We have the following equality given (derived in the paper itself "A Donsker theorem for Levy measures")$(p.16,4.9)$:
$$ \mathcal F^{-1}\Big[\varphi^{-1}(-u)\Big]= \frac1{i\bullet}\mathcal F^{-1}\Big[\big(\varphi^{-1}(-u)\big)'\Big]\tag1 $$ where $\mathcal F$ denotes the inverse Fourier transform and $\varphi$ is the characteristic function of a probability law, with $\varphi^{-1}\equiv\frac1{\varphi}$.
Unfortunately it is not explained in the text, but I am rather sure, the $\bullet$ is just a place holder for the argument of the function - so we actually have $$ \mathcal F^{-1}\Big[\varphi^{-1}(-u)\Big](x)= \frac1{ix}\mathcal F^{-1}\Big[\big(\varphi^{-1}(-u)\big)'\Big](x)\tag2 $$ which also blends in with the known "derivative of the Fourier argument".
Now comes the term which seems to vanish, though this can't be, since otherwise the whole paper would be pointless - therefore from now on I assume I am wrong, we have:
In the first step we have: $(\varphi^{-1}(-u)\big)'=-(\varphi^{-1})' (-u)$ and hence from $(2)$ we get $$ \mathcal F^{-1}\Big[\varphi^{-1}(-u)\Big](x)= \frac1{ix}\mathcal F^{-1}\Big[\big(\varphi^{-1}(-u)\big)'\Big](x)=-\frac1{ix}F^{-1}\Big[(\varphi^{-1})'(-u)\Big](x)\tag3 $$ and it follows then by plugging in ($p.13,(4.2)$) \begin{align} &\frac1{ i\Delta}\int_{\mathbb R}\big( \mathcal{F}^{-1}\big[\varphi^{-1}(-u)\mathcal F[g](u)\mathcal{F}[K_h](-u) \big](x)ix + \mathcal{F}^{-1}\big[ (\varphi^{-1})'(-u)F[g](u)\mathcal{F}[K_h](-u) \big](x) \big)\;P(dx)\\ &=\frac1{ i\Delta}\int_{\mathbb R}\Big(\mathcal{F}^{-1}\big[\varphi^{-1}(-u)\big]*g*K_h)(x)ix+\mathcal{F}^{-1}\big[(\varphi^{-1})'(-u)\big]*g*K_h)P(x)\Big)P(dx)\\ &=\frac1{ i\Delta}\int_{\mathbb R}\Big(\mathcal{F}^{-1}\big[\varphi^{-1}(-u)\big]*g*K_h)(x)ix+(-ix)\mathcal{F}^{-1}\big[\varphi^{-1}(-u)\big]*g*K_h)(x)\Big)P(dx)\\ &=0 \end{align} Where am I wrong? It seems to just cancel out - maybe it's more than simple algebraic manipulations. Anyone?