Proposition 30.3.3.label Let $\fU \subset 2^{B_\complex(0, 1)}$ be an ultrafilter and
\[\phi_{\fU}: H^{\infty}(D) \to \complex \quad f \mapsto \lim_{x, \fU}f(x)\]
then:
- (1)
If $\fU \to x_{0} \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_{0})$.
- (2)
$\phi_{\fU}$ is a multiplicative linear functional on $H^{\infty}(D)$.
Proof. Let $f \in H^{\infty}(D)$, then by Proposition 5.2.4, $f(\fU)$ is an ultrafilter base. Since $f$ is bounded, $f(\fU)$ converges to exactly one element of $\complex$. Hence the limit is well-defined.
(2): By Proposition 10.13.8.$\square$
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