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Write $r=|\mathbf{x}|$, $\theta =\mathbf{x}/r$, and choose $R$ so that the sphere $r=R$ strictly encloses the scatterer. Normalize the spherical harmonics to be an orthonormal basis of $L^2(S^2)$, with surface measure on the unit sphere. The outgoing spherical Hankel function is denoted here by $h_n^{(1)}$, corresponding to the printed $H_n^{(1)}$. For fixed $n$, its large-argument behavior isConsequently the coefficient of $Y_n^m$ in the outgoing expansion is asymptotic to $a_n^m{e}^{ikr}/r$: the factors satisfy $i^{n+1}(-i{)}^{n+1}=1$. With the far-field pattern defined by $u_s(r\theta )=e^{ikr}{u}_{\infty }(\theta )/r+O(r^{-2})$, we obtainOne can first project onto each spherical harmonic and then take the large-$r$ limit. This avoids assuming that the fixed-order asymptotic is uniform in $n$. The Sommerfeld radiation condition selects the outgoing spherical Hankel functions rather than their incoming counterparts.

Let $g(\theta )=u_s(R\theta )$. Its coefficients are $g_n^m=d_n{a}_n^m$, where $d_n=k{i}^{n+1}{h}_n^{(1)}(kR)$. Since the field is smooth on a sphere outside the scatterer, $g\in L^2(S^2)$; Parseval's identity givesThe required coefficient condition uses fixed-argument growth of spherical Hankel functions, a different limit from the far-field limit. For fixed $t>0$,For completeness, the series of the ordinary Bessel function gives $J_{-n-1/2}(t)\sim (t/2{)}^{-n-1/2}/\Gamma (1/2-n)$. The identities $y_n(t)=(-1{)}^{n+1}\sqrt{\pi /(2t)}{J}_{-n-1/2}(t)$ and $\Gamma (1/2-n)=(-4{)}^{n}n!\sqrt{\pi }/(2n)!$ give the first formula; the corresponding positive-order series gives the estimate for $j_n$. The successive series corrections at fixed $t$ are $O(1/n)$. Applying Stirling's approximation to the factorial ratio yieldsIn particular there are positive constants giving upper and lower bounds by the last weight for all sufficiently large $n$. The finite trace norm is therefore equivalent to the spherical far-field range conditionwhere the $n=0$ weight is defined to be one. The same conclusion holds at every larger sphere on which the outgoing field is defined.

The printed comparison with $\infty$ must mean a finite sum: a literal non-strict inequality allows a divergent sum. Also, the supplied one-sided big-$O$ bound alone does not imply this necessary condition from a finite trace norm; a lower bound is needed in that direction. The two-sided asymptotic above supplies it. Conversely, its upper bound makes the weighted condition sufficient for an $L^2$ trace at the chosen sphere. This is a condition on exterior-field continuation, not a sufficient condition that the coefficients come from a particular Dirichlet obstacle.