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I want to understand a bit about the weak derivative of singular function Let u be singular. Using wiki to define the concept of singular functions I denote by $N⊂R$ a set such that $u$ has a classical derivative $u'=0$ on $N^c$ and $N$ has measure zero. To compute the weak derivative let $ψ$ be a test function: $$(u',ψ)=-(u,ψ' )=-∫_R uψ' dx=-∫_{N^c} uψ' dx-∫_N uψ' dx$$ **Assume that $N^c$ is just made of decreasingly small intervals so i can integrate by parts again to get: $$(u',ψ)=∫_{N^c} u'ψdx-∫_N uψ' dx$$ Now by definition $u'=0$ on $N^c$ an thus the first integral is zero. The second Is zero to since we integrate a continues function over a set of measure zero. Thus in the weak sense: $$(u',ψ)=0⇒u'=0$$ I already have read this but I just want clarification. Clearly I am doing something wrong since $$(T',ψ)=0⇒T=c$$ Q1. I suspect that the mistake may be at ** but I have no idea if I’m right or not. The best I can come up with is: $N^c$ is obtained by some limiting process which renders integration by parts wrong.

Q2. what is the actual derivative of a Cantor function in the weak sense ? Is it like a Dirac-delta ?

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    $\begingroup$ Q1: mistake is by forgetting boundary values when integrating by parts ... $\endgroup$ Commented May 13 at 16:05
  • $\begingroup$ Q2: the derivative of the Cantor function is singular with respect to Dirac measures as well as to the Lebesgue measure ... It is something else! In my mind, I imagine it as a lot of very small deltas on the Cantor set $\endgroup$ Commented May 13 at 16:10
  • $\begingroup$ The cantor function does not have a weak derivative. $\endgroup$ Commented May 13 at 16:36
  • $\begingroup$ @JonathanBeer. Why doesn't the Cantor function have a weak derivative? $\endgroup$ Commented May 13 at 17:15
  • $\begingroup$ @md2perpe perhaps we are using terms differently, but weak derivatives need to be L1 functions. There's something called the distributional derivative that lets them be distributions too. Iirc the distributional derivative of the cantor function is the $\log_3(2)$ dimensional Hausdorff measure (maybe restricted to the cantor set? Something like that), but that's not the weak derivative. $\endgroup$ Commented May 13 at 18:23

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