# Prove a function is constant under certain conditions

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.

• 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 $0$ – gt6989b Jan 21 at 4:19

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 $$a Taking $$x=a,$$ the integral of interest is at least

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

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 $$f'(x)\neq 0$$?
• @Vincent: just click the edit button, you'll see the raw markdown (>! in this case) – Mat Jan 21 at 12:36