have a modified linear superfield.

Substituting into we get the (modified) linear multiplet formulation of the above action $$\begin{aligned}
{\cal A}_{LMF} & = & -3\int d^{8}z{\bf E}e^{-K(\phi,\bar{\phi},U)/3}F(\phi,\bar{\phi},U)\label{eq:ALMF}\\
 &  & +(\int d^{8}z\frac{{\bf E}}{2R}W(\phi)+h.c.).\nonumber \end{aligned}$$ Note that there is no explicit gauge kinetic term in this form of the action. It is however implicit because $U$ satisfies the constraint . To see this we first rewrite this as an equation for $-\frac{1}{4}\bar{\nabla}^{2}U$ to get (keeping only terms which contain ${\cal W}^{2}$) $$-\frac{1}{4}\bar{\nabla}^{2}U=\frac{e^{K/3}}{1-\frac{1}{3}UK_{U}}\frac{k}{3\times4}{\cal W}^{2}+\ldots.\label{eq:del2U}$$ Then we have from the first line of , using $\int d^{8}z{\bf E}\bar{\nabla}^{\dot{\alpha}}v_{\dot{\alpha}}=0$ and the above result, $$\begin{aligned}
-3\int d^{8}z\frac{{\bf E}}{2R}(-\frac{1}{4}\bar{\nabla}^{2}+2R)\left(e^{-K/3}F(\phi,\bar{\phi},U)\right) & =\\
-3\int d^{8}z\frac{{\bf E}}{2R}e^{-K/3}(-\frac{K_{U}}{3}F+F_{U})(-\frac{1}{4}\bar{\nabla}^{2}U)+\ldots & = & -\int d^{8}z\frac{{\bf E}}{2R}\frac{k}{4}\Gamma(U,\phi,\bar{\phi}){\cal W}{\cal W}+\ldots,\end{aligned}$$ so that the effective gauge coupling function in the linear multiplet formulation is (the lowest component of) $$\Gamma=-F_{U}+\frac{N}{3}K_{U},\, N\equiv\frac{F-UF_{U}}{1-\frac{1}{3}UK_{U}}\label{eq:Gamma}$$ in agreement with BGG.

Now let us see what we would get in the chiral field formulation of the gauge coupling function. Varying the action with respect to $U$ we get $$(S+\bar{S})(1-\frac{1}{3}UK_{U})=\frac{1}{3}FK_{U}-F_{U}.\label{eq:Ueqn}$$ This equation determines $U=U(S+\bar{S},\phi,\bar{\phi})$. In the chiral field formulation we need to have in addition the normalization condition $$F(\phi,\bar{\phi},U)+U(S+\bar{S})=1.\label{eq:norm}$$ This is just the condition that in the chiral formulation the pre-factor of $e^{-K/3}$ should 