 Solution

A hydrostatic pressure interval contains $dp/(gM)$ moles of background gas per area. Its absorbing column is $x(p)dp/(gM)$, so for a nadir rayThe assumed Planck-temperature power and dry adiabat give $B(p)/B_0=[T(p)/T_0{]}^{3.5}=p/p_0$. To first order in optical depth, $-d\mathfrak{T}\simeq d\tau$, and the radiance solution becomesFor a uniform mixing ratio this also evaluates to $L\simeq B_0[1-k_{0}x{p}_0/(6gM)]$. Its total optical depth is ${\tau }_0=k_{0}x{p}_0/(2gM)$, so the deficit is $B_0{\tau }_0/3$.

A pressure-resolved perturbation defines the infrared absorber sensitivity kernel by $\delta L=\int K_p(p)\delta x(p)dp$, withIt is zero at the top, negative in the interior, and zero again at the surface. The magnitude is largest where $d{K}_p/dp=0$, namelySign and coordinate qualification. Literally the maximum of the signed kernel is zero at the endpoints; $p_0/2$ is its minimum and the maximum of its absolute value. The derivative with respect to a single vertically uniform $x$ is the integrated constant $-k_0{B}_0{p}_0/(6gM)$ and has no pressure location. For equal-width logarithmic pressure perturbations the kernel is instead $K_{\log p}=p{K}_p$ and its magnitude peaks at $2p_0/3$. The stated pressure-interval convention is therefore essential.

The low top-of-atmosphere sensitivity reflects the small pressure-broadened absorption coefficient and small molecular column per $dp$ there; the low surface sensitivity reflects small thermal contrast. The printed phrase that temperature decreases with pressure reverses the actual stipulated relation: with positive exponent, temperature decreases as pressure decreases upward. The derivation uses the explicit adiabatic formula.