Let $K$ be a compact subset of $\mathbb{R}^n$. Choose a finite open cover $\{B_i\}_{i=1,\dots,k}$ so that $K\subset \bigcup_{i=1}^k B_i$. Take a partition of unity $\{\zeta_1,\dots,\zeta_k\}$ so that $\zeta_i \in C^\infty_c(B_i)$, $\mathrm{supp}\, \zeta _i \subset B_i$ for all $1\leq i\leq k$ and $\sum_{i=1}^k \zeta_i =1$ on $K$.
Here is my question. Suppose that $\zeta_i u \in W^{2,2}(B_i)$. Can we say that $u\in W^{2,2}(K)$? It seems to be wrong, but I cannot find exact counterexamples.
Thanks for your time.
It is true that $u\in W^{2,2}(K)$ due to the simple fact that $u=\sum \zeta_i u$ since $\sum \zeta_i=1$ on $K$. This is different from being in $W^{2,2}_{loc}$ however; that would mean that for any compactly contained subset of $K$, say $V$, then the restriction of $u$ to $V$ lies in the Sobolev space.
An example of a function in $W^{2,2}_{loc}$ but not in $W^{2,2}$ can be seen by e.g. $f(x)=x^{-2}$ on $K=[0,1]$. This is not even in $L^2$, but $f$ and its derivatives are easily seen to be in $L^2_{loc}$ since any compactly contained subset is bounded away from 0.