Carleson embedding theorem

Remarks

To show that 1 implies 2, let $f^*$ be the nontangential maximal function of $f$ on the circle. For $\lambda > 0$, the open set $\set{f^* > \lambda}$ of the circle is a union of disjoint arcs $I_k$. Up to a fixed change of aperture, the set $\set{z \in \mathbb{D} : \abs{f(z)} > \lambda}$ is covered by the Carleson boxes over the enlarged arcs $3 I_k$. Hence $\mu(\set{\abs{f} > \lambda}) \le C \abs{\set{f^* > \lambda}}$, and integrating in $\lambda$ gives $\int \abs{f}^p \, d\mu \le C \norm{f^*}_{L^p}^p \le C' \norm{f}_{H^p}^p$. To show that 2 implies 1, test the inequality on $f(z) = \left( (1 - \abs{a}^2)/(1 - \bar a z)^2 \right)^{1/p}$ for $a \in \mathbb{D}$, which has $H^p$ norm $1$ and is large on the Carleson box near $a/\abs{a}$ of size $1 - \abs{a}$.

The theorem is due to Carleson, who used it in his solution of the interpolation problem for bounded holomorphic functions and in his proof of the corona theorem. The condition does not depend on $p$, so a measure that works for one $p$ works for all $p$.

Example: area measure on $\mathbb{D}$ is a Carleson measure, so $\int_{\mathbb{D}} \abs{f}^p \, dA \le C_p \norm{f}_{H^p}^p$, a weak form of the fact that $H^p$ is contained in the Bergman space $A^p$.

Going deeper

For a student

The theorem turns a statement about holomorphic functions into a geometric condition on boxes, which is often easy to check. Let $(z_k)$ be a sequence of points of $\mathbb{D}$. The measure $\sum (1 - \abs{z_k}) \delta_{z_k}$ is a Carleson measure exactly when the map $f \mapsto ((1 - \abs{z_k})^{1/p} f(z_k))_k$ is bounded from $H^p$ to $\ell^p$. This is a key step in interpolation problems. In the upper half-space of $\R^{n+1}$ the analogous theorem holds for Poisson integrals of $L^p$ functions, $1 < p < \infty$. Carleson measures are also linked with BMO. A function $b$ is in BMO exactly when $t \abs{\nabla u(x, t)}^2 \, dx \, dt$ is a Carleson measure, where $u$ is the Poisson integral of $b$.