Carleson measure

Remarks

Carleson's embedding theorem: for $0 < p < \infty$, $\mu$ is a Carleson measure if and only if there is $C_p$ with $\int_{\mathbb{D}} \abs{f}^p \, d\mu \le C_p \norm{f}_{H^p}^p$ for all $f$ in the Hardy space $H^p$.

Example: area measure $A$ on $\mathbb{D}$. In polar coordinates, $A(S(I)) \le \ell(I) \int_{\max(0, 1 - \ell(I))}^1 r \, dr \le \ell(I) \min(\ell(I), 1) \le \ell(I)$. So $A$ is a Carleson measure with $C = 1$.

Non-example: let $z_n = 1 - 2^{-n}$ and $\mu = \sum_{n \ge 1} n^{-2} \delta_{z_n}$, a finite measure. Let $I_k$ be the arc of length $2^{-k}$ centred at $1$. Then $S(I_k)$ contains $z_n$ for all $n \ge k$, so $\mu(S(I_k)) \ge \sum_{n \ge k} n^{-2} > 1/k$. The ratio $\mu(S(I_k)) / \ell(I_k) > 2^k / k$ is unbounded, so $\mu$ is not a Carleson measure.