Does any real function have a Lipschitzian restriction on $D$, where $D$ is an infinite subset of $\Bbb R$ with an accumulation point?
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$\begingroup$ For example, for $D$ we can take any convergent sequence (together with its limit point)? $\endgroup$– Taras BanakhNov 25 '19 at 10:24

2$\begingroup$ So the problem boils down to: GIven an arbitrary real function $f$, find a point $a$ and a sequence $(a_n)_{n=1}^\infty$ converging to $a$, $a_n \ne a$, so that $f$ is Lipschitzian on $D = \{a_n\}_{n=1}^\infty \cup \{a\}$. $\endgroup$– Gerald EdgarNov 25 '19 at 11:55

2$\begingroup$ Consider the Cantor set $C=\big\{\sum_{n=1}^\infty\frac{2x_n}{3^n}:(x_n)_{n\in\mathbb N}\in\{0,1\}^{\mathbb N}\big\}$ and the nondecreasing continuous function $f:C\to[0,1]$, $f:\sum_{n=1}^\infty\frac{2x_n}{3^n}\mapsto \sum_{n=1}^\infty\frac{x_n}{2^n}$. It is easy to check that for any nondiscrete subspace $D\subset C$ the restriction $f{\restriction}D$ is not Lipschitz. $\endgroup$– Taras BanakhNov 26 '19 at 8:15

1$\begingroup$ I suspect that, at least, it is consistent that the answer is no. When you force with finite partial functions $\mathbb{R} \to 2$, you add a function $f$ from the former $\mathbb{R}$ to $2$ that is a counterexample. However $f$ is not defined on the new real numbers. But wouldn't it be possible to iterate this process $\omega$ times to get a function defined on the whole new $\mathbb{R}$? I have no experience in iterated forcing so it's difficult for me to answer... $\endgroup$– N. de RancourtNov 30 '19 at 13:01

2$\begingroup$ What do you mean by "Lipschitzian restriction" on $D$? Do you mean that there is a constant $C$ such that for every $x, y \in D$, $f(x)  f(y) \leq C x  y$? Then the answer is yes (by DenjoyYoungSaks theorem) and, in fact, much stronger results hold. See for example, webhome.auburn.edu/~brownj4/tatras.pdf $\endgroup$– AshutoshDec 1 '19 at 9:07
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