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Let $ A $ be a countable subset of the set of real numbers and $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f'$ is constant on $\mathbb R \setminus A$ , then I know that $f'$ is constant on $\mathbb R$ . My question is ; is it true that for every $c \in \mathbb R$ and uncountable set $B \subseteq \mathbb R$ , there exists a differentiable function $f:\mathbb R \to \mathbb R$ such that $f'(x)=c , \forall x \in \mathbb R \setminus B$ but $f'$ is not constant on $\mathbb R$ ?

  • I'm pretty sure there must be some topological restriction on $B$, not just a cardinality restriction. Maybe think about the case where $B$ is dense and see where that leads. – Ian Jun 30 '15 at 12:53
  • @Ian: What do you mean ? –  Jun 30 '15 at 13:11
  • I am mainly thinking of the fact that a set of continuity points is always a $G_\delta$ set, while a set of finite differentiability points is always a $F_{\sigma \delta}$ set. (For the latter point, cf. http://math.stackexchange.com/questions/905083/characterization-of-sets-of-differentiability ) Returning to your context, if $B$ is dense and $f'(x) \neq c$, then $f'$ is not continuous at $x$. So $B$ must necessarily be a $G_\delta$ set for your situation to occur. I do not know whether $B$ can be any $G_\delta$ set, though. – Ian Jun 30 '15 at 13:14
  • @Ian Why must $B$ be a $G_\delta$? – zhw. Jun 30 '15 at 19:34
  • @zhw. Actually I think that wasn't quite right. First I should change definitions a little bit. Call $C={ x : f'(x)=c }$. Now if $C$ is dense and $x \not \in C$ then $f'$ is not continuous at $x$. But then there might be some points in $C$ where $f'$ is also not continuous. So $C$ itself might not be a $G_\delta$. Hmm... – Ian Jun 30 '15 at 19:49

1 Answers1

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Let $K$ be the Cantor set. Then $K$ is uncountable. Suppose $f:\mathbb {R}\to \mathbb {R}$ is differentiable on $\mathbb {R}$ and $f'(x) = 0$ for all $x\in\mathbb {R}\setminus K.$ Then $f$ is constant on $\mathbb {R}.$

Proof: Clearly $f' = 0$ outide of $[0,1].$ Recall how the Cantor set is constructed: At the $n$th stage, $2^n$ open intervals of length $1/3^n$ are removed. Suppose $(a,b)$ is such an interval. Then $f'=0$ on $(a,b),$ hence $f$ is constant on $(a,b).$ By continuity, $f$ is constant on $[a,b].$ It follows that $f'(a)=f'(b)=0$ as well.

Below all expansions will be ternary. Recall that an open interval of the form

$$\tag 1(.x_1 \dots x_{n-2}2\, 0 \overline 2 , .x_1 \dots x_{n-2} 2 2 \overline 0)$$

is an interval removed at the $n$th stage.

Let $K= A \cup B,$ where $A$ is the set of endpoints of open intervals removed to form $K,$ and $B = K \setminus A.$ Then $B$ is the set of points in $K$ whose expansions have infinitely $0$'s and infinitely many $2$'s.

Suppose now $x = .x_1 x_2\dots \in B.$ Then there will be infinitely many $n$ such that $x_{n-1} = 2, x_n=0.$ For such $n$ define the interval $(a_n,b_n)$ as in $(1).$ Since this is one of the removed intervals, $f(a_n)= f(b_n).$ We get

$$ \frac{f(b_n)-f(x)}{b_n-x}= \frac{f(a_n)-f(x)}{b_n-x}$$

But $a_n-x = 1/3^n,b_n-x = 2/3^n.$ As $n\to \infty$ through the appropriate sequence of $n$'s, we conclude $f'(x) = 2f'(x).$ It follows that $f'(x)=0.$ This shows $f'\equiv 0$ on $\mathbb {R}.$

zhw.
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