Proposition 18.34.1.A1D3D26087E7 For $v,f\in C(X,\mathbb R)$, the measure $\beta _ m^{(r)}$ at the weight $v$ satisfies
\[ \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\mathcal F_ m^{(r)}(v+tf) =\int _ X f\, \mathrm{d}\beta _ m^{(r)}. \]
The functional $\mathcal F_ m^{(r)}$ is concave on $C(X,\mathbb R)$, and for $v,w\in C(X,\mathbb R)$ and $c\in \mathbb R$,
\[ \mathcal F_ m^{(r)}(v+c)=\mathcal F_ m^{(r)}(v) +\frac{r!N_ m}{m^ r}c, \qquad |\mathcal F_ m^{(r)}(v)-\mathcal F_ m^{(r)}(w)| \leq \frac{r!N_ m}{m^ r}\| v-w\| _\infty . \]
If $N_ m>0$, the second derivative of $t\mapsto \mathcal F_ m^{(r)}(v+tf)$ is strictly negative unless $f$ is constant.
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
\[ A_{ij}=\int _ X f\langle s_ i,s_ j\rangle _{h^ m}\mathrm{e}^{-mv}\, \mathrm{d}\nu , \qquad C_{ij}=\int _ X f^2\langle s_ i,s_ j\rangle _{h^ m}\mathrm{e}^{-mv}\, \mathrm{d}\nu . \]
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
\[ \left.\frac{\mathrm{d}^2}{\mathrm{d}t^2}\right|_{t=0}\mathcal F_ m^{(r)}(v+tf) =-\frac{r!}{m^{r-1}} \left(\int _ X f^2B_{m,v,\varphi ,\nu }\, \mathrm{d}\nu -\sum _{i,j}|A_{ij}|^2\right). \]
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
\[ \sum _ j\| (I-P_ v)(fs_ j)\| _{m,v,\nu }^2\geq 0. \]
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.
$\square$
Pome
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