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QUESTION

Let $f$ be an entire function such that $|f(z)| \le A|z|$ for all
$z$, where $A$ is a fixed positive number. Show that $f(z)=az$,
where $a$ is a complex constant. (Hint: Use the Cauchy
inequalities to show that that the second derivative of $f$ is
everywhere zero.)


ANSWER

If $D$ denotes the open unit disc, then if $f({\bf C})=D$, we have
$|f(z)|&lt;1$, for all $z\in {\bf C}$, so that $f$ is a bounded
analytic function and hence by Liouville's Theorem, $f$ is
constant and hence $f$ cannot map ${\bf C}$ {\it onto} $D$.

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