Question

If f1(z) and f2(z) are two entire functions of order r1 and r2 respectively ,then show that (i) order of f1(z) - f2(z) <= max{r1, r2 } (ii) f1(z) . f2(z) <= max{r1 , r2}

Answer 1

Answer on Question #84514 – Math – Complex Analysis

Question

If $f_{1}(z)$ and $f_{2}(z)$ are two entire functions of order $r_1$ and $r_2$ respectively, then show that order of

(i) $f_{1}(z)\pm f_{2}(z)\leq \max \{r_{1},r_{2}\}$

(ii) $f_{1}(z)*f_{2}(z)\leq \max \{r_{1},r_{2}\}.$

Solution

By definition, an entire function $f$ is of finite order $r > 0$ if for (every positive but not negative) $0 \ll e > 0$ there is $w > 0$ such that $|f(z)| < \exp(|z|^{r + e})$ for $|z| > w$.

We have that

(i)

$$\begin{array}{l} | f _ {1} (z) \pm f _ {2} (z) | \leq | f _ {1} (z) | + | f _ {2} (z) | < 2 \exp \big (| z | ^ {\max \{r _ {1} + e, r _ {2} + e \}} \big) = \exp \big (\ln 2 + | z | ^ {\max \{r _ {1}, r _ {2} \} + e} \big) < \\ \exp \big (2 * | z | ^ {\max \{r _ {1}, r _ {2} \} + e} \big) < \exp \big (| z | ^ {\max \{r _ {1}, r _ {2} \} + 2 e} \big) \text{ for } | z | > \max \{w _ {1}, w _ {2}, w (\ln 2, r _ {1}, r _ {2}, e) \} \\ \end{array}$$

(ii)

$$\begin{array}{l} | f _ {1} (z) * f _ {2} (z) | = | f _ {1} (z) | * | f _ {2} (z) | = \exp \left(| z | ^ {r _ {1} + e} + | z | ^ {r _ {2} + e}\right) < \exp \left(2 * | z | ^ {\max \{r _ {1}, r _ {2} \} + e}\right) \\ \exp \left(| z | ^ {\max \{r _ {1}, r _ {2} \} + 2 e}\right) \text{ for all } e > 0 \text{ and } | z | > \max \{w _ {1}, w _ {2}, w (e) \} \\ \end{array}$$

The proof depends on the definition. We may assume equality in (i) if $r_1 \neq r_2$.

References:

Holland, Anthony S B (1973). An introduction to the theory of entire functions. pp 59-61.

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