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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be smooth and such that for every $x$ $$\int_{-\infty}^{\infty}\frac{|f(x)-f(y)|}{|x-y|^2}dy < \infty$$ Prove that $f$ is constant.

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    $\begingroup$ Two observations, not sure if useful: the integrand is even, so this is equivalent to saying $$\int_x^\infty \frac{|f(x)-f(y)|}{(x-y)^2} dy < \infty.$$ That $f$ is constant forces the integral on the LHS to be always $0$ $\endgroup$ – gt6989b Jan 21 at 4:19
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If $f$ is not constant, then $f'(a)\ne 0$ for some $a.$ Thus for some $\delta >0,$ $|f(y)-f(a)|/|y-a|>|f'(a)|/2$ for $a<y<a+\delta.$ Taking $x=a,$ the integral of interest is at least

$$\int_{a}^{a+\delta}\frac{|f(a)-f(y)|}{|a-y|^2}dy \ge \frac{|f'(a)|}{2}\int_{a}^{a+\delta}\frac{1}{y-a}dy =\infty,$$

contradiction.

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Sketch for a proof: from the Taylor's theorem we have that $$ f(x)=f(y)+f'(y)(x-y)+O(|x-y|^2),\quad \text{ when }|x-y|\to 0 $$ And because the integral converges then for any chosen $\epsilon >0$ there is some $\delta >0$ such that $$ \int_{x-\delta }^{x+\delta }\frac{|f(x)-f(y)|}{(x-y)^2}\,\mathrm d y=\int_{x-\delta }^{x+\delta }\left|\frac{f'(y)}{x-y}+ \frac{O(|x-y|^2)}{(x-y)^2} \right|\,\mathrm d y<\epsilon $$ Because $f$ is smooth then $f'$ is continuous in $[x-\delta ,x+\delta ]$ so $f'$ is bounded here, and also its bounded the term $O(|x-y|^2)/(x-y)^2$ (check the definition of big Oh, or instead of big Oh notation use the Taylor's remainder).

If you show that $f'(x)=0$ we are done. Can you continue from here?

What happen if $f'(x)\neq 0$?

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  • $\begingroup$ Wow, how do you make the spoiler thingy? $\endgroup$ – Vincent Jan 21 at 12:32
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    $\begingroup$ @Vincent: just click the edit button, you'll see the raw markdown (>! in this case) $\endgroup$ – Mat Jan 21 at 12:36

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