Proposition 18.34.581FC6A666EC9

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. $\square$