tq$, $\mu _ t=\mu _{v_ t}$ and $\mathcal Q_ t=\mathcal Q_{v_ t}$ for $0\leq t\leq 1$, and put $a=m\operatorname {osc}_ Xq$. For a Borel set $E$, the function $p_ t=\mu _ t(E)$ is differentiable, because integration against $\mathbf1_ E$ is a bounded linear functional for the total variation norm, and selfadjointness gives

The function $(I-\mathcal Q_ t)\mathbf1_ E$ has positive and negative parts whose integrals both equal $p_ t-\langle \mathbf1_ E,\mathcal Q_ t\mathbf1_ E\rangle $. Bounding $q$ above and below on the supports of these parts, and applying positivity to $\mathbf1_ E-p_ t$ as before, we obtain

All the kernels $B_{v_ t}$ are positive exactly off the common zero set $Z_ m$, so the probabilities $\mu _ t$ are mutually absolutely continuous. Hence, if $p_0$ is zero or one, then $p_ t=p_0$ for all $t$. Otherwise $0<p_ t<1$ for all $t$, and integration gives

Let $\sigma (z)=\mathrm{e}^ z/(1+\mathrm{e}^ z)$ be the logistic function. If $p_1\geq p_0$ and $z=\log (p_0/(1-p_0))$, then

The case $p_1\leq p_0$ is symmetric. Since $\mu _ w-\mu _ v$ has total mass zero, its total variation norm is $2\sup _ E|\mu _ w(E)-\mu _ v(E)|$, which proves (18.34.20). Under the hypothesis of the last assertion the right-hand side tends to zero, which gives the final statement. $\square$

Proof. The selected space consists of the constant functions, so the normalized kernel probability at a weight $u$ is $\mathrm{e}^{-mu}\nu /\int _ X\mathrm{e}^{-mu}\, \mathrm{d}\nu $. Choose a spherical cap $E$ with $\nu (E)=p=(1+\mathrm{e}^{a/2})^{-1}$; its boundary circle has zero area. Choose smooth functions $0\leq \chi _ j\leq 1$, equal to one on an inner cap and zero outside an outer cap, converging almost everywhere to $\mathbf1_ E$; each has oscillation one. Put $w_ j=-a\chi _ j/m$. The densities $\mathrm{e}^{a\chi _ j}$ converge boundedly almost everywhere, and their normalizing constants converge to a positive limit, so $\mu _{w_ j}$ converges in total variation to

This limit gives $E$ the mass $1-p$, and its density with respect to $\nu $ is greater than one on $E$ and less than one outside $E$. Its total variation distance from $\nu =\mu _0$ is therefore $2(1-2p)=2\tanh (a/4)$. The triangle inequality transfers this limit to the measures $\mu _{w_ j}$. This shows that the bound is optimal as a supremum; we do not claim that it is attained by a smooth weight. $\square$

Proof. On $V_ m$ use $T(f)=P_ v(\mathrm{e}^{-f}\, \cdot \, )$, the operator in the proof of proposition 18.34.5 for the perturbation $f/m$ of the weight $v$. That proof gives $\| T(f)^{-1}\| \leq 2$ on the ball. The extended measure has the density

with respect to $\nu $. By the Riesz identity at the weight $v$, its absolute value is at most $2\mathrm{e}^{\| f\| _\infty }B_ v(x)/N_ m\leq 3B_ v(x)/N_ m$, so its total variation is at most three. Holomorphy follows from the same finite-dimensional construction and the bounded linear passage from densities to measures. For a complex constant $c$ we have $T(f+c)=\mathrm{e}^{-c}T(f)$, and the resulting factor $\mathrm{e}^{c}$ cancels the factor $\mathrm{e}^{-c}$ in the density; thus adding constants to $f$ does not change the extension. For nonconstant real $q$, replace $q$ by $q-c$ with $c=(\sup _ Xq+\inf _ Xq)/2$, so that $\| q-c\| _\infty =\operatorname {osc}_ Xq/2$. Then $t\mapsto \mu _{v+tq}$ extends holomorphically to the disk $|t|<\rho /(m\| q-c\| _\infty )$, with total variation at most three. The Cauchy estimates for Banach-valued holomorphic functions, applied on circles of radius less than $\rho /(m\| q-c\| _\infty )$, followed by letting the radius increase to that value, prove (18.34.23). If $q$ is constant, the measure does not depend on $t$, and all derivatives vanish. $\square$

Proof. Every section descends to $\mathbb P^1$, because the fibers $\{ b\} \times Y$ are compact and connected, as in proposition 18.36.1. Hence the selected space is $H^0(\mathbb P^1,\mathcal O(m))$, of dimension $m+1$. Its unweighted Bergman kernel for $\nu _{\mathrm{FS}}$ is constant, since the unitary group acts transitively on $\mathbb P^1$ and preserves the metric and the volume; its integral $m+1$ determines the constant. For a section pulled back from $\mathbb P^1$, the norm on $X$ at the weight $v_ m$ is $Z_ q$ times the norm on $\mathbb P^1$, which proves (18.34.25). At the weight zero the normalized kernel probability is $\nu _{\mathrm{FS}}\otimes \nu _ Y$. These probabilities differ: equality would force $\mathrm{e}^{-q}/Z_ q=1$ almost everywhere, and then everywhere, because $q$ is continuous and every nonempty open set has positive volume; this contradicts the choice of $q$. The kernel measures with $r=1$ are $(m+1)/m$ times these probabilities, and the normalization in dimension $1+\dim Y\geq 2$ gives measures tending to zero in total variation. Finally, $m^{-1}\log B_{m,v_ m}=m^{-1}\bigl (\log (m+1)-q-\log Z_ q\bigr )$ and $m^{-1}\log B_{m,0}=m^{-1}\log (m+1)$ both tend uniformly to zero. $\square$