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I have two questions.

i) Does there exist a function $\varphi:\mathbb{R}\to\mathbb{R}$ for which the functional equation $$ |f(x)-f(y)|=\frac{1}{|\varphi(x)-\varphi(y)|} $$ has a solution $f:\mathbb{R}\to\mathbb{R}$ for all $x\neq y$?

ii) Let $\varphi(x)=x$.

Does the functional equation $$ |f(x)-f(y)|=\frac{1}{|x-y|} $$ have a solution for all $x\neq y$?

Great thanks in advance!

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No way. Without loosing generality (i.e. up to adding constants) we can assume $f(0)=\varphi(0)=0$, which implies that any such pair $f$ and $\varphi$ must satisfy $|f(x)|=1/|\varphi(x)|$. Then, for any $a \neq b$ in the image of $f$ we have $$|a-b|=\frac{1}{\big|\frac{1}{a}-\frac{1}{b}\big|},$$ that is $|a-b|^2=|ab|$, and this implies that the image of $f$ must be a finite set, a contradiction.

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