Suppose f is Complex differentiable everywhere and that Lim|z|→∞}f(z)=1

Prove f(z) is constant

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- Oct 27th 2009, 09:56 PMthespianLimit question
Suppose f is Complex differentiable everywhere and that Lim|z|→∞}f(z)=1

Prove f(z) is constant - Oct 28th 2009, 01:49 AMEnrique2
You have to use Liouville theorem. Find a closed ball B such that outside the modulus of f is smaller than 2, and apply that the closed ball is compact to conclude that f is bounded-

- Oct 28th 2009, 02:35 AMtonio

First proof: Liouville's theorem states that a holomorphic (analytic) function which is bounded must be constant, and that's exactly what happens with our function: in some circle $\displaystyle |f(z)|<K\,,\;\;\forall\;|z|<R\,,\,\,and\;\;|f(z)|< 1.5\;\forall |z|>R,because \;f(z)\xrightarrow [|z| \to \infty] {} 1$

Second proof: By Cauchy's integral representation and by the estimation lemma: $\displaystyle |f'(z)|=\left|\frac{1}{2\pi i}\oint\limits_{|z|=R}\frac{f(w)\;dw}{(w-z)^2}\right|\leq \frac{1}{2\pi} 2\pi R \frac{\max\limits_{|z|=R} (|f(z)|)}{R^2}=\frac{\max\limits_{|z|=R} (|f(z)|)}{R} \xrightarrow [R \to \infty] {} 0$, so $\displaystyle f'(z)=0 \Longrightarrow f(z)$ constant.

Tonio