The action variable of a two-simple-turning-point family is $C^1$ with derivative $T(E)\mathrm{/}(2\pi )$, the period quadrature over $2\pi$

Fix $E_{\ast }\in (E_0,E_1)$. Since $U^{\mathrm{\prime }}(a(E_{\ast }))\mathrm{\ne }0$ and $U^{\mathrm{\prime }}(b(E_{\ast }))\mathrm{\ne }0$, the inverse function theorem [L1] applied to $U$ at $a(E_{\ast })$ and at $b(E_{\ast })$ gives $\delta >0$ and $C^3$ (hence $C^1$) functions $A,B:(E_{\ast }-\delta ,E_{\ast }+\delta )\to J_0$ with $U(A(E))=U(B(E))=E$, $A(E_{\ast })=a(E_{\ast })$ and $B(E_{\ast })=b(E_{\ast })$. Shrinking $\delta$, continuity gives $A<B$ on the smaller interval. For every such $E$, the two real numbers $A(E)$ and $B(E)$ are distinct solutions of $U(q)=E$; by hypothesis the only two solutions in $J_0$ are $a(E)<b(E)$, so $\{A(E),B(E)\}=\{a(E),b(E)\}$; since $A<B$ and $a<b$, this forces $a=A$ and $b=B$ on that neighbourhood. Hence $a$ and $b$ are $C^1$ near the arbitrary point $E_{\ast }$, which is clause 1.

Fix $E_{\ast }$ again. Since $U^{\mathrm{\prime }}$ is continuous and $U^{\mathrm{\prime }}(a(E_{\ast }))<0<U^{\mathrm{\prime }}(b(E_{\ast }))$, choose disjoint neighborhoods of $a(E_{\ast })$ and $b(E_{\ast })$ on which $U^{\mathrm{\prime }}<0$ and $U^{\mathrm{\prime }}>0$, respectively. Choose $q_1,q_2$ in these neighborhoods with $a(E_{\ast })<q_1<q_2<b(E_{\ast })$. By continuity of $a$ and $b$ from step 1.1, shrink $\delta >0$ so that for $\mathrm{\mid }E-E_{\ast }\mathrm{\mid }<\delta$, $a(E)$ and $q_1$ remain in the first neighborhood, $q_2$ and $b(E)$ remain in the second, and $a(E)<q_1<q_2<b(E)$. Thus $U$ is strictly decreasing on $[a(E),q_1]$ and strictly increasing on $[q_2,b(E)]$. On this energy interval write $M(E):={\int }_{q_1}^{q_2}\sqrt{2m(E-U(q))}\mathrm{d}q,L(E):={\int }_{a(E)}^{q_1}\sqrt{2m(E-U(q))}\mathrm{d}q,R(E):={\int }_{q_2}^{b(E)}\sqrt{2m(E-U(q))}\mathrm{d}q,$ so that $J_{\mathrm{a}}=(L+M+R)\mathrm{/}\pi$ there.

The function $u(E,q):=\sqrt{2m(E-U(q))}$ is continuous on the compact rectangle $[E_{\ast }-\delta ,E_{\ast }+\delta ]\times [q_1,q_2]$ and ${\mathrm{\partial }}_{E}u(E,q)=m\mathrm{/}\sqrt{2m(E-U(q))}$ is continuous there, because $U(q)<E_{\ast }$ for every $q\in [q_1,q_2]$ and $U$ is continuous on the compact interval, so that $E-U(q)\ge ϵ>0$ on the rectangle after shrinking $\delta$ [L4, L5]. Hence by [L2] the middle term is $C^1$ with $M^{\mathrm{\prime }}(E)={\int }_{q_1}^{q_2}\frac{m\mathrm{d}q}{\sqrt{2m(E-U(q))}}(\mathrm{\mid }E-E_{\ast }\mathrm{\mid }<\delta )\mathrm{.}$

For the left leg, use the open interval about $a(E_{\ast })$ chosen in step 2.1, containing every segment $[a(E),q_1]$ and having $U^{\mathrm{\prime }}<0$. On this interval $U$ is strictly monotone; its local $C^3$ inverses from [L1] agree on overlaps by injectivity, giving a $C^3$ inverse $\psi$ on the whole image interval. For $E$ near $E_{\ast }$ the substitution $s=U(q)$ [L3] gives $L(E)={\int }_{U(q_1)}^E\sqrt{2m(E-s)}w(s)\mathrm{d}s,w(s):=-\frac{1}{U^{\mathrm{\prime }}(\psi (s))}>0,$ and $w$ is $C^2$ on the image interval; here $U(q_1)<E$ for $E$ near $E_{\ast }$ since $U(q_1)<E_{\ast }$.

