About functions in fractional Sobolev spaces $W^{s,p}(\mathbb{R})$ and $\lim_{x\rightarrow a^{+}}\int_{a}^{x}\frac{f(y)}{(x-y)^{s}}dy=0$

Let $$1\leq p<+\infty$$, $$0 and $$f\in W^{s,p}\left(\mathbb{R}\right)$$, where $$W^{s,p}\left(\mathbb{R}\right):=\left\{ u\in L^{p}\left(\mathbb{R}\right):\frac{\left|u\left(x\right)-u\left(y\right)\right|}{\left|x-y\right|^{\frac{1}{p}+s}}\in L^{p}\left(\mathbb{R}\times\mathbb{R}\right)\right\}.$$ Furthermore, fix $$[a,b]\subset\mathbb{R}$$. I would like to prove, if possible, that $$\lim_{x\rightarrow a^{+}}\int_{a}^{x}\frac{f\left(y\right)}{\left(x-y\right)^{s}}dy=0\tag{1}$$ without any other assumption on $$p$$. For example, it is quite simple to prove that $$(1)$$ holds if $$p^{*}<\frac{1}{s}$$, where $$p^{*}$$ is the Hölder conjugate of $$p$$, but this results holds for a generic $$f\in L^{p}(\mathbb{R})$$. I tried some manipulations to exploit the convergence of the double integral, for example writing $$\frac{1}{\left(x-y\right)^{s}}=\frac{\Gamma\left(1-\alpha\right)}{\Gamma\left(s\right)\Gamma\left(1-\alpha-s\right)}\int_{y}^{x}\frac{1}{\left(t-y\right)^{s+\alpha}}\frac{1}{(x-t)^{1-\alpha}}dt$$ for some $$\alpha$$ but, again, in the end I had to assume the condition on $$p^*$$. If my idea was wrong, I would appreciate an explicit counterexample. Thank you.

1 Answer

Yes, this is true at least when $$1. By Hölder's inequality $$I_x = \int_a^x \frac{f(y)}{|x-y|^s}\,\mathrm d y \leq \|f\|_{L^q} \,\varepsilon(x-a)$$ where $$\varepsilon(z)\underset{z\to 0}{\to} 0$$ as soon as $$s \,q' < 1$$ ($$q'$$ is the Hölder conjugate), that is as soon as $$0\leq \frac{1}{q}<1-s$$. In particular, one can take $$\frac{1}{q} = \frac{1}{p} - s$$ (or $$q=\infty$$ if $$sp>1$$), so that by Sobolev's inequality, $$\|f\|_{L^q} \leq C\, \|f\|_{W^{s,p}}$$.

• No, I did not assume anything on $p$ ... I am just doing Hölder's inequality with $q = p/(1-sp)$ and then use Sobolev embedding. I can give more details but I need to know which part is unclear then? Jan 31 at 19:18
• If $f\in W^{s,p}$ then it is justified to use $\|f\|_{L^q}$ by Sobolev inequalties ... If $p=1/s$, this is the critical case in Sobolev embeddings, you can always take any other $q\geq p$ such that $\frac{1}{p}-s = 0 <\frac{1}{q} < 1-s$. Jan 31 at 21:56
• Wait, the Sobolev inequality holds for $f$ with compact support, what happen if $f$ does not have compact support? See theorem 6.5 of arxiv.org/pdf/1104.4345.pdf
– User
Feb 1 at 8:50
• This is not a good reference, just sn introduction. Sobolev inequalities work for functions converging to $0$ at infinity such as $L^p$ functions. See Feb 1 at 8:57
• Can you give me a reference?
– User
Feb 1 at 9:01