Proposition 18.47.4AD8335312F45

Proof. At $p_+$ the monomial with exponents $(a,b)$ has local exponents $(a,m-b)$, and at $p_-$ it has local exponents $(m-a,b)$. By lemma 18.47.1, both total orders must be at least $m-1$. Distinct exponent pairs cannot cancel, so these conditions select exactly the monomials with $|a-b|\leq 1$. The three bands $a=b$, $a=b+1$ and $a=b-1$ have $m+1$, $m$ and $m$ elements. The bound $|\sigma |_{h^ m}^2\leq C\mathrm e^{m\psi }$ would require total order $m$ at both poles and leave only the diagonal. The product of two sections with offset $a-b=1$ has offset two and lies outside the selected space of the sum degree; such products fail for every pair of positive degrees. Thus the linear growth of $N_ m$ is not the growth of a graded section ring.

Integration over the angles makes all global monomials orthogonal, including those outside the selected space. For $t\in [0,1]$ set

The squared norm of the monomial with exponents $(a,b)$ is

With $c=\int _{-1}^1\mathrm e^{-y^2/2}\, \mathrm dy$, uniformly in $t\in [0,1]$,

For the lower bound, integrate over the interval of radius $m^{-1/2}$ centered at $t$; for the last assertion, note that each of the two omitted Gaussian tails is at most $\mathrm e^{-m/2}/m$. Hence

For $x\in [-1,2]^2$, completing squares bounds the weighted contribution of each normalized monomial by $9m/c^2$. Summing over all $(m+1)^2$ global monomials and using the evaluation inequality gives

This polynomial bound proves the Bernstein–Markov property. On the selected space the same argument gives the factor $Cm$. $\square$