In many ways $H^1$ is just a natural substitute for $L^1$. A typical $H^1$ function is a $1$-atom, i.e. a function $\phi\in L^1(\mathbb R^n)$ such that the support of $\phi$ is contained in some ball $B(a,r)$, the bound $$\sup|\phi(\cdot)|\leq \frac{1}{m(B(a,r))}$$ holds true, and $$\int_{\mathbb R^n} \phi(x)dx=0.$$
In fact, following a Fefferman suggestion, one can use $1$-atoms to characterize the whole Hardy space $H^1(\mathbb R^n)$. More precisely, the following is true.
Theorem. A function $f$ belongs to $H^1(\mathbb R^n)$ if and only if there exist $1$-atoms $\phi_k$ and complex numbers $c_k$ such that $$f=\sum_{k=1}^{\infty}c_k\phi_k,$$ where the convergence is in $L^1$ and $\sum_{k}|c_k|<\infty.$
One should note that the mean-value condition in the definition of $1$-atom plays a key role here. If it is removed from the definition, the above description with sums would result in the standard Lebesgue space $L^1(\mathbb R^n)$.