as logarithmic cutoff dependence with a high order operator $${\cal L} = -\frac{1+a\kappa^2 \Lambda^2}{4e_0^2}F_{\mu\nu}F^{\mu\nu} + b \ln \Lambda^2 ~F_{\mu\nu}\partial^2 F^{\mu\nu} \,.
\label{cutoffL}$$ This demonstrates that, in contrast with dimensional regularization, the lowest order charge $e_0$ does get renormalized (quadratically) by graviton loops when using a dimensionful cutoff as a regularizer.

The incorrect interpretation of this is to identify the cutoff dependence with the running of the coupling. Specifically, by writing $$e^2(\Lambda) =\frac{ e_0^2}{1+ a \kappa^2 \Lambda^2}\,,$$ the authors of [@does2] identify a beta function $$\beta (e) = \Lambda \frac{\partial e}{\partial \Lambda} =-  a e\kappa^2 \Lambda^2$$ for the running of the coupling.

However this interpretation is incorrect, because the quadratic $\Lambda$ dependence disappears from physical observables in the process of renormalization. For example, the Coulomb potential at low energy, calculated from Eq. \[cutoffL\] is $${ V}(r) = \frac{ e_0^2}{4\pi(1+ a \kappa^2 \Lambda^2)} \frac1{r}\,.$$ If we use this to identify the electric charge we obtain $$\alpha = \frac{e^2}{4\pi} =\frac{ e_0^2}{4\pi(1+ a \kappa^2 \Lambda^2)}  =\frac1{137}\,.$$ When expressing predictions in terms of the measured value of $\alpha$, the quadratic $\Lambda$ dependence is removed from all observables at all energies. It does not indicate the running of the electric charge. This analysis is supported by 