PomeProposition 18.34.5.81FC6A666EC9 The map $v\mapsto B_ v$ from $C(X,\mathbb R)$ to $C(X,\mathbb R)$ is real analytic. The map $v\mapsto \beta _ v$ is real analytic with values in the finite signed measures on $X$, equipped with the total variation norm $\| \cdot \| _{\mathrm{TV}}$. If $V_ m=\{ 0\} $, both maps vanish identically.
18.34.3 Dependence on the continuous weight6D1351EF2D7C
When only the continuous weight changes, the selected space stays the same, and the two norms are related by a multiplication operator compressed to that space. Inverting this operator expresses the new reproducing vectors through the old ones. As a consequence the kernel itself, and not only the determinant functional, depends analytically on the weight.
Keep $m\geq 1$, the selecting potential $\varphi $, the metric $h$ and the probability measure $\nu $ fixed as in section 18.30, and write $B_ v=B_{m,v,\varphi ,\nu }$ and $\beta _ v=\beta _ m^{(r)}$ for the kernel and the kernel measure at the weight $v$. Let $P_ v$ be the orthogonal projection of the weighted $L^2$ space of measurable sections of $L^ m$, with the norm $\| \cdot \| _{m,v,\nu }$, onto $V_ m$, and let
\[ Z_ m=\{ x\in X:s(x)=0\text{ for every }s\in V_ m\} \]
be the common zero set of the selected sections. At each $x\in X$ choose a covector of unit length on the fiber $(L^ m)_ x$; it defines a complex-linear evaluation $\ell _ x:V_ m\to \mathbb C$. Let $k_{v,x}\in V_ m$ be its Riesz vector, so that $\ell _ x(s)=\langle s,k_{v,x}\rangle _{m,v,\nu }$ for all $s\in V_ m$. Then
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18.34.4
\begin{equation} B_ v(x)=\mathrm{e}^{-mv(x)}\ell _ x(k_{v,x}) =\mathrm{e}^{-mv(x)}\| k_{v,x}\| _{m,v,\nu }^2. \end{equation}
These values do not depend on the phase of the chosen covector.
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
\[ T(f)s=P_{v_0}(\mathrm{e}^{-mf}s). \]
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,
\[ \| T(f)-I\| \leq \mathrm{e}^{m\| f\| _\infty }-1. \]
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:
\[ \langle s,u\rangle _{m,v_0+f,\nu } =\langle T(f)s,u\rangle _{m,v_0,\nu }. \]
In particular it is positive definite, bounded below by $\mathrm{e}^{-m\| f\| _\infty }I$. The Riesz identity therefore gives
\[ k_{v_0+f,x}=T(f)^{-1}k_{v_0,x}. \]
For complex $f$ the right-hand side is still a holomorphic function of $f$. Consequently
\[ f\longmapsto \mathrm{e}^{-m(v_0(x)+f(x))} \ell _ x\bigl (T(f)^{-1}k_{v_0,x}\bigr ) \]
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. $\square$
Now let $I\subset \mathbb R$ be an open interval and let $t\mapsto v_ t$ be a smooth curve in $C(X,\mathbb R)$. The complete exponential Bell polynomials are
\[ \mathcal B_ j(x_1,\ldots ,x_ j) =\sum _{\sum _{\ell =1}^ j\ell n_\ell =j} j!\prod _{\ell =1}^ j \frac{(x_\ell /\ell !)^{n_\ell }}{n_\ell !}, \qquad j\geq 1, \]
where the sum runs over nonnegative integers $n_1,\ldots ,n_ j$. Write $\mathcal B_ j(-mv_ t)$ for $\mathcal B_ j$ evaluated at the first $j$ derivatives of $-mv_ t$ with respect to $t$, so that $\partial _ t^ j\mathrm{e}^{-mv_ t}=\mathrm{e}^{-mv_ t}\mathcal B_ j(-mv_ t)$ by Faà di Bruno's formula. Define operators on $V_ m$ by
\[ \mathcal T_{t,j}u=-P_{v_ t}\bigl (\mathcal B_ j(-mv_ t)u\bigr ), \qquad u\in V_ m. \]
For $k\geq 1$, let $\Lambda _ k$ be the set of compositions of $k$, that is, of ordered tuples $\alpha =(\alpha _1,\ldots ,\alpha _ p)$ of positive integers with sum $k$, and set $\alpha !=\prod _{j=1}^ p\alpha _ j!$.
