Lemma: Constant-cone mollification

Let $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^d)$ and let $A\subset {\mathbb{R}}^d\setminus 0$ be a closed angular cone with $\mathrm{WF}(u)\subset {\mathbb{R}}^d\times A$. Then

in the normal topology of ${{\mathcal{D}}^{\prime }}_{{\mathbb{R}}^d\times A}({\mathbb{R}}^d)$.

Proof. Strong convergence follows by transposing convolution:

The expression converges in smooth seminorms uniformly on bounded test sets.

Compact support gives polynomial Fourier growth, and the wavefront condition gives rapid decay on every closed cone separated from $A$. Since

a frequency ball–tail split gives weighted Fourier convergence on such cones.

For localized Fourier seminorms, convolve this identity with $(2\pi {)}^{-d}\hat{\chi }$. In a slightly enlarged tested cone, use rapid decrease of $\hat{u}$. In its complement, angular separation forces the Schwartz argument of $\hat{\chi }$ to control the total frequency, dominating the polynomial bound of $\hat{u}$. This gives uniformly small weighted tails. Strong convergence controls the bounded frequency region. All normal seminorms therefore converge.