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Suppose that for some positive integer n and some positive constant M an entire function $f(z)$ satisfies the inequality:$$ |f(z)|\leq M.|z|^n$$ for all $ z\in \mathbb{C}$ with sufficiently large modulus. Show that $f(z)$ must be a polynomial whose degree does not exceed n.

and find all entire functions that are uniformly continuous on the complex plane.

I tried this by using the Cauchy estimates: $$ |f^k(0)|\leq \frac{M |z|^n}{R^k}$$. Where R is the radius of sufficiently large disk such that $|z|> R$. For upto $(k=0,1,......n)$
$|z|^n/R^k$ grater than $1$ as $k>n$ we the term $|z|^n/R^k$ is less than $1$ and we can make as small as we like for sufficiently large n. Thus after big $n$ all my $f^k(0)$ are all most zero thus $f(z) = \sum_{n=0}^n a_n z^n$ which is a polynomial of degree n.

and I do not have any idea about the another part.

Does someone have suggestion and hints?

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2 Answers 2

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$$f^k(0)=\dfrac{k!}{2\pi}\int_0^{2\pi}\frac{f(Re^{it})}{R^ke^{ikt}}dt$$ Hence for $R>>$ we have

$$|f^k(0)|\leq \dfrac{k!}{2\pi}\int_0^{2\pi}|\frac{f(Re^{it})}{R^ke^{ikt}}|dt \leq\dfrac{k!}{2\pi}\int_0^{2\pi}\frac{R^n}{R^k}dt\leq k!R^{n-k}$$ Now for $k>n$, and passing to limit as $R\to \infty$ we get $$|f^k(0)|=0$$.

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Hint: You can show that any polynomial of degree $> 1$ is not uniformly continuous, by appealing directly to the definition.

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  • $\begingroup$ $f:x\mapsto \sqrt x$ is uniformly continuous (on $\Bbb R_+^*$) but $f'$ is note bounded. $\endgroup$
    – Hamou
    Oct 3, 2014 at 11:07
  • $\begingroup$ Thanks, Hamou, I'd temporarily confused uniformly continuous with Lipschitz. $\endgroup$ Oct 3, 2014 at 11:24

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