Let X be a compact set in
the complex plane C. Denote by R(X) the closure in the supremum norm of the
rational functions with poles off X and by A(X) the set of continuous functions,
which are analytic on the interior of X. The analytic capacity of a set S is denoted
by γ(S). For the definition of γ see below. Let B_{z}(δ) = {ζ ∈ C;z − ζ < δ} and let
∂X denote the boundary of X. Vitushkin has proved that R(X) = A(X)
if
Let ψ be a function from R^{+} to R^{+}, where R^{+} = {x ∈ R;x ≧ 0}. We now ask
the following questions. If lim_{δ→0}ψ(δ) = 0, is it possible to find a compact set X such
that R(X)≠A(X) and such that γ(B_{z}(δ) ∖ X) ≧ δψ(δ) for all z ∈ ∂X and for all
δ, 0 < δ < δ_{z}? If the answer is yes, can the answer still be yes, if lim_{δ→0}ψ(δ) = 0 is
replaced by _{δ→0}ψ(δ) > 0? The answers of these questions can be found in
Theorem 1 and Theorem 2.
