 Solution

In local thermodynamic equilibrium, thermal intensity samples the Planck function near an optical depth of order unity. The Eddington-Barbier relation makes this explicit: $I_{\lambda }(0,\mu )\simeq B_{\lambda }[T({\tau }_{\lambda }=\mu )]$. A molecular band has greater opacity than its adjacent continuum and therefore samples a higher layer. A band in emission relative to the continuum implies that this higher layer is hotter: the line-forming region has an atmospheric thermal inversion under the assumed LTE, thermal interpretation.

The continuum is thermal radiation from an optically thick, deeper photosphere, with comparatively smooth opacity. In an H/He hot Jupiter, collision-induced absorption by H2-H2 and H2-He collisions supplies an important continuum; weak overlapping molecular lines and opaque exoplanet clouds can contribute too. It is not a separate blackbody emitter floating above the gas. A strongly isothermal layer would erase LTE molecular contrast rather than generate emission peaks.

In the emitting inversion, $dT/dz>0$, whereas the dry adiabatic lapse rate has $dT/dz=-g/C_p<0$. Thereforewhich lies on the stable side of the Schwarzschild criterion. An upward-displaced parcel cools and becomes denser than the ambient hot upper gas. The region can thus carry and redistribute thermal energy by radiative transfer, not by unstable thermal convection. Winds may transport energy horizontally; stability rules out the specified buoyant vertical convection, not every possible motion.