IR cutoff of the UV tail as $\kappa_0\sim T$.

Let us check whether sum rule 2 is satisfied. From Eqs.(\[eq:T-sumrule2-transport\]-\[eq:T-sumrule2-UVtail\]), we see that for the ${{\mathbf{p}}}$-independent part, the contributions from the continuum and the UV tail have the same order of magnitude ($\sim {C_{\text{em}}}T^4$) while the contribution from the transport peak is much smaller, $\sim {C_{\text{em}}}T^2 \tau^{-1}\varLambda\sim {C_{\text{em}}}g^5 T^4$. These contributions are found to agree with the left-hand side of Eq. (\[eq:sumrule-2-T\]) by using $\delta \langle T^{00}\rangle={N_c}\pi^2T^4(8{C_F}+7{N_f})/60$, which is obtained from Eqs. (\[eq:T00f-free\]) and (\[eq:T00g-free\]). For the ${{\mathbf{p}}}^2$ term, the contributions from the transport peak and the continuum cancel, while the UV tail contribution is much smaller. Therefore, we have confirmed that sum rule 2 is satisfied up to order ${{\mathbf{p}}}^2$, in the chiral and weak coupling limits.

### Sum rule 3

The derivation of the third sum rule turns out to be somewhat more tricky. Equation (\[eq:residue-theorem\]) for $\mu=\nu=1$ can be rewritten as $$\begin{aligned}
\begin{split}
&\delta G_{T}(i\omega,{{\mathbf{p}}})-\delta G_{T}(\infty,{{\mathbf{p}}}) 
= \frac{1}{\pi }\int^\infty_{0} d\omega' \\
&~~~\times\frac{\omega' \delta \rho_{T}(\omega',{{\mathbf{p}}})+\omega[{\mathrm{Re}}\delta G_{T}(\omega',{{\mathbf{p}}})-\delta G_{T}(\infty,{{\mathbf{p}}})]}{\omega'{}^2+\omega^2}\\
&~~~= \frac{2}{\pi }\int^\infty_{0} d\omega' 
\frac{\omega' \delta \rho_{T}(\omega',{{\mathbf{p}}})}{\omega'{}^2+\omega^2} ,
\end{split}\end{aligned}$$ where we have used the relation $$\begin{aligned}
&0 = \nonumber \\
& \int^\infty_{-\infty} d\omega' 
\frac{\omega' \delta \rho_{T}(\omega',{{\mathbf{p}}})-\omega[{\mathrm{Re}}\delta G_{T}(\omega',{{\mathbf{p}}})-\delta G_{T}(\infty,{{\mathbf{p}}})]}{\omega'{}^2+\omega^2},\end{aligned}$$ in the last 