Forward positivity for a gapped identical scalar

Apply [F3] to $B$ on any circle centered at $0$ with radius $r<{\nu }_0$. Expanding $(\zeta -v{)}^{-1}={\sum }_{j\ge 0}{v}^j{\zeta }^{-j-1}$ uniformly on that circle for $\mathrm{\mid }v\mathrm{\mid }<r$ gives the Taylor series of $B$ near $0$. By [F2], every odd coefficient vanishes. Hence $H(v):=(B(v)-B(0))\mathrm{/}{v}^2$ has a holomorphic extension through $v=0$ with $H(0)=B^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{/}2!$, and $H$ is real on the real gap.

Fix a real $v\in (-{\nu }_0,{\nu }_0)$, $R>{\nu }_0+\mathrm{\mid }v\mathrm{\mid }$, $0<\eta <\min (1,({\nu }_0-\mathrm{\mid }v\mathrm{\mid })\mathrm{/}3)$, and $\delta >0$. Choose $ϵ>0$ so the closed radius-$ϵ$ discs about $\pm {\nu }_0$ are disjoint, avoid the radius-$\eta$ disc about $v$, and lie strictly inside the radius-$R$ circle about $v$. Let $U_{\delta }=\{z:\mathrm{Im}z>-\delta \mathrm{/}2\}$, which is star-shaped by [F6]. The contour ${\mathrm{\Gamma }}_{R,\eta ,ϵ}$ follows the real line left-to-right from $v-R$ to $v+R$, replacing the intervals about $v$ and $\pm {\nu }_0$ by upper semicircles traversed left-to-right, and closes by the radius-$R$ upper semicircle traversed right-to-left. The filled function $G_{\delta }(z)=(H(z+i\delta )-H(v+i\delta ))\mathrm{/}(z-v)$ is continuous on $U_{\delta }$ and holomorphic away from $v$ by [F4].

Apply [F5] to $G_{\delta }$ on ${\mathrm{\Gamma }}_{R,\eta ,ϵ}$. The integral of $dz\mathrm{/}(z-v)$ over the contour without threshold detours is zero: the real-segment logarithms cancel, and the small and large semicircles contribute $-i\pi$ and $+i\pi$. Replacing a real interval about either threshold by its semicircle preserves that integral, because $1\mathrm{/}(z-v)$ is holomorphic on a disc containing the interval and semicircle but not $v$; [F5] applied there gives equality of the two path integrals. Thus ${\int }_{{\mathrm{\Gamma }}_{R,\eta ,ϵ}}dz\mathrm{/}(z-v)=0$, and the definition of $G_{\delta }$ yields ${\int }_{{\mathrm{\Gamma }}_{R,\eta ,ϵ}}H(z+i\delta )\mathrm{/}(z-v)dz=0$.

First hold $R,\eta ,ϵ$ fixed and let $\delta \downarrow 0$. The detoured contour avoids the threshold points and $v$. On each retained real cut segment $K$, the new continuous-bank hypothesis in [F1] gives uniform convergence of $H(x+i\delta )$ to $H_{+}(x)$. Explicitly, the integral error is at most $\mathrm{\mid }K\mathrm{\mid }{\sup }_K\mathrm{\mid }H(x+i\delta )-H_{+}(x)\mathrm{\mid }\mathrm{/}\mathrm{dist}(K,v)$, which tends to zero. Compact gap segments and semicircles have the same uniform convergence by holomorphy, with their length and minimum distance from $v$ giving the analogous bound. Therefore the entire fixed contour integral converges to its boundary-contour value. Take imaginary parts at these fixed radii: every retained real-gap segment contributes zero because $H$ is real there. Now let $ϵ\downarrow 0$: the threshold semicircle contributions vanish by exactly the endpoint-arc hypothesis in [F1], and the retained imaginary cut integrals converge by the specified improper-bank limits. This step does not require convergence of separate real-part integrals on the gap side of a threshold. Finally let $R\to \mathrm{\infty }$ and $\eta \downarrow 0$. The outer integral has modulus at most $\pi {\sup }_{\mathrm{\mid }z-v\mathrm{\mid }=R,\mathrm{Im}z\ge 0}\mathrm{\mid }H(z)\mathrm{\mid }\to 0$ by the uniform high-energy bound. On the small semicircle the difference from $-i\pi H(v)$ is bounded by $\pi {\sup }_{\mathrm{\mid }z-v\mathrm{\mid }=\eta }\mathrm{\mid }H(z)-H(v)\mathrm{\mid }\to 0$. Since $H(v)$ is real, the resulting imaginary-part identity is $H(v)={\pi }^{-1}{\int }_{\mathbb{R}}\mathrm{Im}{H}_{+}(x)\mathrm{/}(x-v)dx$, with the integrand set to zero throughout the gap, including at $x=v$. Only the cut integrals remain; a full complex real-line principal value is not asserted. No dominated-convergence theorem or unasserted interchange of pointwise boundary limits is used.

