Lemma: Normal convergence

For every fixed $f$, choose a compact $K$ containing a neighborhood of $\mathrm{supp}{b}_f$. Then, for all sufficiently small $\epsilon >0$,

with its normal topology.

Proof. Smoothness and the fixed compact support follow from convolution with a compact approximate identity.

Strong distribution convergence follows from the finite order of $b_f$ and the convergence

in all smooth seminorms, uniformly for $\psi$ in a bounded set of test functions.

For completeness, the microlocal estimates can be checked directly because ${\Gamma }_B$ has a translation-invariant angular cone. Compactness and $\mathrm{WF}(b_f)\subset (M\times M)\times {\Lambda }_B$ imply rapid decrease of ${\hat{b}}_f$ in every closed cone separated from ${\Lambda }_B$. Moreover,

The multipliers are uniformly bounded and tend to one on compact frequency sets.

On a closed cone separated from ${\Lambda }_B$, split frequency space into a fixed ball and its complement. On the ball, the multiplier converges uniformly to one. On the complement, arbitrarily high rapid-decay bounds for ${\hat{b}}_f$ give a uniformly small tail. Thus the weighted Fourier seminorms converge to zero.

For a localized seminorm, use

A slightly larger cone still disjoint from ${\Lambda }_B$ controls the nearby-frequency part. On the complementary part, the angular separation forces the argument of $\hat{\chi }$ to be large; its arbitrary rapid decrease dominates the uniform polynomial growth of ${\hat{b}}_f$. This proves convergence of every normal-topology seminorm.