A fixed partial-wave cutoff misses the forward high-frequency plane wave

Given: $d\in S^2$, $k>0$, and integer $L\ge 0$.

[F1] The addition kernel for degree $\mathrm{\ell }$ is $Z_{\mathrm{\ell }}(\omega ,\eta )=\frac{2\mathrm{\ell }+1}{4\pi }{P}_{\mathrm{\ell }}(\omega \cdot \eta )$. Expanding the finite orthogonal projection in an orthonormal basis gives its kernel $Z_{\mathrm{\ell }}$, so $({\mathrm{\Pi }}_{L}g)(d)={\sum }_{\mathrm{\ell }=0}^L{\int }_{S^2}{Z}_{\mathrm{\ell }}(d,\omega )g(\omega )d\sigma (\omega )$ (Spherical harmonics form a complete basis with the addition formula).

[F2] In the coordinates $\omega (t,\varphi )=(\sqrt{1-t^2}\cos \varphi ,\sqrt{1-t^2}\sin \varphi ,t)$, the map $T(r,t,\varphi )=r\omega (t,\varphi )$ is a $C^1$ diffeomorphism from $(0,1)\times (-1,1)\times (0,2\pi )$ onto the open unit ball with its meridional half-plane removed, and $\mathrm{\mid }\det DT\mathrm{\mid }=r^2$. The omitted half-plane is null by the singular clause of the linear change-of-variables theorem. The $C^1$ change-of-variables theorem and the radial-cone definition of $d\sigma$ therefore give $d\sigma =dtd\varphi$ on this chart. Rotations preserve $d\sigma$ because they preserve radial cones and have determinant one (The polar surface set function on the unit sphere, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions, A linear map $T$ of ${\mathbb{R}}^n$ sends Lebesgue measurable sets to Lebesgue measurable sets, with ${\lambda }_n(T[E])=\mathrm{\mid }\det T\mathrm{\mid }{\lambda }_n(E)$ when $T$ is invertible and $T[E]$ Lebesgue null when it is not). Choosing a rotation taking $d$ to the north pole yields ${\int }_{S^2}h(d\cdot \omega )d\sigma (\omega )=2\pi {\int }_{-1}^{1}h(t)dt$ for every continuous $h$.

[F3] For real $t$, the complex derivative theorem gives $\frac{d}{dt}{e}^{ikt}=ik{e}^{ikt}$; this follows from the difference quotient with complex increment $ikh$ (The complex exponential is entire and its complex derivative is itself).

[F4] For real $k$, $\mathrm{\mid }{e}^{ik}\mathrm{\mid }=1$ ($\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$).