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

Question

Let $1\leq p<+\infty$, $0<s<1$ 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.

Answer 1

Yes, this is true at least when $1<p\leq\infty$. 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, $\|f\|_{L^q} \leq C\, \|f\|_{W^{s,p}}$.