The goal of the problem is to prove the weak* compactness of $\mathbf{H}_r^1$. First a few definitions.
A bounded measurable function $\mathfrak{a}$ on $\mathbb{R}^d$ is an atom associated to a ball $B \subset \mathbb{R}^d$, if:
- $\mathfrak{a}$ is supported in $B$, with $|\mathfrak{a}(x)| \le 1/m(B)$, for all $x$; and,
- $\int_{\mathbb{R}^d} \mathfrak{a}(x) dx = 0$.
The space $\mathbf{H}_r^1 (\mathbb{R}^d)$ consists of all $L^1$ functions $f$ that can be written as $$f=\sum_{k=1}^\infty \lambda_k \mathfrak{a}_k,$$ where the $\mathfrak{a}_k$ are atoms and the $\lambda_k$ are scalars with $$\sum_{k=1}^\infty |\lambda_k| < \infty.$$
The infimum of the values $\sum_{k=1}^\infty |\lambda_k|$, taken over all possible decompositions of $f$, by definition, the $\mathbf{H}_r^1$ norm of $f$, written as $\|f\|_{\mathbf{H}_r^1}$.
Now a description of the problem itself.
Suppose $\{f_n\}$ is a sequence in $\mathbf{H}_r^1$ with $\|f_n\|_{\mathbf{H}_r^1} \le A$. Then we can select a subsequence $\{ f_{n_k} \}$ and find an $f \in \mathbf{H}_r^1$ so that $\int f_{n_k}(x) \phi(x) dx \rightarrow \int f(x) \phi(x) dx$, as $k \rightarrow \infty$, for every $\phi$ that is a continuous function of compact support.
[Hint: Apply known fact #1 (see below) to obtain a subsequence $\{ f_{n_k} \}$ and a finite measure $\mu$ so that $f_{n_k} \rightarrow \mu$ in the weak* sense. Next use the fact that if $\sup_{\epsilon>0} |µ ∗ \phi_\epsilon| \in L^1$, for an appropriate $\phi$, then $\mu$ is absolutely continuous.]
If we follow the hint, we can get a measure $\mu$ that gives us the desired weak* convergence. However, we need to prove that (a) $\mu$ is absolutely continuous and thus $d\mu = f dx$ for an appropriate $f$, and (b) such an $f$ is in $\mathbf{H}_r^1$. Also, I don't know how to prove the so-called "fact" that if $\sup_{\epsilon>0} |µ ∗ \phi_\epsilon| \in L^1$, for an appropriate $\phi$, then $\mu$ is absolutely continuous.
For reference, we have the following known facts, that have been proved elsewhere.
- Suppose $X$ is a $\sigma$-compact measurable metric space, and $C_b(X)$ is separable, where $C_b (X)$ denotes the Banach space of bounded continuous functions on $X$ with the sup-norm. If $\{f_n\}$ is a bounded sequence of functions in $L^1(\mu_0)$, then there exist a $\mu \in M(X)$ and a subsequence $\{ f_{n_j} \}$ such that the measures $f_{n_j} d \mu_0$ converge to $\mu$ in the weak* sense, i.e. $$\int_X g(x) d\mu_{n_j}(x) \rightarrow \int_X g(x) d\mu(x),$$ for all $g \in C_b(X)$.
- Suppose $\Phi$ is a $C^1$ function with compact support on $\mathbb{R}^d$. Define $$M(f)(x)=\sup_{\epsilon>0} |(f*\Phi_\epsilon)(x)|,$$ where $\Phi_\epsilon = \epsilon^{-d} \Phi(x/\epsilon)$. We have that $M(f) \in L^1(\mathbb{R}^d)$, whenever $f \in \mathbf{H}_r^1$. Moreover $$\|M(f)\|_{L^1(\mathbb{R}^d)} \le A \|f\|_{\mathbf{H}_r^1(\mathbb{R}^d)}$$