ONE–TO–ONE CONTINUOUS EXTENSIONS OF ANALYTIC FUNCTIONS

Ramesh Garimella Northwest Missouri State University

Let U be the open unit disc in the complex plane. Let H(U) stand for the space of functions analytic on U. Let

A = {gεH(U) : g′(0) ≠ 0} .

For gεA,

g(z) = ∑_{n=0}^{∞} a_{n}z^{n} ,

following the lead of Walter Rudin (see [1] problem E3325 p.445), we say g has the property P_{t} if

∑_{n=2}^{∞} |a_{n}|n ≤ t .

In this short note we prove the following result which is an extension of the problem E3325 of [1]. \underline{Theorem}: Let gεA have the property P_{t} for some t > 0. Then g is one–to–one and admits a one–to–one continuous extension to the closed unit disc if t ≤ |g′(0)|. First we prove a lemma.

\underline{Lemma}: Assume gεA and has the property P_{t} for some t > 0. Then g has a continuous extension to the closure of U.

\underline{Proof}: Let

f(z) = (g(z) − g(0))(g′(0))^{−1} .

Obviously fεA, f(0) = 0 and f′(0) = 1 and has the property P_{s} where s = t|g′(0)|^{−1}. Let

f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n} .

Since f has the property P_{s}, the series

z + ∑_{n=2}^{∞} a_{n}z^{n}

is absolutely convergent for every z in the unit circle. Now we show that f is continuous on the closed unit disc. Let z, w be in the closed unit disc. We have,

|f(z) − f(w)| = |(z − w) + ∑_{n=2}^{∞} a_{n}(z^{n} − w^{n})|

= |z − w||1 + ∑_{n=2}^{∞} a_{n}(z^{n−1} + z^{n−2}w + ⋯

+ zw^{n−2} + w^{n−1})|

≤ |z − w|[1 + ∑_{n=2} |a_{n}|(|z|^{n−1} + |z|^{n−2}|w| + ⋯ + |w|^{n−1})]

≤ |z − w|[1 + ∑_{n=2}^{∞} n|a_{n}|]

≤ |z − w|(1 + s) .

The next–to–last inequality is true since |z| ≤ 1, and |w| ≤ 1. From the above it follows that f has a continuous extension to the closed unit disc. Hence, g has a continuous extension to the closed unit disc. Q.E.D. \underline{Proof of the Theorem} : Let

f = (g(z) − g(0))^{−1}(g′(0))^{−1} .

Then fεA, f(0) = 0, f′(0) = 1 and has the property P_{t} where t ≤ 1. Since f has continuous extension to the closed unit disc (by the lemma), it is enough to show that f is one–to–one and the extension of f is one–to–one on the closed unit disc. Let

f(z) = z + ∑_{n=2}^{∞} a^{n}z^{n}.

For z, w in the open unit disc, since

\begin{aligned}|f(z) − f(w)| & = |(z − w) + ∑_{n=2}^{∞} a_{n}(z^{n} − w^{n})| \\ & = |z − w||1 + ∑_{n=2}^{∞} a_{n}(z^{n−1} + z^{−2}w + ⋯ + w^{n−1})| , \end{aligned}

and since

| ∑_{n=2}^{∞} a_{n}(z^{n−1} + z^{n−2}w + ⋯ + w^{n−1})| < ∑_{n=2}^{∞} n|a_{n}| ≤ t ≤ 1

it follows that f is one–to–one on the open unit disc. Also from the above it follows that f(z) ≠ f(w) if one of z, w is in the open unit disc and the other on the unit circle. Now we show that f is one–to– one on the unit circle. By the lemma, f has continuous extension to the closed unit disc. Let f(U) = Ω. If possible assume that z, w are distinct points on the unit circle such that f(z) = f(w). Let z_{n} = (1 − n^{−1})z, w_{n} = (1 − n^{−1})w. Since f is continuous on the closed unit disc, f(z_{n}) → f(z) and f(w_{n}) → f(w). Since f is a homeomorphism of U onto Ω, f(z)(= f(w)) is a boundary point of Ω. Now for each n ≥ 1, let

s_{n} = \{ \begin{matrix}f(z_{n}) & if n is even \\ f(w_{n}) & if n is odd. \end{matrix}

Clearly s_{n} is a sequence in Ω converging to the boundary point of Ω. Since f^{−1} from Ω onto U is a homeomorphism, the sequence f^{−1}(s_{n}) must converge to a point of the unit circle. This is a contradiction because by definition of s_{n}, the sequence f^{−1}(s_{n}) has two subsequences converging to two different limits. Hence f is one–to–one on the unit circle. It follows that g is one– to–one in the unit disc and has a one–to–one continuous extension to the closed unit disc. Q.E.D. \underline{Remark}: For any t > 1 there exists a function gεA having the property P_{t} and fails to be one–to–one (refer to (c) of problem E3325 of [1]). Let t > 1. Write t = 1 + c. Define

g(z) = z − \frac{1}{2}z^{2} + \frac{α}{3}z^{3} + \frac{β}{4} z^{4}

where

α = \frac{−72 − 3c}{5} and β = \frac{8c + 72}{5} .

Since

∑_{n=2}^{4} n|a_{n}| = 1 + α + β = 1 + c = t ,

g has the property P_{t}. Clearly gεA. Now by the choice of α, β it is easy to verify that g(1/2) = 0. Also g(0) = 0. Hence g is not one–to–one on the unit disc.

\underline{References}

1. The American Mathematical Monthly, Vol. 96, No. 5, May 1989.