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Let $f: \left [ a,b \right ] \to \mathbb{R}^2$. And suppose that for all $\varepsilon>0$ there exists $\delta > 0$ such that $\forall x_1, x_2\in \left [ a,b \right ]: \left|x_1 -x_2\right| < \delta \Rightarrow \left \|f(x_1) -f(x_2)\right \|<\varepsilon \left|x_1 -x_2\right|$.

Show that $f \left ( \left [ a,b \right ] \right )$ is of Lebesgue measure zero.

They call this form of continuity "Equicontinuity", but I have never heard this term used for a single function, only for a family of functions.

Me and a friend tried this for some time and couldn't solve it. It probably involves finding a rectangular cover of $f \left ( \left [ a,b \right ] \right )$, but I don't know enough properties of this "Equicontinuity" nor do I have enough intuition to work with it. My friend says he didn't even find non-constant functions that satisfy it. Any ideas?

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  • $\begingroup$ Is it $f:[a,b]\rightarrow \mathbb{R}$ or $\mathbb{R}^2$? The title and question do not match. $\endgroup$ – Mathematician 42 Dec 15 '17 at 10:09
  • $\begingroup$ $\mathbb{R}^2$. fixed. $\endgroup$ – Bary12 Dec 15 '17 at 10:10
  • $\begingroup$ Hmm, good question, intuitively this continuity condition will imply that the image is at most 'one-dimensional', by that I mean something that looks like a curve, which obviously has zero measure as it has no area. But I need to think about the details of showing that. $\endgroup$ – Mathematician 42 Dec 15 '17 at 10:13
  • $\begingroup$ Are you sure the absolute value behind the epsilon belongs there? $\endgroup$ – Bananach Dec 15 '17 at 10:24
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    $\begingroup$ @G.S. I am aware of that, but I suspect all these 'space-filling' curves are not equicontinuous. As shown in the answer below, this condition is quite restrictive. But you are right, my comment is perhaps misleading. $\endgroup$ – Mathematician 42 Dec 15 '17 at 10:39
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Let $f=(f_1,f_2)$ . Show that the $ \epsilon - \delta$ - property gives that $f_1$ and $f_2$ are differentiable and $f_1'=f_2'=0$ on $[a,b]$.

Hence $f$ is constant.

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  • $\begingroup$ Nice one, didn't realize that! Clearly equicontinuity is quite a strong condition. $\endgroup$ – Mathematician 42 Dec 15 '17 at 10:19
  • $\begingroup$ Wow, very nice!! (+1) $\endgroup$ – Shashi Dec 15 '17 at 10:20

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