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Results & Lemmas (4)
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Theorem 1
Theorem 1. Suppose is connected. Let F be a subset of with zero harmonic measure for each point of U (with respect to U). Then F is a peak…
Theorem 1. Suppose $C \setminus U$ is connected. Let F be a subset of $\partial U$ with zero harmonic measure for each point of U (with respect to U). Then F is a peak interpolation set for $H_F^{\infty}(U)$ .
The proof follows from the following lemma.
LEMMA. With U and F as in the theorem, let X be a compact
subset of $\overline{U}$ , W a neighborhood of X, and $\varepsilon > 0$ . Then we can find $f \in H_F^{\infty}$ with $||f|| \leq 2$ , $|1 - f| < \varepsilon$ on $F \cap X$ , and $|f| < \varepsilon$ on $U \setminus W$ .
Theorem 2
Theorem 2. Let and. Suppose there is a decreasing sequence of open sets containing F, such that uniformly for. Then F is a peak…
Theorem 2. Let $y \in U$ and $F \subseteq \partial U$ . Suppose there is a decreasing sequence $\{V_n\}$ of open sets containing F, such that $\mu(V_n) \to 0$ uniformly for $\mu \in M_y$ .
Then F is a peak interpolation set for $H_F^{\infty}(U)$ .
Lemma 2
Lemma 2. Let F be a subset of such that for each there exists such that is a peak interpolation set for, then F is a peak interpolation set…
Lemma 2. Let F be a subset of $\partial U$ such that for each $z \in F$ there exists $\delta > 0$ such that $F_z = F \cap \{w : |w-z| \le \delta/2\}$ is a peak interpolation set for $H_{F \cap A(z,\delta)}^{\infty}(U \cap \Delta(z,\delta))$ , then F is a peak interpolation set for $H_F^{\infty}(U)$ .
Theorem 3
Theorem 3. Let S be a subset of such that U is locally simply connected at each point of. Then S is an interpolation set for if and only…
Theorem 3. Let S be a subset of $\overline{U}$ such that U is locally simply connected at each point of $S \cap \partial U$ . Then S is an interpolation set for $H_{S \cap \partial U}^{s}(U)$ if and only if:
- (i) $U \cap S$ is an interpolating sequence for $H^{\infty}(U)$ ,
- (ii) $S \cap \partial U$ has zero harmonic measure for each point of U, with respect to U.