Let $E\subset [0,1]$ with $\mu(E)=0$. Does there exist a continuous, strictly increasing function $f$ on $[0,1]$ so that $f'(x)=\infty$ for all $x\in E$ (in Lebesgue sense)?
I think there exist such a function, but I don't know how to construct.
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$\begingroup$ Is E given and you want to construct f or do you just need an example of such an E and f? Do you expect something from f? Differentiable everywhere away feom E ...? $\endgroup$– YannickSSECommented May 10, 2016 at 15:10
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$\begingroup$ @user3808066 E is given and I need to construct an f for it. f is differentiable at every points since it is a strictly increasing function. $\endgroup$– Peter LiuCommented May 10, 2016 at 15:12
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$\begingroup$ Strictly increasing only implies allmost everywhere differentiability. You can have edges. $\endgroup$– YannickSSECommented May 10, 2016 at 15:14
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$\begingroup$ And what does the in "Lebesque sense" mean? It is allready a condition on a zero set. $\endgroup$– YannickSSECommented May 10, 2016 at 15:16
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1$\begingroup$ See my 18 April 2007 sci.math post (follow-up comment here) and the stackexchange question Absolute continuous function with given set of discontinuities of derivatives.. $\endgroup$– Dave L. RenfroCommented May 12, 2016 at 14:39
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1 Answer
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I couldn't figure how the general case works but if $E$ is countable consider $E=\{q_n\;:\;n\in\mathbb{N}\}$ and take $f$ a function which is $0$ on $(-\infty,-1/2]$, $1$ on $[1/2,\infty)$, monotonic and has a derivative $f':\mathbb{R} \to [0,\infty]$ with $f'(0)=\infty$, additionally take a sequence $(a_n)_{n\in\mathbb{N}}\in \ell^1$, $a_n>0$. Then the function $$ g(x) = \sum_{n\in\mathbb{N}} a_nf(x-q_n) $$ is continuous and monotone. Further since for all $n$ $a_nf'(x-q_n)$ is a non negative function $$ g(x) = \int_{-1}^x \sum_n a_n f'(s-q_n)\mathrm{d}s $$ and hence at every $q_m$ $$ g'(x) = f'(0) + \sum_{n\not= m} f'(q_m-q_n) =\infty. $$
