Exercises on analytic preparation

These two exercises and complete solutions use the proved analytic preparation and Puiseux treatment. The inverse-power example comes from Guillaume Valette’s On subanalytic geometry; its finite-refinement obstruction and the parameter/parity exercise are expanded here. The Valette component and these identified teaching additions are available under CC BY 4.0. Adapted by GPT-6.1 Sol (OpenAI), Ultra, October 2026; no endorsement is implied.

15. A bounded function that needs a negative preparation exponent

Level: advanced. On

$$C=\{(x,y):0<x<\epsilon,\ x^2<y<x\},\qquad 0<\epsilon<1,$$

consider $f(x,y)=x^3/y$. It is positive and bounded. Show that a finite preparation of $f$ cannot have only nonnegative last-coordinate exponents, even if translations and analytic units are allowed. Use the convergent Puiseux theorem proved in the analytic preparation treatment.

Solution. We have $0<f<x<\epsilon$. The formula $f=x^3y^{-1}$ is already a reduction with translation zero, unit one and exponent $-1$.

Suppose there is a finite cell preparation with all exponents nonnegative. Refine its one-dimensional base and shrink $\epsilon$ so its base interval next to zero is $(0,\epsilon)$. Over it, list the finitely many endpoints within the original band, including $x^2$ and $x$:

$$x^2=\alpha_0(x)<\alpha_1(x)<\cdots<\alpha_N(x)=x.$$

Duplicate endpoints are removed; every open interval between consecutive endpoints is a preparation cell. Each positive endpoint has a convergent Puiseux expansion with leading term $c_jx^{\nu_j}$, where $c_j>0$. The bounds $x^2\le\alpha_j\le x$ force $1\le\nu_j\le2$. Since $\nu_0=2$ and $\nu_N=1$, some consecutive pair $\alpha<\beta$ has strictly decreasing leading order. Thus $\alpha/\beta\to0$.

On the corresponding band, write the supposed reduction as

$$f(x,y)=a(x)|y-\theta(x)|^rU(x,y),\qquad r\ge0, \qquad 0<m\le|U|\le M.$$

The coefficient cannot vanish, since $f>0$. The center is outside the band, hence either $\theta\le\alpha$ or $\theta\ge\beta$ throughout this connected base interval. After shrinking it, take

$$y_\ell=2\alpha,\qquad y_h=\beta/2, \qquad \alpha<y_\ell<y_h<\beta.$$

The actual ratio is

$$\frac{f(x,y_h)}{f(x,y_\ell)} =\frac{4\alpha(x)}{\beta(x)}\longrightarrow0.$$

If $\theta\le\alpha$, the distance ratio is at least one, so the prepared ratio is at least $m/M$. If $\theta\ge\beta$, then

$$\frac{\theta-y_h}{\theta-y_\ell}\ge\frac12,$$

so the prepared ratio is at least $2^{-r}m/M$. Both contradict its limit zero. Some negative exponent is therefore necessary. Boundedness of the whole function does not force nondegenerate preparation; its base coefficient can compensate for an inverse power on a narrowing band.

16. Even substitutions and parameter-dependent poles

Level: intermediate. Explain why the substitution $t=\tau^2$ does not make $\sqrt t$ analytic on a two-sided neighborhood of zero, while $t=\tau^4$ does. Then analyze the globally subanalytic function

$$f(x,t)=xt^{-1/2}+t^{1/3}, \qquad x\in\mathbb R,\quad 0<t<1.$$

Give one common root-variable Laurent expansion and one even substitution that agrees with the actual function for both signs of the new variable. Determine the parameter pieces on which a continuous analytic extension at zero is possible.

Solution. The two compositions are $|\tau|$ and $\tau^2$. A two-sided analytic substitution must make every cleared exponent even, not merely make the substitution exponent even.

With $t=\tau^6$ on the positive $\tau$-axis the Laurent expansion is

$$f(x,\tau^6)=x\tau^{-3}+\tau^2,\qquad \tau>0.$$

For both signs, choose $t=\tau^{12}$. Then

$$f(x,\tau^{12})=x\tau^{-6}+\tau^4,\qquad \tau\ne0.$$

For $x\ne0$ the pole remains; no continuous extension is possible there. On the parameter cell $\{0\}$, the function becomes $\tau^4$, analytic across zero. The cells $(-\infty,0),\{0\},(0,\infty)$ give one finite analytic parameter partition.

Even at $(x,t)=(0,0)$, continuity fails if one keeps all parameters together. Along $x=t^{1/4}$, the first summand equals $t^{-1/4}$ and diverges. The continuous-parameter theorem requires joint continuity on its stated neighborhood; continuity on the one slice $x=0$ does not supply that hypothesis.