The right leg is treated identically: near $b(E_{\ast })$ one has $U^{\mathrm{\prime }}>0$, the $C^3$ local inverse $\psi$ of $U$ gives $R(E)={\int }_{U(q_2)}^E\sqrt{2m(E-s)}w(s)\mathrm{d}s$ with $w:=1\mathrm{/}{U}^{\mathrm{\prime }}(\psi (s))>0$ of class $C^2$, the substitution $s=E-z^2$ and the variable-limit Leibniz rule give $R^{\mathrm{\prime }}(E)={\int }_{q_2}^{b(E)}\frac{m\mathrm{d}q}{\sqrt{2m(E-U(q))}}\mathrm{.}$

Substituting $s=E-z^2$ in the last integral [L3] turns the left leg into $L(E)=2\sqrt{2m}{\int }_0^{z(E)}{z}^{2}w(E-z^2)\mathrm{d}z,z(E):=\sqrt{E-U(q_1)}>0,$ with $z$ of class $C^1$ and the integrand $G(E,z):=2\sqrt{2m}{z}^{2}w(E-z^2)$ of class $C^1$ on a compact rectangle. By the variable-limit Leibniz rule [L2], $L^{\mathrm{\prime }}(E)=G(E,z(E))z^{\mathrm{\prime }}(E)+{\int }_0^{z(E)}{\mathrm{\partial }}_{E}G(E,z)\mathrm{d}z=\sqrt{2m}z(E)w(E-z(E{)}^2)+2\sqrt{2m}{\int }_0^{z(E)}{z}^2{w}^{\mathrm{\prime }}(E-z^2)\mathrm{d}z\mathrm{.}$

Integration by parts, equivalently the product rule applied to $z\mapsto zw(E-z^2)$, gives ${\int }_0^{z(E)}w(E-z^2)\mathrm{d}z=z(E)w(E-z(E{)}^2)+2{\int }_0^{z(E)}{z}^2{w}^{\mathrm{\prime }}(E-z^2)\mathrm{d}z;$ substituting this into step 4.1 yields $L^{\mathrm{\prime }}(E)=\sqrt{2m}{\int }_0^{z(E)}w(E-z^2)\mathrm{d}z={\int }_{a(E)}^{q_1}\frac{m\mathrm{d}q}{\sqrt{2m(E-U(q))}},$ the last equality by the same substitutions read backwards [L3]: $\sqrt{2m}{\int }_0^{z(E)}w(E-z^2)\mathrm{d}z={\int }_{a(E)}^{q_1}u(E,q{)}^{-1}m\mathrm{d}q$.

Adding the three derivatives of steps 3.1, 5.1 and 3.3 gives $J_{\mathrm{a}}^{\mathrm{\prime }}(E)=\frac{1}{\pi }{\int }_{a(E)}^{b(E)}\frac{m\mathrm{d}q}{\sqrt{2m(E-U(q))}}=\frac{\sqrt{2m}}{2\pi }{\int }_{a(E)}^{b(E)}\frac{\mathrm{d}q}{\sqrt{E-U(q)}}=\frac{T(E)}{2\pi }$ for $\mathrm{\mid }E-E_{\ast }\mathrm{\mid }<\delta$, with the middle integral understood improperly at $a(E)$ and $b(E)$. The endpoint pieces converge by steps 3.2–5.1 and 3.3; additivity on interior truncations and passage to the endpoint limits justify splitting this improper integral at $q_1,q_2$. The second equality uses $m\mathrm{/}\sqrt{2m}=\sqrt{2m}\mathrm{/}2$, and the last is the definition of $T(E)$.

The integrand $q\mapsto 1\mathrm{/}\sqrt{E-U(q)}$ is continuous and strictly positive on the nonempty open interval $(a(E),b(E))$, so $T(E)>0$ by [L4]; and the derivative computed in step 6.1 is independent of the auxiliary $E_{\ast }$, so the formula holds on all of $(E_0,E_1)$ and $J_{\mathrm{a}}$ is $C^1$ there. This is clause 2.