On the gap, $\mathrm{Im}{H}_{+}(x)=0$. For $\nu >{\nu }_0$ on the right cut, $\mathrm{Im}{H}_{+}(\nu )=\rho (\nu )\mathrm{/}{\nu }^2$. On the left cut, evenness and real analyticity give $\mathrm{Im}{H}_{+}(-\nu )=-\rho (\nu )\mathrm{/}{\nu }^2$. Substituting these boundary values into step 3.1 and changing variables $x=-\nu$ on the left cut gives $H(v)={\pi }^{-1}{\int }_{{\nu }_0}^{\mathrm{\infty }}\rho (\nu ){\nu }^{-2}((\nu -v{)}^{-1}+(\nu +v{)}^{-1})d\nu =(2\mathrm{/}\pi ){\int }_{{\nu }_0}^{\mathrm{\infty }}\rho (\nu )\mathrm{/}[\nu ({\nu }^2-v^2)]d\nu$. Both integrals converge: on each compact interval, $\rho$ is Riemann integrable and each kernel is continuous, so their products are Riemann integrable. Explicitly, for bounded $f$ and continuous $g$, ${\mathrm{osc}}_J(fg)\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\mathrm{osc}}_{J}f+\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\mathrm{osc}}_{J}g$ on every 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 this product claim. The kernels are bounded near threshold and at infinity they are bounded by a constant times ${\nu }^{-3}$; [F7, F10] apply. This same compact-interval argument applies to every continuous kernel multiplying $\rho$ below. This is the twice-subtracted forward dispersion relation, with subtraction constants $B(0)$ and $B^{\mathrm{\prime }}(0)=0$ and with all specified subthreshold poles removed.

Fix an integer $N\ge 1$. For real $\mathrm{\mid }v\mathrm{\mid }<{\nu }_0$, expand $1\mathrm{/}(1-v^2\mathrm{/}{\nu }^2)$ through order $N$; its remainder is bounded uniformly in $\nu \ge {\nu }_0$ by $q^{N+1}\mathrm{/}(1-q)$, where $q=v^2\mathrm{/}{\nu }_0^2<1$. Using [F8] for the finite sum and [F9, F10] for the nonnegative remainder gives $B(v)=B(0)+{\sum }_{n=1}^{N+1}((2\mathrm{/}\pi ){\int }_{{\nu }_0}^{\mathrm{\infty }}\rho (\nu ){\nu }^{-2n-1}d\nu )v^{2n}+o(v^{2N+2})$ as $v\to 0$ through real values: the integrated remainder is bounded by $(2\mathrm{/}\pi )v^2{q}^{N+1}(1-q{)}^{-1}{\int }_{{\nu }_0}^{\mathrm{\infty }}\rho (\nu ){\nu }^{-3}d\nu =O(v^{2N+4})$. The Taylor series of $B$ from step 1.1 is unique: if a first coefficient differed, dividing the difference by its least differing power and taking $v\to 0$ would give a nonzero constant equal to $0$. Since $N$ is arbitrary, $B^{(2n)}(0)\mathrm{/}(2n)!=(2\mathrm{/}\pi ){\int }_{{\nu }_0}^{\mathrm{\infty }}\rho (\nu ){\nu }^{-2n-1}d\nu$ for every $n\ge 1$.

For each finite $R>{\nu }_0$, the integrand $\rho (\nu )\mathrm{/}{\nu }^{2n+1}$ is locally Riemann integrable by the compact-kernel argument in step 4.1 and is nonnegative, so [F9] gives ${\int }_{{\nu }_0}^R\rho (\nu ){\nu }^{-2n-1}d\nu \ge 0$. Its improper limit is finite by [F10], since $0\le \rho (\nu ){\nu }^{-2n-1}\le {\nu }_0^{-2n+2}\rho (\nu ){\nu }^{-3}$. Taking that limit and multiplying by $2\mathrm{/}\pi >0$ proves the asserted nonnegativity. If $\rho$ vanishes, equality can occur; strict positivity is not concluded.