Let $f$ be an entire function (i.e. analytic everywhere, i.e. holomorphic) such that $\left\vert f(z) \right\vert \leq A \left\vert z \right\vert$, $\forall z \in \Bbb{C} $, where $A$ is a fixed positive constant. Show that $f(z) = \alpha z $ where $\alpha$ is a complex Constant.
I'm going to begin by applying Cauchy's Inequality. Namely, given that some function $f(z)$ is complex analytic inside and on some contour $C$ centered at $z_0$ with radius $R$,
$$ \left\vert f^{n}(z_0) \right\vert \leq \frac{n!M_R}{R^n}$$
where $M_R = \max_{z \in C_R} f(z)$. In this case:
$$ \begin{aligned} \left\vert f^{2}(z_0) \right\vert &\leq \frac{2M_R}{R^2} \end{aligned}$$
In order to force the 2nd derivative to be 0, we must force $\frac{2M_R}{R^2} = 0$, where $M_R \leq A(|z_0| + R) $, thus $\left\vert f^{2}(z_0) \right\vert \leq 0$, Since it is a magnitude, $\left\vert f^{2}(z_0) \right\vert = 0$. How do I show this?
Note: Correct me if I'm wrong, the term $M_R \leq A(|z_0| + R) $ comes from the fact that the max achieved must be less than or equal to the max of the function multiplied by the length of the contour? But isn't $|z_0 + R| $ the distance of the contour to the origin?
$\endgroup$1 Answer
$\begingroup$We know that $\left|f(z)\right| \le A|z|$. Let $z \in D(0, R)$. Within this disk, $\left|f(z)\right| \le AR$. It follows from Cauchy's estimates that:
$$ \left|f^{(n)}(0)\right| \le \frac{n! A R}{R^n} $$
If $n > 1$, the RHS goes to $0$ as $R \to \infty$. This forces all $f^{(n)}(0)$ to be $0$ for $n > 1$. It's also clear that $f(0) = 0$. We conclude that $f(z) = az$ for some constant $a$.
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