Embedding of Hardy space $\mathcal{H}^1$ Does local (real) Hardy space $\mathcal{H}^1_{loc}(\mathbb{R}^2)$ embed compactly into $W^{-1,2}_{loc}(\mathbb{R}^2)$? It looks possible, but I cannot figure out either a proof or a counterexample. 
For the definition of local Hardy space, we follow Goldberg (1979, Duke Math. Jour.), also cf. Stein, Harmonic Analysis, Chapter III, 5.17 (P134): $\mathcal{H}^1_{loc}$ can be defined via truncated maximal functions: say $f\in \mathcal{H}^1_{loc}(\mathbb{R}^n)$ iff $M^{(1)}\Phi(f):=\sup_{0<t\leq 1}|\Phi_t\ast f|\in L^1(\mathbb{R}^n)$, where $\Phi$ is in the Schwarz class, $\int\Phi dx \neq 0$, $\Phi_t(x):=t^{-n}\Phi(x/t)$.
 A: Not compact because the norms scale in the same way. Indeed, let $u_\lambda=u (\lambda x)$ (with large $\lambda$). Then $\|u_\lambda\|_{\mathcal H^1}=\lambda^{-2}\|u\|_{\mathcal H^1}$ and $\|u_\lambda\|_{W^{-1,2}}=\lambda^{-2}\|u\|_{W^{-1,2}}$ (kind of; to make it really true consider the homogeneous form of the space). So, starting with one compactly supported $u$ with unit $\mathcal H^1$ norm, we  can produce infinitely many rescaled copies with disjoint supports (all within the unit ball) that have the same $\mathcal H^1$ norm. Their images will be an infinite, uniformly separated, subset of $W^{-1,2}$. 
The scaling argument may be easier to see by dualizing the problem. The embedding of $W^{1,2}$ into $BMO$ is not compact for the same reason: both norms are invariant under scaling (kind of; the homogeneous part of $W^{1,2}$ norm is). 
This is why the compact embeddings $X\to Y$ involve functions spaces $X$ and $Y$ such that under "shrinking the support" scaling, the $X$ norm  grows faster than the $Y$ norm.
