Hardy spaces of harmonic functions

Remarks

Each $h^p$ with $1 \le p \le \infty$ is a Banach space for the norm above. Write $P(x, \zeta) = (1 - \abs{x}^2) / \abs{x - \zeta}^n$ for the Poisson kernel of $B$. For $1 < p \le \infty$, $u \in h^p$ if and only if $u(x) = \int_S P(x, \zeta) f(\zeta)\, d\sigma(\zeta)$ for some $f$ in the Lp space $L^p(\sigma)$, and then $\norm{u}_{h^p} = \norm{f}_{L^p}$. For $p = 1$, $u \in h^1$ if and only if $u$ is the Poisson integral of a finite signed Borel measure on $S$. Example: fix $\eta \in S$ and let $u = P(\cdot, \eta)$. It is positive and harmonic, and by the mean value property its mean over each sphere $\abs{x} = r$ is $u(0) = 1$, so $u \in h^1$ with norm $1$. It is not in $h^p$ for $p > 1$, since it is the Poisson integral of the point mass at $\eta$, which has no $L^p$ density. Non-example: in the plane, let $u = \operatorname{Re}\left( \frac{1 + z}{1 - z} \right)^2$ on the unit disc. For $e^{i\theta} \ne 1$, $u(r e^{i\theta}) \to -\cot^2(\theta / 2)$ as $r \to 1$, and this function is not integrable on the circle. By Fatou's lemma the means of $\abs{u}$ over circles are unbounded, so $u \notin h^1$. In the plane, if $1 < p < \infty$ and $f = u + iv$ is holomorphic with $v(0) = 0$, then $u \in h^p$ if and only if $f$ is in the holomorphic Hardy space $H^p$, by the M. Riesz theorem. This fails for $p = 1$ and for $p = \infty$.