Hardy space ($H^p$)

For $0<p<\infty$, an analytic Hardy space controls the integral means of a holomorphic function as it approaches a domain boundary. On the disk, $H^p$ requires ${\sup }_{0<r<1}\int |F(r{e}^{i\theta }){|}^{p}d\theta <\infty$; on the upper half-plane it requires ${\sup }_{y>0}{\int }_{\mathbb{R}}|f(x+iy){|}^{p}dx<\infty$. For $p=\infty$ use a supremum norm. These analytic spaces differ from the harmonic Hardy space of the upper half-space; the Hardy space of the circle is the boundary Hilbert-space realization of the disk $H^2$ space.

This Hardy space consists of holomorphic functions with $\Vert f{\Vert }_1={\sup }_{r<1}{\int }_{\partial \mathbb{D}}|f(r\xi )|dm(\xi )<\infty$, where $m$ is normalized angular measure. The H1 non-tangential maximal inequality on the disk controls their approach to the boundary, and the Carleson embedding theorem on the disk characterizes measures against which all these functions remain integrable.

For an analytic H1 space on the unit disk function, its fixed-aperture non-tangential maximal function obeys the displayed strong $L^1$ bound. First replace $f$ by $f_R(z)=f(Rz)$, $R<1$. The fact that modulus powers of holomorphic functions are subharmonic gives $|f_R{|}^{1/2}\le P[|f(R\xi ){|}^{1/2}]$. Since a Poisson maximal function is dominated by the Hardy-Littlewood maximal function, the strong $L^2$ Hardy-Littlewood maximal inequality yields $\int N_A{f}_R\le C_A\int |f(R\xi )|\le C_A\Vert f{\Vert }_1$. Pointwise interior convergence and the Fatou lemma remove $R$. Analyticity permits this endpoint estimate; general $L^1$ boundary data have only a weak $L^1$ maximal bound.

For $a\in \mathbb{D}$, this holomorphic function has H1 space on the unit disk norm one: its boundary modulus is the unit-mass Poisson kernel on the circle. The radial integrals are bounded by that boundary integral by subharmonic comparison and tend to it by continuity. Choosing $a$ at depth comparable to an arc's length makes $|f_a|$ bounded below by the reciprocal length throughout its hyperbolic geodesic cap. It tests necessity in the Carleson embedding theorem on the disk.

This Hardy space consists of bounded holomorphic functions on the unit disc, with norm $\Vert f{\Vert }_{\infty }={\sup }_{z\in \mathbb{D}}|f(z)|$. Its almost-everywhere boundary realization follows from the Fatou theorem for bounded holomorphic functions.

A bounded holomorphic function on the unit disc has finite nontangential limits almost everywhere on its boundary. To outline a proof, bounded real and imaginary harmonic parts have $L^{\infty }$ boundary representations: apply the Banach-Alaoglu theorem to their circular traces and pass to the limit in the Poisson integral on the unit disk. Almost every boundary point is a Lebesgue point. The fact that a Poisson integral converges nontangentially at Lebesgue points gives the desired limits. Their $L^{\infty }$ norm equals the interior supremum norm, and the Poisson integral recovers the original function. Exceptional points can lack even radial limits; $\exp (i\mathrm{Log}(1-z))$ is a bounded example at the boundary point one.

If a finite complex Borel measure $\mu$ on the circle has Fourier coefficients $\hat{\mu }(n)=0$ for all $n<0$, then it is absolutely continuous with respect to arc length. The reflected Fourier convention gives the equivalent theorem with positive coefficients vanishing. The Poisson integral on the unit disk is analytic because of the one-sided Fourier spectrum. Subharmonic comparison for its square root and an $L^2$ maximal estimate produce an integrable radial maximum. This maximum dominates the weak-star boundary measure, forcing absolute continuity of measures. The same proof on the real line says that a finite measure with Fourier transform of a finite measure supported in $[0,\infty )$ has a Lebesgue density.