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Suppose that $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function, and that on some interval $(a,b)$, $|f'|\leq1$. Is it true that for all measurable sets $E\subset(a,b)$, $\lambda(f(E))\leq\lambda(E)$? (Here $\lambda$ denotes Lebesgue measure.) The intuition is that a small derivative means that $f$ is "shrinking sets" in some sense.

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  • $\begingroup$ The answer is yes, but I can't write a detailed explanation right now. You can probably find a writeup if you look up something like "Lipschitz functions are absolutely continuous", or perhaps "Lipschitz functions have the Lusin N property". $\endgroup$ – Ian Jul 20 '15 at 3:32
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    $\begingroup$ Since $E\subset\mathbb{R}$ is measurable, we can suppose that for any given $\epsilon > 0$, $E\subset U_\epsilon$ for some open set $U_\epsilon$ such that $\lambda(U_\epsilon \setminus E) < \epsilon$. While we can write $U_\epsilon$ as a disjoint union of open intervals, $$U_\epsilon = \bigcup_{n=1}^\infty I_n$$ therefore, $$\lambda\big( f(E) \big) \leq \lambda(U_\epsilon) \leq \sum_{n=1}^\infty \lambda\big( f(I_n) \big) \leq \sum_{n=1}^\infty \lambda (I_n) = \lambda(U_\epsilon) $$ (f is Lip-conti) Then $\lambda\big( f(E) \big)\leq \lambda(E) + \epsilon$. But $\epsilon > 0$ is arbitrary. $\endgroup$ – Chival Jul 20 '15 at 3:47
  • $\begingroup$ Thanks @Chival, A little too long since measure theory for me I guess. $\endgroup$ – Trevor J Richards Jul 20 '15 at 4:04
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    $\begingroup$ @Chival: It might be good to post your comment as answer. $\endgroup$ – Prahlad Vaidyanathan Jul 20 '15 at 4:56
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I don't remember much about neasure theory. This is a proof when $E$ is a closed interval. It should be helpful, together with Chival's solution, that assumes this one proved.

So, let $E=[c,d]\subset(a,b)$. Since $f$ is differentiable, it is also continuous. Weierstrass' extreme value theroem says that $f$ obtains its maximum $M$ and its minimum $m$ on $[c,d]$. Say, $f(r)=M$ and $f(s)=m$, where $r$ and $s$ are in $[c,d]$. Continuity also guarantees that $f$ reaches every point in $[m,M]$.

Then, $\lambda(f(E))=M-m$ and mean value theroem says that there exists some $t\in[c,d]$ such that $f'(t)(r-s)=M-m$. Therefore

$$\lambda(f(E))=|f'(t)|\cdot|r-s|\le1\cdot|d-c|=\lambda(E)$$

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  • $\begingroup$ Thanks for the special case of a closed interval, interesting to see the application of the MVT. It looks like Chival's comment covers the general case. $\endgroup$ – Trevor J Richards Jul 20 '15 at 4:01
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Yes, the (absolute value/ norm) Jacobian , here the derivative gives you a measure of the scaling, or change of volume of sets resulting from a transformation. Since ||J(f)||=|f'(x)|<1 , the sets are non-expanding.

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