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.

$\blacksquare$


Also see