Measuring almost-critical values of smooth functions. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:38:24Z http://mathoverflow.net/feeds/question/109799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109799/measuring-almost-critical-values-of-smooth-functions Measuring almost-critical values of smooth functions. Vidit Nanda 2012-10-16T09:33:47Z 2012-10-16T11:17:58Z <p>Consider a compact sub-manifold $X \subset \mathbb{R}^n$ of Euclidean space and let $f:X \to \mathbb{R}$ be any smooth function. Recall that $x \in X$ is a <em>critical point</em> of $f$ if the gradient $\nabla_x f$ is identically zero, and in this case $f(x) \in \mathbb{R}$ is called a <em>critical value</em> of $f$. It is <a href="http://en.wikipedia.org/wiki/Sard%27s_theorem" rel="nofollow">well-known</a> by Sard's theorem that the set of critical values of $f$ has Lebesgue measure zero as a subset of $\mathbb{R}$. </p> <blockquote> <p>Is there a "fuzzy" version of this theorem for a suitable class of smooth functions?</p> </blockquote> <p>More precisely, define the monotone function $M_f : \mathbb{R}^+ \to \mathbb{R}^+$ as follows: $M_f(\epsilon)$ is the Lebesgue measure of the set $$\lbrace p \in \mathbb{R} \mid p = f(x) \text{ for some } x \in X \text{ with }\|\nabla_x f \| &lt; \epsilon\rbrace.$$ By Sard's theorem, $M_f(0) = 0$ for any smooth $f$.</p> <p>Here's the question:</p> <blockquote> <p>Can we classify those smooth $f:X \to \mathbb{R}$ for which $M_f$ is continuous on its domain of definition in general (and in particular, continuous from above at $0$)?</p> </blockquote> <p>Essentially, we know that the measure of values of $f$ where the gradient of the pre-image equals zero is zero. When the gradient comes within $\epsilon$ of zero, can we bound the measure by a continuous function of $\epsilon$?</p> http://mathoverflow.net/questions/109799/measuring-almost-critical-values-of-smooth-functions/109806#109806 Answer by Pietro Majer for Measuring almost-critical values of smooth functions. Pietro Majer 2012-10-16T11:17:58Z 2012-10-16T11:17:58Z <p>It true that the function $M_f: \epsilon\mapsto \operatorname{meas} f\{\|\nabla f(x)\| &lt; \epsilon\}$ is left-continuous (that is, since it is in any case an increasing function, lower semi-continuous); in particular continuous at $\epsilon=0$.</p> <p>Indeed, if $\epsilon_n\ge0$ is an increasing sequence converging to $\epsilon\ge0$ one has </p> <p>$$f\{\|\nabla f(x)\| &lt; \epsilon\}=\bigcup_{n\in\mathbb{N}} f\{\|\nabla f(x)\| &lt; \epsilon _ n\}\ ,$$ so $\operatorname{meas} f\{\|\nabla f(x)\| &lt; \epsilon _ n\}\to \operatorname{meas} f\{\|\nabla f(x)\| &lt; \epsilon \}$. However it is not right-continuous in general. Think of the case $X:=\mathbb{S}^1=\mathbb{R}/\mathbb{Z}$, with a 1-periodic function $f:\mathbb{R}\to\mathbb{R}$ with a minimum value $f(0)=0$, and a maximum value $f(1/2)$, and such that $|f'(x)|=1$ everywhere but in a nbd $U$ of small measure $\delta &lt; f(1/2)/2$ of the maximum and minimum points, where $|f'(x)|&lt;1$. In this situation, $f\{| f'(x)| &lt; 1\}=f(U)$ has measure at most $2\delta$, whereas $f\{| f'(x)| \le 1\}=f([0,1])=f(1/2) .$ Therefore $M_f$ can't be continuous at $\epsilon=1$.</p>