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Let $f : \mathbb{R}\rightarrow \mathbb{R}$ be a function such that $| f(x)-f(y)| \leq 4321|x-y|$ for all $x,y \in \mathbb{R}$. Then which one of the following is true.

  1. $f$ is always differentiable.
  2. There exists at least one such $f$ that is continuous and satisfies $\displaystyle{\lim_{x \to \pm\infty}\frac{f(x)}{|x|}} = \infty$.
  3. There exists at least one such $f$ that is continuous but not differentiable at exactly $2018$ points and satisfies $\displaystyle{\lim_{x \to \pm\infty}\frac{f(x)}{|x|}} = 2018$.
  4. It is not possible to find a sequence $\{x_n\}$ of reals such that $\displaystyle{\lim_{n \to \infty} x_n = \infty}$ and further satisfying $\displaystyle{\lim_{n \to \infty}|\frac{f(x_n)}{x_n}|} \leq 10000$.

My argument: It is known that a Lipschitz map is uniformly continuous and differentiable almost everywhere. Thus option one is eliminated. Also since $\frac{f(x)}{x}$ is bounded, option 2 and 4 can be eliminated. But I cannot establish the necessary argument for option 3.

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  • $\begingroup$ For (3), get a function $g(x) = 2018 |x| $. The behavour on infinity is ok, now you are free to make whatever you want with this function on any compact. E.g. look what happens with the function $| g(x) -1|$. $\endgroup$ Oct 27, 2021 at 9:55

1 Answer 1

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To elimnate 1) you have to give an example. $f(x)=|x|$ will do.

$f(x)=c \sum\limits_{k=1}^{2018} \frac {|x-k|} {2^{k}}$ satsifies 3) if $c$ is chosen so that $c \sum\limits_{k=1}^{n} \frac {1} {2^{k}}=2018$.

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  • $\begingroup$ It will be further helpful if you kindly elaborate on choosing this function. $\endgroup$ Oct 27, 2021 at 12:26
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    $\begingroup$ @debabratachakraborty I just built $f$ on the fact that $|x-c|$ is differentiable at points except $c$. So our $f$ is differentiable at all points except $1,2,...,2018$. We can make the limit of $\frac {f(x)} x$ equal to any number by choosing $c$ appropriately. $\endgroup$ Oct 27, 2021 at 12:40

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