Solution
= Solution
When $h_0=0$, the cubic contribution is absent and
$$
\boxed{\Delta S=S_{0,>}[\phi^+]
+\frac{g_0}{4!}\int d^4x\,
\left[4\phi^3\phi^++6\phi^2(\phi^+)^2
+4\phi(\phi^+)^3+(\phi^+)^4\right].}
$$
Although terms odd in $\phi^+$ occur for a fixed low field, the simultaneous transformation $(\phi,\phi^+)\mapsto(-\phi,-\phi^+)$ shows that integrating out the shell preserves the original $\mathbb Z_2$ <discrete symmetry> of the effective action.