Rellich's theorem fails at the end point. If $\{f_k\}\subseteq H^s(\Bbb{R}^d)$ is a bounded sequence and support of each $f_k$ is contained in a common compact set, then there exists a subsequence that converges in $L^q$ for $\forall$ $q, 2\leq q<\frac{2d}{d-2s}$. In this corollary of the Rellich's theorem, why is it necessary that $q<\frac{2d}{d-2s}$ ? I can't immediately see what goes wrong for when $q = \frac{2d}{d-2s}$. 
Also, why is the existence of compact $K$ that contains all of the supports of $f_k$-s is crucial? 
It is of course assumed that $0<s<\dfrac{d}{2}$. 
 A: Take a positive, nonzero $u \in C^{\infty}_c(\mathbb{R}^d)$ and consider $\{u_{\epsilon}\}$, where $u_{\epsilon}(x) = u(x\epsilon^{-1})$. For simplicity, let $s = 1$ and notice that (by a change of variables)
$$\|u_{\epsilon}\|_{L^{2^*}} = \Big(\int |u_{\epsilon}|^{2^*}\Big)^{\frac{1}{2^*}} = \epsilon^{\frac{d}{2^*}}\|u\|_{L^{2^*}}$$
 $$\|\nabla u\|_{L^2} = \epsilon^{\frac{d - 2}{2}}\|\nabla u\|_{L^2}.$$ The crucial observation here is that $\frac{d}{2^*} = \frac{d - 2}{2}$, which implies that the two norms have the same scaling. This allows us to define $$u^{\epsilon}(x) : = \epsilon^{-\frac{d}{2^*}}u_{\epsilon}(x).$$ 
By construction, $\{u^{\epsilon}\}$ is bounded in $H^1$ with uniformly bounded support. Assume by contradiction that there is a subsequence that converges in $L^{2^*}$, then it would converge to $0$ since $u^{\epsilon}(x) \to 0$ a.e. as $\epsilon \to 0^+$. This contradicts the fact that, by construction, $\|u^{\epsilon}\|_{L^{2^*}} = \|u\|_{L^{2^*}} > 0$.
This, together with Ian's comment, should clarify all the assumptions in the Theorem.
