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The following result is well known:

If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$ and $f'$ is non-zero on $(a, b)$ then $f$ is strictly monotone on $[a, b]$.

However if the derivative vanishes at a finite number of points in $(a, b)$ and apart from these derivative maintains a constant sign in $(a, b)$ then also the function is strictly monotone on $[a, b]$ (just split the interval into finite number of intervals using these points where derivative vanishes and $f$ is strictly monotone in same direction in each of these intervals).

Let's suppose now that $f$ is strictly monotone and continuous in $[a, b]$ and differentiable in $(a, b)$. What can we say about set of points $$A = \{x \mid x \in (a, b), f'(x) = 0\}$$ Can it be infinite? Can it be uncountable? How large the set $A$ can be?

Paramanand Singh
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3 Answers3

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It's clear that $A$ must have empty interior. At least for closed sets that's a characterization:

Theorem If $A\subset[0,1]$ is compact with empty interior then there exists a strictly increasing $f\in C^1([0,1])$ such that $A$ is the zero set of $f'$.

Exercise Modify the following proof to show that we can even get $f$ infinitely differentiable.

Proof. Say the connected components of $[0,1]\setminus A$ are $I_1,\dots$. Each $I_n$ is a relatively open interval; there exist $a_n$, $b_n$ with $$I_n\cap(0,1)=(a_n,b_n).$$

Choose $f_n\in C^1(\Bbb R)$ so that $f_n(t)=0$ for all $t\le a_n$, $f(t)=1$ for all $t\ge b_n$, and $f_n'(t)>0$ for all $t\in(a_n,b_n)$. Choose $c_n>0$ so that $$\sum_n c_n\sup_t(|f_n(t)|+|f_n'(t)|)<\infty,$$and let $$f=\sum_n c_nf_n.$$The hypothesis implies that $$f'=\sum_nc_nf_n',$$so $f$ is continuously differentiable and $$\{t\in[0,1]:f'(t)=0\}=A.$$And $f$ is strictly increasing: Say $0\le x<y\le 1$. Since $A$ has empty interior there exists $t\in(x,y)$ with $f'(t)>0$. QED

David C. Ullrich
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  • Very nice theorem and the proof. Since cantor set also has empty interior, it means we can have $A$ uncountable. +1 – Paramanand Singh Jul 01 '16 at 16:02
  • @ParamanandSingh And $A$ could be a "fat Cantor set", so that $A$ can have positive measure. – David C. Ullrich Jul 01 '16 at 16:04
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    I was also looking for $A$ with positive measure and found this nice article https://golem.ph.utexas.edu/category/2010/08/the_difference_between_measure.html which gives the same fat cantor set as example. – Paramanand Singh Jul 01 '16 at 16:07
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On $[0,1]$, consider $$ \frac{\mathrm{d}}{\mathrm{d}x}\frac x{1-x\sin\left(\frac1x\right)}=\frac{1-\cos\left(\frac1x\right)}{\left(1-x\sin\left(\frac1x\right)\right)^2} $$

robjohn
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Consider the function $f(x) = \int_x^1 (\sin(\frac{1}{t})+1) dt$, with $f(0)$ defined so that $f$ is continuous at $0$. This function is $C^1$ on $(0, 1)$, strictly decreasing and continuous on $[0, 1]$, and has zero derivative for every zero of $\sin(\frac{1}{t})+1$, of which there are infinitely many.

Further, consider an arbitrary strictly monotone $C^1$ function $g$ on $[0, 1]$. Sard's theorem tells us that the set of critical values of $g$ has measure zero. And if I'm not mistaken, if we have Lipschitz continuity for $g^{-1}$, we can conclude that the set of critical points must also have measure zero.

I don't know about having uncountably many critical points, though.

florence
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    No, Lipschitz does not imply measure zero. – David C. Ullrich Jul 01 '16 at 15:45
  • @DavidC.Ullrich Does it work if $g$'s inverse is Lipschitz? According to this thread (http://math.stackexchange.com/questions/139883/why-does-a-lipschitz-function-f-mathbbrd-to-mathbbrd-map-measure-zero-s) Lipschitz functions map measure zero sets to measure zero sets, and since $g$ is strictly monotonic, it's invertible. – florence Jul 01 '16 at 15:53
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    If $g$ is increasing and the inverse of $g$ is Lipschitz then $|g(s)-g(t)|\ge c|s-t|$ for some $c>0$, hence $g'\ge c$ everywhere and $A=\emptyset$. – David C. Ullrich Jul 01 '16 at 16:02
  • @DavidC.Ullrich Great observation, thank you. – florence Jul 01 '16 at 16:07