Results & Lemmas (8)
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Lemma 2.1.
Lemma 2.1. For f = αz + a2z2 + a3z3 + · · · ∈H, ℜa3 ≤α(1 −α2)(2 −α2)
Lemma 2.1. For f = αz + a2z2 + a3z3 + · · · ∈H, ℜa3 ≤α(1 −α2)(2 −α2)
Lemma 2.2.
Lemma 2.2. If f(z) = αz + a2z2 + · · · ∈H is rotated so that a2 is real, then 2α −a2 ≥0.
Lemma 2.2. If f(z) = αz + a2z2 + · · · ∈H is rotated so that a2 is real, then 2α −a2 ≥0.
Lemma 4.1.
Lemma 4.1. Lt(f) is rotationally invariant.
Lemma 4.1. Lt(f) is rotationally invariant.
Lemma 4.1.
Lemma 4.1. (The Step Down Lemma) If f is extremal, then it can have at most two proper sides
Lemma 4.1. (The Step Down Lemma) If f is extremal, then it can have at most two proper sides
Lemma 6.1.
Lemma 6.1. For symmetric maps in H2 having non-zero second coefficient, the image of ∂D under the kernel K(ξ) = 2tαξ2 + 4a2(t + α)ξ + 3ta3 +…
Lemma 6.1. For symmetric maps in H2 having non-zero second coefficient, the image of ∂D under the kernel K(ξ) = 2tαξ2 + 4a2(t + α)ξ + 3ta3 + 4a2 2 is (Euclidean) convex for t ∈ · − αa2 2α + a2 , αa2 2α −a2 ¸ .
Theorem 6.1.
Theorem 6.1. For f(z) = αz + a2z2 + a3z3 + · · · ∈H2 (a2 ̸= 0) the image of ∂D under the kernel K(ξ) = 2tαξ2 + 4a2(t + α)ξ + 3ta3 + 4a2 2…
Theorem 6.1. For f(z) = αz + a2z2 + a3z3 + · · · ∈H2 (a2 ̸= 0) the image of ∂D under the kernel K(ξ) = 2tαξ2 + 4a2(t + α)ξ + 3ta3 + 4a2 2 is (Euclidean) convex for t ∈ · − αa2 2α + a2 , αa2 2α −a2 ¸ .
Theorem 6.1.
Theorem 6.1. Suppose f ∈H is such that a2 ̸= 0 and f is not the half-plane map- ping. Then there exists a range of t such that f is not…
Theorem 6.1. Suppose f ∈H is such that a2 ̸= 0 and f is not the half-plane map- ping. Then there exists a range of t such that f is not extremal for Lt, in particular, for t in the interval of convexity: t ∈ · − αa2 2α + a2 , αa2 2α −a2 ¸ . As an example, figure 6.5 shows the kernel of hα with α = 0.42 and b = 0.5 (a2 ≈0.2004 and a3 ≈0.1959).
theorem 6.1.
theorem 6.1. 26
theorem 6.1. 26