Is there any function $$f:\mathbb R \rightarrow \mathbb R$$ such that it has right derivative but not left derivative at every point?

  • $\begingroup$ What are your thoughts? What have you tried? What is the context of this problem? It reads remarkably similar to the last one that was answered here. $\endgroup$ – Sloan Jul 10 '15 at 15:20
  • $\begingroup$ not that similar in my very own very humble very opinion $\endgroup$ – Jorge Fernández Hidalgo Jul 10 '15 at 15:20
  • 2
    $\begingroup$ This seems relevant. $\endgroup$ – David Mitra Jul 10 '15 at 15:24
  • $\begingroup$ First one was easier than this. I thought same in the answer but couldnt be sure. I also wonder what people think about them. I am learning in basic. These are good for me. But for this one, it's seen that there is no such function. But this is not mathematical. $\endgroup$ – 153 Jul 10 '15 at 15:25

Let $q_1,q_2,\ldots$ be an enumeration of the rational numbers in $[0,1]$, and

$$ f_n(x) = \left\{\begin{array}{rl}1 & \text{if } x\geq q_n,\\ 0 & \text{otherwise}.\end{array}\right.$$ If we consider $$ f(x) = \sum_{n\geq 1}\frac{1}{2^n}\,f_n(x) $$ then $f(x)$ has a right derivative for every $x\in[0,1]$ but no left derivative exists if $x\in\mathbb{Q}\in[0,1]$.

Now we may try to condensate singularities, but due to Denjoy-Young-Saks theorem, we cannot achieve the non-existence of the left derivative over a subset of $[0,1]$ with positive measure.

  • $\begingroup$ According to this (as noted in my link above), you won't get positive measure. I wonder if you can get uncountable. $\endgroup$ – David Mitra Jul 10 '15 at 15:53

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