Abstract
We investigate the relationship between the univalence of $f$ and of $h$ in the decomposition $f=h+\bar{g}$ of a sense-preserving harmonic mapping defined in the unit disk $\mathbb{D}\subset\mathbb{C}$. Among other results, we determine the holomorphic univalent maps $h$ for which there exists $c>0$ such that every harmonic mapping of the form $f=h+\bar{g}$ with $|g'|< c|h'|$ is univalent. The notion of a linearly connected domain appears in our study in a relevant way.
Results & Lemmas (4)
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Theorem 1
Theorem 1: Let h: D →C be a holomorphic univalent map. Then there exits c > 0 such that every harmonic mapping f = h + g with dilatation…
Theorem 1: Let h : D →C be a holomorphic univalent map. Then there exits c > 0 such that every harmonic mapping f = h + g with dilatation |ω| < c is univalent if and only if h(D) is a linearly connected domain. The proof will show that c may be taken equal to 1 when h is convex, and we will show that c = 1 only in this case.
Theorem 2
Theorem 2: Let f = h + g be a sense-preserving univalent harmonic mapping defined on D, and suppose that Ω= f(D) is linearly connected with…
Theorem 2: Let f = h + g be a sense-preserving univalent harmonic mapping defined on D, and suppose that Ω= f(D) is linearly connected with constant M. If |ω| < 1/(1 + M) then h is univalent.
Theorem 3
Theorem 3: Let f = h + g be a sense-preserving univalent harmonic mapping defined on D, and suppose that Ω= f(D) is linearly connected with…
Theorem 3: Let f = h + g be a sense-preserving univalent harmonic mapping defined on D, and suppose that Ω= f(D) is linearly connected with constant M. If |ω| < 1/(1 + 2M) then F = h + eiθg is univalent for every θ.
Theorem 3
Theorem 3 can also be deduced from Theorem 1 and the remarks to Theorem 2, to draw a stronger conclusion. In effect, let f = h + g satisfy…
Theorem 3 can also be deduced from Theorem 1 and the remarks to Theorem 2, to draw a stronger conclusion. In effect, let f = h + g satisfy the hypotheses of Theorem 3. Because 4