If we consider the function $f(x) = x^{\alpha}$ for $\alpha \in (1/2,1)$ and $x \in \Omega = (0,1)$, this function should be in the Sobolev-Slobodeckij space $H^{1+\sigma}(\Omega) (= W^{1+\sigma,2}(\Omega))$ for some $\sigma \in (0,1)$, where the norm is given by $$ ||u||_{H^{1+\sigma}(\Omega)} = \left( ||u||_{H^{1}(\Omega)}^{2} + \int_{\Omega}\int_{\Omega} \frac{|u'(x)-u'(y)|^{2}}{|x-y|^{1+2\sigma}}dxdy\right)^{1/2}. $$
A text I'm reading (p and hp Finite Element Methods by Schwab) simply states that this function should be in $H^{k}(\Omega)$ for $k < \alpha+1/2$. I tried to reason out when the Slobodeckij seminorm would be finite, which I assume would show that we need $\sigma < \alpha-1/2$ for the singular integral to converge, but wasn't able to get the result that way. I'm just wondering if I'm missing something really obvious, or if perhaps there's a simpler way to see this.
If you differentiate $f(x)=x^\alpha$ $(\alpha+1/2)$-many times, then you have $x^{-1/2}$ and this is the critical polynomial that is just not anymore $L^2(\Omega)$ (you can check this), but $x^{-1/2+\epsilon}$ is in $L^2(\Omega)$ for any $\epsilon>0$.