Order of Shifted Entire Function
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Theorem
Let $f: \C \to \C$ be an entire function of order $\alpha$.
Let $a \in \C$.
Then $\map f {z + a}$ has order $\alpha$.
Proof
We shall verify Definition 3 of Order of Entire Function.
Let $\map g z := \map f {z + a}$ for $z \in \C$.
Then for all $R > \cmod a$:
- $\ds \max_{\cmod z \mathop \le R \mathop - \cmod a} \cmod {\map f z} \le \max_{\cmod z \mathop \le R} \cmod {\map g z} \le \max_{\cmod z \mathop \le R \mathop + \cmod a} \cmod {\map f z}$
Thus:
- $\paren 1 : \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R \mathop - \cmod a} \cmod {\map f z} }{\ln R} \le \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R} \cmod {\map g z} }{\ln R} \le \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R \mathop + \cmod a} \cmod {\map f z} }{\ln R}$
On the other hand, for each $c \in \R$:
- $\ds \paren 2 : \lim_{R \mathop \to +\infty} \dfrac {\map \ln {R + c} } {\ln R} = 1$
Thus:
\(\text {(3)}: \quad\) | \(\ds \) | \(\) | \(\ds \limsup_{R \mathop \to +\infty} \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R \mathop - \cmod a} \cmod {\map f z} }{\ln R}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{R \mathop \to +\infty} \dfrac {\map \ln {R - \cmod a} }{\ln R} \limsup_{R \mathop \to +\infty} \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R \mathop - \cmod a} \cmod {\map f z} }{\map \ln {R - \cmod a} }\) | Product Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \cdot \alpha\) | by $\paren 2$ and by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha\) |
Similarly, we have:
- $\ds \paren 4 : \limsup_{R \mathop \to +\infty} \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R + \cmod a} \cmod {\map f z} }{\ln R} = \alpha$
In view of $\paren 3$ and $\paren 4$, the claim follows from $\paren 1$ by Sandwich Rule.
$\blacksquare$