Let $f$ be an entire function such that
$$\left| f(z) \right| \leq \log \left( 1 + \left| z \right| \right) \quad \text{forall} \; z \in \mathbb{C}$$
Show that $f(z)=0 $ forall $z \in \mathbb{C}$.
Solution
The key here is actually that $f$ is entire.
Since $f$ is entire it can be expressed as a power series that converges everywhere ,i.e :
$$f(z) = \sum_{m=0}^{\infty} a_m z^m$$
For $z=0$ we get that $\left | f(0) \right | \leq 0$ implying that $a_0=0$. This means that the function $\displaystyle g(z) = \frac{f(z)}{z}$ can be continued to an entire function satisfying:
$$\left | g(z) \right | \leq \frac{\log \left ( 1+\left | z \right | \right )}{\left| z \right|} \quad \text{forall} \; z \neq 0$$
Since the right hand side converges to $0$ as $\left|z \right| \rightarrow +\infty$ (that is , it is bounded) we conclude from Liouville's theorem that $g$ is eventually constant (and specifically $0$) and so is $f$.
$$\left| f(z) \right| \leq \log \left( 1 + \left| z \right| \right) \quad \text{forall} \; z \in \mathbb{C}$$
Show that $f(z)=0 $ forall $z \in \mathbb{C}$.
Solution
The key here is actually that $f$ is entire.
Since $f$ is entire it can be expressed as a power series that converges everywhere ,i.e :
$$f(z) = \sum_{m=0}^{\infty} a_m z^m$$
For $z=0$ we get that $\left | f(0) \right | \leq 0$ implying that $a_0=0$. This means that the function $\displaystyle g(z) = \frac{f(z)}{z}$ can be continued to an entire function satisfying:
$$\left | g(z) \right | \leq \frac{\log \left ( 1+\left | z \right | \right )}{\left| z \right|} \quad \text{forall} \; z \neq 0$$
Since the right hand side converges to $0$ as $\left|z \right| \rightarrow +\infty$ (that is , it is bounded) we conclude from Liouville's theorem that $g$ is eventually constant (and specifically $0$) and so is $f$.
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