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1) Find all entire functions that are uniformly continuous on $\mathbb{C}$.

2) Find all entire functions $f(z)$ such that such that for every integer $n \geq 1$,

$$\oint_{\partial\mathbb{D}} f(z)\bar{z}^ndz = 0,$$ where $\mathbb{D}$ is the unit disk.

I'm a bit shaky on the first one, but I think it's that an entire function has an infinite radius of convergence, so is everywhere normally convergent. So if each term in it's power series is uniformly continuous on $\mathbb{C}$, then the function will be uniformly continuous on $\mathbb{C}$. Am I on the right track?

For the second, I'm not sure how to use the Cauchy Integral Formula since $f(z)\bar{z}^n$ isn't holomorphic.

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$z \mapsto z^2$ is entire, but not uniformly continuous. – copper.hat Oct 1 '12 at 0:41
Is $\mathbb{D}$ the unit disk? – user17794 Oct 1 '12 at 0:42
In general for differentiable functions, uniform continuity does not imply that the derivative is bounded, but I would suggest starting there. For the second question, I suggest writing $f$ as a power series, writing the integral in terms of the standard parametrization $z=e^{i t}$, and computing. – Jonas Meyer Oct 1 '12 at 0:44
Hint for #1: $f(z+\delta)-f(z)$ is an entire function. – user31373 Oct 1 '12 at 1:11
@KannaguchiO. Having $|f(z+\delta)-f(z)|<\epsilon$ for all small $\delta$ does not imply $|f'(z)|<\epsilon$. The definition of derivative involves divided difference. – user31373 Oct 1 '12 at 2:46
up vote 1 down vote accepted

Definition of uniform continuity: for any $\epsilon$ there is a $\delta$ such that $|x-y| < \delta \implies\ |f(x)−f(y)|<\epsilon$ for any choice of $x, y$.

A trascendental entire function cannot be uniformly continuous, having an essential singularity at $\infty$ (so you can find arithmetical sequences $z_i$ with exponentially growing values of $f(z_i)$).

So we are left with polynomial, which are not uniformly continuous even in the real numbers, except for the linear case $z \to az+b$ which is uniformly continuous almost by definition.

For the second question, the line integral is exactly the Cauchy integral over the unit cicle, as $z^{-1} = \bar{z}$. So the definition is equivalent to saying that every derivative (starting from $n=0$ which is the actual value of the function) is zero, so the only solution is the constant $f:z \to 0$

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And how exactly you are going to construct $z_j$ and guarantee exponential growth? – leshik Oct 1 '12 at 2:54
Yes, it was a stupid idea. A trascendent function's derivative is trascendent too, so it takes arbitrarily high values. This is probably the key to proving the non-uniformity. – rewritten Oct 1 '12 at 3:33
In fact, @LVK's comment on the question is a plainly better route. Just take $g(z) = f(z+\delta/2)-f(z)$. This is entire, and bounded by $\epsilon$, so it's a constant. So the function is linear. – rewritten Oct 1 '12 at 3:37

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