Showing atomic $H^{1,p}$ is a Banach Space Consider $1<p\leq\infty$ and let $H^{1,p}=H^{1,p}(\mathbb{R}^n)$ be an atomic Hardy space, that is $H^{1,p}\subset L^1(\mathbb{R}^n)$, and $f\in H^{1,p}$ if there exists a sequence of $p$-atoms $\{a_i\}_{i\in \mathbb{N}}$ and a sequence $(\lambda_i)\in \ell^1(\mathbb{N})$ such that 
$f=\sum_{i=0}^\infty \lambda_i a_i $
Recall a $p$-atom $a$ satifies:


*

*$supp(a)\subset Q$ for some cube $Q\subset\mathbb{R}^n$

*$\int_Q a =0$

*$||a||_p\leq \frac{1}{|Q|^{1-\frac{1}{p}}}$
The norm on $H^{1,p}$ is given by
$||f||_{H^{1,p}}=\inf\{\sum_i |\lambda_i| : f=\sum_i\lambda_i a_i\}$
I am trying to prove that $H^{1,p}$ is a Banach space. To start with, it is not hard to see that for any $f\in H^{1,p}$:
$||f||_{L^1}\leq||f||_{H^{1,p}}$
One can see this by considering
$\int |f|=\int |\sum_i \lambda_i a_i| \leq \sum_i |\lambda_i|\int|a_i|\leq \sum_i|\lambda_i|$
Where we have used dominated convergence and the property that for any $p$-atom $||a||_1 \leq 1$.
Using this, if we have $f_n$ a Cauchy sequence in $H^{1,p}$ then we have a Cauchy sequence in $L^1$, so it converges to some $f\in L^1$. I am now stuck on how to prove that $f\in H^{1,p}$.
I have attempted to construct a representation of $f$ in terms of $\lambda_i$ and $a_i$ where $\lambda_i=\lim_{n\rightarrow \infty} \lambda_i^n$ for representatives $f_n=\sum_i \lambda_i^n a_i^n$ (similarly for $a_i$). But this hasn't succeeded.
Thanks for any help
 A: I think I figured it out, for anyone who is interested: In general to prove a normed vector space $X$ is a banach space it is enough to prove:
Given a sequence $(x_n)$ such that $\sum_i ||x_i||<\infty$ then $\sum_i x_i \in X$.
This is my technique for proving $H^{1,p}$ is complete:
Suppose that we have $(f_n)_{n\in\mathbb{N}}$ such that $\sum_{n=0}^\infty||f_n||_{H^{1,p}}<\infty$ then using the estimate shown in the OP, we have $\sum_{n=0}^\infty||f_n||_{L^1}<\infty$, so in fact
$f:=\sum_{n=0}^\infty f_n \in L^1$ using the completeness of $L^1$. We want to show that $f\in H^{1,p}$ and $\sum_{n=0}^N f_n\rightarrow f$ as $N\rightarrow\infty$ with $H^{1,p}$ convergence.
To do this choose an "efficient" representative of $f_n$, that is choose sequence's and $p$-atoms so that $f_n=\sum_{i=0}^\infty \lambda_i^n a_i^n $ and
$\sum_{i=0}^\infty |\lambda_i^n| \leq ||f_n||_{H^{1,p}} + \epsilon 2^{-n} $
Such expressions can always be chosen because the norm is defined in terms of an infimum. It is clear that $f$ has an atomic decomposition of the form
$f=\sum_{n=0}^\infty f_n = \sum_{n=0}^\infty \sum_{i=0}^\infty \lambda_i^na_i^n$
We need to check that our sequence $(\lambda_i^n)$ is in $\ell^1(\mathbb{N})$:
$\sum_{n=0}^\infty \sum_{i=0}^\infty |\lambda_i^n|\leq \sum_{n=0}^\infty (||f_n||_{H^{1,p}} +\epsilon 2^{-n}) = \sum_{n=0}^\infty ||f_n||_{H^{1,p}} +\epsilon <\infty$
Lastly, to check $H^{1,p}$ convergence $f-\sum^N_{n=0}f_n$ has the representation $f-\sum^N_{n=0}f_n=\sum^\infty_{n=N+1}f_n = \sum^\infty_{n=N+1}\sum_{i=0}^\infty \lambda_i^na_i^n$. Hence we have:
$||f-\sum^N_{n=0}f_n||_{H^{1,p}}\leq \sum^\infty_{n=N+1}\sum_{i=0}^\infty |\lambda_i^n|\rightarrow 0$ as $N\rightarrow \infty$
Because we have an "efficient" cover, meaning $\sum_{n=0}^\infty \sum_{i=0}^\infty |\lambda_i^n|<\infty$.
