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Today,my frend ask this follow question:and I consider sometime,and I can't solve it.

I hope see someone can help me

Question:

let $f$ is continuous strictly increasing function,

prove or disprove :the $f$ Non-differentiable points countably infinite？

since the contably infinite define:

http://en.wikipedia.org/wiki/Countable_set

and I find somebook,and I found this enter image description here

This follow famous example(Pringsheim) is $f$ is continuous strictly increasing function,and there exsit a point not differentiable $$\begin{cases} x(1+\dfrac{1}{3}\sin{(\ln{x^2})}&x\neq 0\\ 0&x=0 \end{cases}$$

we can show $f$ is strictly increaing and continuous on $R$,but $x=0$,$f$ is not differentable.

But My question:

there exsit $f$ such strictly increaing and continuou,but have infinite point not differentable?

I can't have coumexample,Thank you very much!

math110
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2 Answers2

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Take a look at this Essay on non-differentiability points of monotone functions, by Dave L Renfro, for a wealth of excellent information on the topic.

In particular, Renfro mentions that if $f$ is monotone, then the set of points where $f'(x)$ does not exist has Lebesgue measure zero, and this is sharp even if $f$ is continuous, in the sense that for any set $E\subseteq[a,b]$ having measure zero there is a strictly increasing continuous function $f:[a,b]\to\mathbb R$ with $f'(x)=+\infty$ at all $x$ in $E$.

The example he refers to, from A. M. Bruckner, J. B. Bruckner, and B. S. Thomson, Real analysis (downloadable here), appears as Theorem 7.9 in page 460 in their current edition. Here is a brief sketch: Let $E$ be a set of measure zero, and (for $n=1,2,\dots$) let $G_n$ be open sets covering $E$ with measure $\lambda(G_n)<1/2^n$. If $g_n(x)=\lambda(G_n\cap[a,x])$, then $g_n$ is nondecreasing, continuous, and $0\le g_n\le 1/2^n$. Now let $$ g(x) =\sum_n g_n(x), $$ so $g$ is increasing, $0\le g\le 1$, and $g$ is continuous (by the $M$-test). If $f(x)=g(x)+x$, then $f$ is strictly increasing. Also, at any point of $E$, $f'(x)=\infty$ since $g'(x)=\infty$, because $g_n'(x)=1$ for all $n$.

Andrés E. Caicedo
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The Cantor function has uncountably many points of nondifferentiability. By adding $x$, you get it to be strictly increasing.

Mikhail Katz
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  • No,I remember cantor function differentiability everywhere point.and you link can't open,Thank you – math110 Dec 08 '13 at 16:32
  • It is nondifferentiable at uncountably many points of the Cantor set. The link opens for me. Try https://en.wikipedia.org/wiki/Cantor_function – Mikhail Katz Dec 08 '13 at 16:46
  • @chinamath An excellent reference on the function is The Cantor function, by Dovgoshey, Martio Ryazanov, and Vuorinen, Expo. Math., 24, (2006), 1-37. As user72694 indicates, the function has uncountably many points of nondifferentiability. (Much stronger properties than just nondifferentiability are discussed in the paper, see particularly section 8.) – Andrés E. Caicedo Dec 08 '13 at 16:59