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Suppose $f_1, \dots, f_n$ are real-valued real analytic functions defined on an open set $B=(0,1)^d$ of $\mathbb{R}^d$. For $r \in \mathbb{R}$, let $S_r$ be the sub-level set in $B$ defined by the system of inequalities $$f_1(x) <r, \dots, f_n(x)<r,$$ and denote by $V(r)$ the volume of the set $S_r$. One generally cannot expect $V(r)$ to be a real analytic function of $r$; take, for instance, $B=(0,1)$ and $f_1(x)=x$.

It seems to me that if one somehow avoids "incoming" or "outgoing" conditions as $r$ varies, then $V(r)$ should depend real-analytically on $r$. Can such a statement be precisely formulated and proved, and is there a reference for it?

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    $\begingroup$ The problem should be related to Sard's lemma. The regular values of the combined map $(f_1,\dots,f_n)$ should be ones where the level sets cut out smoothly varying manifolds with corners, so smoothly varying integrals, I guess. $\endgroup$
    – Ben McKay
    Commented Nov 15, 2021 at 13:44

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In your setting, it seems that your function $V(r)$ depends too much on the shape of the domain $B$. Take for instance $B= \{0\leq y \leq h(x)\} \subset \mathbb R^2$ for a continuous but very wild positive function $h$ and $f_1(x,y) = x$. Then $$V(r) = \int_0^r h(x) \mathrm d x$$ which is far from analytic.

So let me change the setting a little bit: assume $B$ is $[0,1]^n$ and the $f_i$ are analytic in a neighbourhood of $B$. Then it seems to me that $o$-minimality (an more precisely, the fact that $\mathbb R_{an}$ is an $o$-minimal structure) might give some answer to your question. I am not really familiar with this, but the argument would go along the following lines:

Fact 1: The function $$(x,r) \mapsto 1_{x\in B \textrm{ and }\forall i, f_i(x)<r}$$ is definable in $\mathbb R_{an}$.

I think this is ok, or I completely misunderstood the notion of definable function.

Guess 2: If $g(x,r): \mathbb R^n \times \mathbb R \to \mathbb R$ is definable in $\mathbb R^{an}$, then $$r\mapsto \int_{\mathbb R^{n}} g(x,r) \mathrm d x$$ is definable in $\mathbb R_{an}$. In particular, $r\mapsto V(r)$ is definable in $\mathbb R^{an}$.

I have no idea if this is true, but this probably has been considered before.

Guess 3: $V:\mathbb R \to \mathbb R$ definable in $\mathbb R_{an}$ implies that $V$ is piecewise analytic.

Again, I'm not sure if this is true, but this is the kind of statement that $o$-minimality implies.

Perhaps someone with more knowledge on $o$-minimality can tell us if this is plausible?

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  • $\begingroup$ Thanks, Nicolas, for the answer and suggested modification. In my setting $B$ is not a product of intervals, but is itself intersection of sub-level sets of real analytic functions. So, I assume if your argument works in the case of the product of intervals, it must work in my setting too. I will add "o-minimal" as a tag, hoping it catches someone's interest. $\endgroup$
    – Keivan Karai
    Commented Nov 15, 2021 at 14:22
  • $\begingroup$ Guess 2 would be false with $\mathbb{R}^{an}$ replaced by $\mathbb{R}^{alg}$, since the integral of an algebraic function may not be algebraic. On the other hand, you might divide the analytic region into cells, and take the integral in each cell, since the integrals of analytic functions are analytic. $\endgroup$
    – user44143
    Commented Dec 1, 2021 at 13:39

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