the spectral representation of $F(s)$ in we can obtain a rigorous lower bound on this exact undershoot. In [@cappelli] it was shown that the spectral density can be written as $$\nonumber
\tilde{c}(\mu^2)=c_{IR}\delta(\mu^2)+\hat{c}(\mu^2),$$ we showed in that $\int d\mu^2 \tilde{c}(\mu^2)=c_{UV}$ hence $\int d\mu^2 \hat{c}(\mu^2)=\Delta c_{exact}$. Using this in we get $$\label{estimate}
\begin{split}
  \Delta c_{est}=\ & c_{UV}-I({\rho'}_m)=c_{UV}-\frac{1}{2\pi
  i}\int_{\tilde{C}}\frac{ds}{s}e^{{\rho'}_m/s}F(s) \\  =\ &
\Delta c_{exact}-\int d\mu^2\
  \hat{c}(\mu^2)\Upsilon(\mu^2, {\rho'}_m) \nonumber
\end{split}$$ where $$\label{psi}
\begin{split}
  \Upsilon(\mu^2,{\rho'}_m)= \ & \frac{1}{2\pi i}\int_{\tilde{C}}
\frac{ds}{s}e^{{\rho'}_m/s}\frac{\mu
  s^{1/4}}{48}\left(((\mu s^{1/4})^3  \right. \\ 
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. +24 \mu
  s^{1/4})K_0(\mu s^{1/4})+(8(\mu s^{1/4})^2+48)K_1(\mu
  s^{1/4}) \right)
\end{split}
\nonumber$$ and $F(s)=\tilde{F}(s^{1/k})$ with $k=4$. Rescaling $s$ allows us to move all $\mu$ dependence into ${\rho'}_m(\mu)={\rho'}_m \mu^4$, so that we can write $\Upsilon(\mu^2,{\rho'}_m)=\Upsilon({\rho'}_m(\mu))$. Unitarity ensures that $\hat{c}(\mu^2)\geq 0$ hence $$\label{bound1}
  \Delta c_{exact}-\Delta c_{est}\geq \min_{\mu} \Upsilon ({\rho'}_m(\mu))
\int_0^{\infty}d\mu^2
  \hat{c}(\mu^2),$$ $\Delta c_{est}>0$ so that $$\label{bound2}
  0 \geqslant \frac{\Delta c_{exact}-\Delta c_{est}}{\Delta c_{exact}}\geqslant
  \min_{\rho>0}\Upsilon (\rho).$$

The lower bound in the relative undershoot of $I({\rho'}_m)$ therefore equals $\min_{\rho}\Upsilon(\rho)$. In figure \[fig:bound\] we plot $\Upsilon (\rho)$ for $k=4$. With this value we get that $\min_\rho \Upsilon (\rho)=-0.0232$ so the relative overshoot in $\Delta c_{est}$ compared with $\Delta c_{exact}$ is maximally 