Finite truncated forward-positivity constraints

For $\nu \ge {\nu }_0$, put $x(\nu )={\nu }_0^2\mathrm{/}{\nu }^2$ and $w(\nu )=(2{\nu }_0^2\mathrm{/}\pi )\rho (\nu )\mathrm{/}{\nu }^3$. Then $0<x(\nu )\le 1$, $w(\nu )\ge 0$, and ${\int }_{{\nu }_0}^{\mathrm{\infty }}w(\nu )d\nu <\mathrm{\infty }$ by [F1] and [F3]. For every $0\le k\le N$, [F2] gives $b_k={\nu }_0^{2k+2}{c}_{k+1}=(2{\nu }_0^{2k+2}\mathrm{/}\pi ){\int }_{{\nu }_0}^{\mathrm{\infty }}\rho (\nu ){\nu }^{-2k-3}d\nu ={\int }_{{\nu }_0}^{\mathrm{\infty }}w(\nu )x(\nu {)}^{k}d\nu$. These moments are finite since $0\le x^k\le 1$.

On each finite interval $[{\nu }_0,R]$, $\rho$ is Riemann integrable by [F1], while ${\nu }^{-3}$, $x$, and every polynomial in $x$ are continuous. Thus $\rho$ times every kernel used below is Riemann integrable: for bounded $f$ and continuous $g$, ${\mathrm{osc}}_J(fg)\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\mathrm{osc}}_{J}f+\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\mathrm{osc}}_{J}g$ for each subinterval $J$, where ${\mathrm{osc}}_{J}f:={\sup }_{J}f-{\inf }_{J}f$. A Darboux-good partition for $f$, refined to small mesh using uniform continuity of $g$, proves the Darboux criterion for $fg$. Thus every finite truncation used below is well-defined.

Fix $r\ge 0$ with $2r\le N$ and real $a_0,\dots ,a_r$. Let $p(x)={\sum }_{i=0}^r{a}_i{x}^i$. Using the finitely many moment identities from step 1.1 and [F4], ${\sum }_{i,j=0}^r{a}_i{a}_j{b}_{i+j}={\int }_{{\nu }_0}^{\mathrm{\infty }}w(\nu )p(x(\nu ){)}^{2}d\nu$. The polynomial $p$ is bounded on $[0,1]$, so the integrand is bounded above by a fixed multiple of the integrable nonnegative function $w$; [F6] gives convergence. Its finite truncations are nonnegative by [F5], and their limit is nonnegative by [F3]. Thus the first family holds.

Fix $r\ge 0$ with $2r+1\le N$ and the same arbitrary vector, and use $p(x)={\sum }_{i=0}^r{a}_i{x}^i$. The moment identities and [F4] give ${\sum }_{i,j=0}^r{a}_i{a}_j{b}_{i+j+1}={\int }_{{\nu }_0}^{\mathrm{\infty }}w(\nu )x(\nu )p(x(\nu ){)}^{2}d\nu$ and ${\sum }_{i,j=0}^r{a}_i{a}_j(b_{i+j}-b_{i+j+1})={\int }_{{\nu }_0}^{\mathrm{\infty }}w(\nu )(1-x(\nu ))p(x(\nu ){)}^{2}d\nu$. The index bound ensures every moment used has index at most $N$. Since $0<x(\nu )\le 1$, both integrands are nonnegative. Each is bounded by $Cw(\nu )$ for $C={\sup }_{x\in [0,1]}p(x{)}^2<\mathrm{\infty }$, so [F6] and [F3] give convergent nonnegative improper integrals, using [F5] on every finite truncation. Both remaining families therefore hold. At threshold $x=1$ the second localizer vanishes, and as $\nu \to \mathrm{\infty }$ one has $x\to 0$; neither endpoint changes the nonnegative-integral argument.

For $N=1$, all three applicable families have $r=0$ and are included in steps 3.1 and 3.2. For every $N\ge 1$, those steps establish every inequality in the defining index ranges, so the vector belongs to ${\mathcal{K}}_N$. Equality may occur when an integrand vanishes; the proof gives no strict inequality. The set ${\mathcal{K}}_N$ was defined only by these necessary tests, so this membership proves no converse or full-amplitude realization.