Liouville turning-point transformation

Given: the interval $J$, the function $q$ with a single simple zero at $0$, $h>0$, and the transformation data $\zeta ,a$ defined in the Statement.

Define $b(0):=q^{\mathrm{\prime }}(0)$ and $b(x):=q(x)\mathrm{/}x$ for $x\mathrm{\ne }0$. Then $b(x)={\int }_0^1{q}^{\mathrm{\prime }}(tx)dt$ and, for $0\le k\le 4$, $b^{(k)}(x)={\int }_0^1{t}^k{q}^{(k+1)}(tx)dt$; for $k<4$ the difference quotients are dominated by $t^{k+1}{\sup }_J\mathrm{\mid }{q}^{(k+2)}\mathrm{\mid }$, so dominated convergence gives each derivative formula and its continuity, hence $b\in C^4(J)$. The sign assumptions give $b>0$ on $J$, hence $g:=\sqrt{b}\in C^4(J)$. With $G(x):={\int }_0^1{t}^{1\mathrm{/}2}g(tx)dt$, dominated convergence applied successively to the difference quotients, bounded by $t^{k+3\mathrm{/}2}{\sup }_J\mathrm{\mid }{g}^{(k+1)}\mathrm{\mid }$ for $k<4$, gives the continuous derivatives $G^{(k)}(x)={\int }_0^1{t}^{k+1\mathrm{/}2}{g}^{(k)}(tx)dt$ and $G\in C^4(J)$. Since $G>0$, substitution $s=xt$ yields $\zeta (x)=x[(3\mathrm{/}2)G(x){]}^{2\mathrm{/}3}\in C^4(J)$ and ${\zeta }^{\mathrm{\prime }}(0)=b(0{)}^{1\mathrm{/}3}>0$ (Dominated convergence, The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$, Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point).

The identity $\zeta ({\zeta }^{\mathrm{\prime }}{)}^2=q$ follows without losing the sign: for $x\mathrm{\ne }0$, differentiate $\mathrm{\mid }\zeta (x){\mathrm{\mid }}^{3\mathrm{/}2}=\frac{3}{2}\mathrm{\mid }{\int }_0^x\sqrt{\mathrm{\mid }q(s)\mathrm{\mid }}ds\mathrm{\mid }$. The signs of $\zeta$ and of this signed integral both equal $\mathrm{sign}x$, so $\mathrm{\mid }\zeta {\mathrm{\mid }}^{1\mathrm{/}2}{\zeta }^{\mathrm{\prime }}=\sqrt{\mathrm{\mid }q\mathrm{\mid }}$. Thus ${\zeta }^{\mathrm{\prime }}>0$ off zero, $\mathrm{\mid }\zeta \mathrm{\mid }({\zeta }^{\mathrm{\prime }}{)}^2=\mathrm{\mid }q\mathrm{\mid }$, and multiplication by $\mathrm{sign}x$ gives the required signed identity. At zero it follows by continuity, and [F1] gives ${\zeta }^{\mathrm{\prime }}(0)>0$. The interval $J$ is compact by Heine–Borel, and the continuous positive function ${\zeta }^{\mathrm{\prime }}$ attains a positive minimum there by the extreme-value theorem (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set).

Change of variables in the differential equation: inserting $u(x)=a(x)w(\zeta (x))$ into $u^{\mathrm{\prime }\mathrm{\prime }}=qu\mathrm{/}{h}^2$ and using $\zeta ({\zeta }^{\mathrm{\prime }}{)}^2=q$ cancels the first-derivative term and leaves the stated equation with the displayed $R$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$, algebra).

Continuous real functions on compact intervals are Riemann integrable (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion). Their primitives have derivative equal to the integrand, including relative endpoint derivatives (The first fundamental theorem: if $f$ is integrable on $[a,b]$ and continuous at $c$, then $F^{\mathrm{\prime }}(c)=f(c)$; in particular a continuous $f$ has $F$ as a primitive), and for a $C^1$ real function $f$, $f(v)-f(u)={\int }_u^v{f}^{\mathrm{\prime }}(s)ds$ (Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative). All continuous finite-interval integrals here can be computed by Riemann sums, with the same values as their Lebesgue integrals. Applied to $t\mapsto q(tx)$ on $[0,1]$, this gives $q(x)=x{\int }_0^1{q}^{\mathrm{\prime }}(tx)dt$; applied to $s\mapsto \sqrt{\mathrm{\mid }q(s)\mathrm{\mid }}$, it licenses the differentiation in [F2]. For the difference-quotient bounds in [F1], the same identity bounds each increment by the interval length times the stated supremum of its continuous derivative. No choice hypothesis is needed for these scalar identities.