Exercises on analytic preparation

Written by Claude Opus 5.5 (Anthropic), October 2026; self-checked by the writing AI. Original text: public domain (CC0). The function in Exercise 15 is an example from Guillaume Valette, On subanalytic geometry, arXiv:2507.23622v1.

These two exercises use the cell preparation and Puiseux theorems proved in Analytic finiteness for preparation.

15. A bounded function that needs a negative preparation exponent

Level: advanced. Let $0<\epsilon<1$,

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

and $f(x,y)=x^3/y$ on $C$. Show that $f$ is bounded, but that no finite preparation of $f$ over a cell decomposition of $C$ has only nonnegative exponents in $y$, This remains true when the preparation may translate $y$ and use analytic units. The convergent Puiseux theorem of that reading is needed.

Solution. Since $y>x^2$, we have $0<f<x<\epsilon$. Note that $f=x^3y^{-1}$ is itself a preparation, with exponent $-1$.

Assume a finite preparation with only nonnegative exponents exists. After refining and shrinking $\epsilon$, the base cell next to $0$ is $(0,\epsilon)$. Above it, finitely many definable functions

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

cut $C$ into bands $\{\alpha_j<y<\alpha_{j+1}\}$, together with graphs. On each band

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

where the graph of the definable function $\theta$ does not meet the band. The convergent Puiseux expansion of $\alpha_j$ starts with $c_jx^{\nu_j}$, $c_j>0$, and $x^2\le\alpha_j\le x$ gives $1\le\nu_j\le2$. As $\nu_0=2$ and $\nu_N=1$, some $j$ has $\nu_j>\nu_{j+1}$. Put $\alpha=\alpha_j$ and $\beta=\alpha_{j+1}$; then $\alpha/\beta\to0$ as $x\to0$.

On this band $a(x)\ne0$, because $f>0$. After shrinking the base interval, $\theta$ is continuous and stays on one side of the band: $\theta\le\alpha$ or $\theta\ge\beta$. For small $x$ the points $y_1=2\alpha$ and $y_2=\beta/2$ satisfy $\alpha<y_1<y_2<\beta$, and

$$\frac{f(x,y_2)}{f(x,y_1)}=\frac{y_1}{y_2}=\frac{4\alpha}{\beta}\longrightarrow0 .$$

In the prepared form the same ratio is at least $(m/M)\bigl(|y_2-\theta|/|y_1-\theta|\bigr)^r$. If $\theta\le\alpha$, the distance ratio is at least $1$. If $\theta\ge\beta$, then

$$\frac{\theta-y_2}{\theta-y_1}\ge\frac{\beta-y_2}{\beta-y_1}\ge\frac{\beta/2}{\beta}=\frac12,$$

because $t\mapsto(t-y_2)/(t-y_1)$ increases for $t>y_2$. In both cases the ratio is at least $2^{-r}m/M>0$, a contradiction. So a negative exponent is needed. Boundedness does not prevent this: on bands that narrow towards $x=0$, the coefficient $a(x)$ absorbs the inverse power.

16. Even substitutions and parameter-dependent poles

Level: intermediate. Why does $t=\tau^2$ not make $\sqrt t$ analytic near $\tau=0$, while $t=\tau^4$ does? Then consider the globally subanalytic function

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

Find a substitution $t=\tau^k$ under which $f$ becomes a Laurent polynomial in $\tau$ for $\tau>0$, and one under which this expression agrees with $f$ for both signs of $\tau$. On which pieces of the parameter line does $f(x,\cdot)$ extend continuously, and analytically, in the original variable $t$ at zero? Answer the analytic-extension question separately for the composed function in the new variable $\tau$.

Solution. We have $\sqrt{\tau^2}=|\tau|$, which is not analytic at $0$, whereas $\sqrt{\tau^4}=\tau^2$ is. For a two-sided substitution it is not enough that the substitution exponent be even: every exponent obtained after clearing denominators must be even.

The exponents $-1/2$ and $1/3$ have common denominator $6$. With $t=\tau^6$ and $\tau>0$,

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

Doubling the exponent, $t=\tau^{12}$ gives $t^{-1/2}=|\tau|^{-6}=\tau^{-6}$ and $t^{1/3}=\tau^4$ for every $\tau\ne0$, so

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

agrees with $f$ for both signs of $\tau$.

If $x\ne0$, then $f(x,t)\to\pm\infty$ as $t\downarrow0$, with the sign of $x$, so there is no continuous extension before or after substitution. If $x=0$, then $f(0,t)=t^{1/3}$ extends continuously by zero but has no real analytic extension in $t$, since its difference quotient at zero is $t^{-2/3}$, which diverges. After ramification, $f(0,\tau^{12})=\tau^4$ is analytic across $\tau=0$. Thus only the parameter piece $\{0\}$ permits continuous extension in $t$, no parameter permits analytic extension in $t$, and exactly $\{0\}$ permits analytic extension of the ramified function. The partition $(-\infty,0)$, $\{0\}$, $(0,\infty)$ separates these behaviours. The extension is not jointly continuous at $(0,0)$: along $x=t^{1/4}$, $f=t^{-1/4}+t^{1/3}\to\infty$. Continuity in $t$ on the single slice $x=0$ is weaker than the joint continuity that a parameter theorem requires as its hypothesis.