Theorem 18.34.6.F82E8E685FEA For every $k\geq 1$,
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18.34.7
\begin{equation} \partial _ t^ k k_{v_ t,x} =\sum _{\alpha \in \Lambda _ k}\frac{k!}{\alpha !} \mathcal T_{t,\alpha _ p}\cdots \mathcal T_{t,\alpha _1}k_{v_ t,x}. \end{equation}
Suppose moreover that $s\geq 1$, that $J\Subset I$ is a compact interval, and that constants $A,R>0$ satisfy
\[ \sup _{t\in J}\| \partial _ t^ jv_ t\| _\infty \leq AR^ j(j!)^ s\qquad (j\geq 1). \]
Then there are constants $C,L>0$, depending only on the fixed data and on $m,r,J,A,R,s$, such that
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18.34.8
\begin{equation} \sup _{t\in J} \left(\| \partial _ t^ kB_{v_ t}\| _\infty +\| \partial _ t^ k\beta _{v_ t}\| _{\mathrm{TV}}\right) \leq CL^ k(k!)^ s\qquad (k\geq 0). \end{equation}
Thus the kernel and the kernel measure are of Gevrey class $s$ in the parameter.
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
\[ H(a)=P_{v_ t}\bigl (\mathrm{e}^{-m(v_{t+a}-v_ t)}\, \cdot \, \bigr )|_{V_ m}. \]
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 $\sum \ell n_\ell =j$,
\[ \prod _{\ell =1}^ j(\ell !)^{(s-1)n_\ell } \leq (j!)^{s-1}. \]
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
\[ \sup _{t\in J}\| \mathcal B_ j(-mv_ t)\| _\infty \leq L_0^ j(j!)^ s \]
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. $\square$
Corollary 18.34.9.7733E1B20E7B Let $v_ t=v+tq$ with $v,q\in C(X,\mathbb R)$. For $x\notin Z_ m$, the probability measure
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18.34.10
\begin{equation} \mathrm{d}\sigma _{t,x}(y) =\frac{|k_{v_ t,x}(y)|_{h^ m}^2\mathrm{e}^{-mv_ t(y)}}{\| k_{v_ t,x}\| _{m,v_ t,\nu }^2}\, \mathrm{d}\nu (y) \end{equation}
satisfies
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18.34.11
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}\log B_{v_ t}(x) =m\left(\int _ Xq\, \mathrm{d}\sigma _{t,x}-q(x)\right). \end{equation}
On $Z_ m$ the kernel and all its $t$-derivatives vanish. Moreover, $\int _ X\partial _ tB_{v_ t}\, \mathrm{d}\nu =0$.
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
\[ \partial _ t\ell _ x(k_{v_ t,x}) =m\int _ Xq|k_{v_ t,x}|_{h^ m}^2 \mathrm{e}^{-mv_ t}\, \mathrm{d}\nu . \]
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. $\square$
The uniform norm on the weights is essential in these results, because it controls the weight on the whole region of integration. The next example shows that real-analytic dependence of the weight on a parameter at each interior point of a domain does not suffice.
Example 18.34.12.5DA140DF64C6 On the unit disk $D\subset \mathbb C$ with Euclidean area $\mathrm{d}A$, let $K_ t$ be the Bergman kernel of the Hilbert space of holomorphic functions $f$ on $D$ with the norm $(\int _ D|f|^2\rho _ t\, \mathrm{d}A)^{1/2}$, where
\[ \rho _ t(z)=1+\frac{t^2}{(1-|z|^2)^2+t^2},\qquad t\in \mathbb R. \]
The weights $-\log \rho _ t$ are bounded and depend real analytically on $(z,t)\in D\times \mathbb R$, but
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18.34.13
\begin{equation} K_ t(0,0)= \begin{cases} [\pi (1+|t|\arctan (1/|t|))]^{-1},& t\ne 0,\\ \pi ^{-1},& t=0. \end{cases} \end{equation}
In particular, $K_ t(0,0)=\pi ^{-1}-|t|/2+O(t^2)$, so $K_ t(0,0)$ is not differentiable at $t=0$.
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 $t\ne 0$,
\[ \int _ D\rho _ t\, \mathrm{d}A =\pi \int _0^1\left(1+\frac{t^2}{a^2+t^2}\right)\mathrm{d}a =\pi \bigl (1+|t|\arctan (1/|t|)\bigr ). \]
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. $\square$
Throughout these estimates the degree $m$ and the selected space were fixed, and the constants depend on $m$. A limit envelope need not even be differentiable in the weight, as the corner in example 16.28.40 shows; differentiating limits would require estimates uniform in $m$.
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