Local vs global Sobolev space

Question

Let \begin{equation} H_{\text{loc}}^{2}(\mathbb{R}^d)=\{u:\mathbb{R}^d\to\mathbb{R}\;|\;u\in H^{2}(V)\text{ for all }V\subset\subset \mathbb{R}^d\}. \end{equation} I was wondering if it is true that \begin{equation*} L^2(\mathbb{R}^d)\cap H^s_\mathrm{loc}(\mathbb{R}^d)=H^s(\mathbb{R}^d). \end{equation*} Intuitively, I was thinking that the local Sobolev regularity should `cure' the local singularities by making them $H^s$, while the decay of derivatives could be controlled by the $L^2$ decay. However, I was not able to come up with a proof or with a counterexample.

Answer 1

No. Take a radially symmetric $\phi\in C^\infty_c(\mathbb R^d)$, fix $k\in \mathbb R$ and let $$ f(x):=\sum_{n=1}^\infty a_n \phi(n^k x-n^{k+1}), $$ where $a_n\in \ell^2$. This function is in $H^1_{\mathrm{loc}}(\mathbb R^d)$, since it reduces, locally, to a finite sum. Moreover, $$ \lVert f\rVert_{L^2(\mathbb R^d)}^2=C_1\sum_{n=1}^\infty n^{-kd} |a_n|^2.$$ On the other hand, since $$ \partial_r f=\sum_{n=1}^\infty n^ka_n \phi'(n^kx-n^{k+1}), $$ we have that $$ \lVert\partial_r f\rVert_{L^2(\mathbb R^d)}^2=C_2\sum_{n=1}^\infty n^{2k-kd} \lvert a_n\rvert^2. $$ Choosing, say, $|a_n|^2=n^{kd-\alpha}$ with $$ kd-\alpha <-1\qquad 2k-kd-\alpha \ge -1$$ produces a counterexample.

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The idea is taking an infinite series of functions, with disjoint supports that stretch more and more. The $L^2$ norm of the function and the one of the derivative behave differently under this stretching, and this can be used to make the derivative blow up.