It is well known that Laplace's fractional operator is defined by
$$(-\Delta)^s u(x) = c_{n,s}\lim_{\varepsilon \to 0^+}\int_{\mathbb{R}^N \backslash B_\varepsilon (x)} \frac{u(x) - u(y)}{|x - y|^{N + 2s}}dy.$$
It is also well known that if $s \in (0,\frac{1}{2})$ and $u \in L^2(\mathbb{R}^N)$ is such that $u$ is bounded and globally lipschitz, then the above integral is finite. The proof essentially consists of showing that the integral
\begin{split} \int_{\mathbb{R}^N}\frac{u(x) - u(y)}{|x - y|^{N + 2s}}dy &\leqslant C\int_{B_1(x)} \frac{|x - y|}{|x - y|^{N + 2s}} + 2\|u\|_{L^\infty}\int_{\mathbb{R}^N \backslash B_1(x)} \frac{1}{|x - y|^{N + 2s}}dy\\ &\leqslant \bar{C}\Big(\int_{B_1(x)} \frac{1}{|x - y|^{N + 2s - 1}} dy + 2\|u\|_{L^\infty}\int_{\mathbb{R}^N \backslash B_1(x)} \frac{1}{|x - y|^{N + 2s}} dy\Big) < \infty. \end{split}
And we say that while the integral in the second term is always finite, the integral in the first term is finite if and only if $s \in (0,\frac{1}{2})$. But how is this possible? For example, if $N = 3$, $s = 1/4$ the integral $\int_{B_1(x)} \frac{1}{|x - y|^{N + 2s - 1}} dy$ does not converge.