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In Bak-Newman's "Complex Analysis", there are two versions of Liouville's Theorem given:

1) A basic version: An entire function bounded by a constant $M$ is constant.
and
2) An extended version: An entire function $f(z)$ bounded by $A + B|z|^k$, for integer $k \ge 0$, is a polynomial of degree at most k.

I am wondering if in the extended version the exponent $k$ really needs to be an integer? Can the conclusion of the theorem be obtained (even with $k$ real but not an integer) by using the Cauchy Integral Formula and the $ML$ inequality for line integrals to show that the power series for $f(z)$ needs to be a polynomial?

Thanks!

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$k$ need not be integer. If $f(z)$ is bounded by $A+B |z|^k$ then $f(z)$ is bounded by $\tilde{A}+\tilde{B} |z|^{ \lceil k \rceil}$ which means $f$ is a polinomial of degree $\leq \lceil k \rceil$

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  • $\begingroup$ OK gotcha - thanks! So if $k$ were not an integer, then $f$ is a polynomial of degree at most $\lceil k \rceil$, so in fact must be degree $ \le \lceil k \rceil-1$, else the original bound wouldn't hold? $\endgroup$ – bryanj Mar 13 '13 at 19:38
  • $\begingroup$ That sounds right (unless $k$ is an integer of course). $\endgroup$ – user1337 Mar 13 '13 at 20:11
  • $\begingroup$ @bryanj, I would emphasize that the polynomial is of degree at most $\lfloor k \rfloor,$ because, for large $|z|,$ the modulus of $a_m z^m + a_{m-1} z^{m-1} + \mbox {more}$ is $|a_m| |z|^m \cdot \left( 1 + \mbox{small} \right).$ If integer $m > \lfloor k \rfloor $ we also have $m > k,$ and eventually $|a_m| |z|^m$ is too large. $\endgroup$ – Will Jagy Mar 13 '13 at 22:32

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