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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.

Kannaguchi O.
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    $z \mapsto z^2$ is entire, but not uniformly continuous. – copper.hat Oct 01 '12 at 00:41
  • Is $\mathbb{D}$ the unit disk? –  Oct 01 '12 at 00:42
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    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 01 '12 at 00:44
  • If $\mathbb{D}$ is the circle of center $0$ and radius $r$, then the integral in then second question is the $n^{\textrm{th}}$ Fourier coefficient of the function $\theta \mapsto f(r e^{i \theta})$ (up to a factor $r^n$). – Joel Cohen Oct 01 '12 at 00:47
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    Hint for #1: $f(z+\delta)-f(z)$ is an entire function. –  Oct 01 '12 at 01:11
  • In the first, $f$ uniformly continuous implies that $|f(z+\delta)-f(z)|< \epsilon$. This in turn implies that $|f'(z)| < \epsilon$. So $f$ must be a constant. – Kannaguchi O. Oct 01 '12 at 01:28
  • My calculation for the second problem implies that $f$ must be such that $f^{(n)}(0) = 0$. So $f$ must be identically zero on the unit disk. – Kannaguchi O. Oct 01 '12 at 01:43
  • @Kannaguchi O: I don't understand your comment about the first question, but I take back my suggestion because LVK's is better. As for your comment about the second, be careful about which $n$. And remember your conclusion is not just about what happens on the disk. – Jonas Meyer Oct 01 '12 at 01:50
  • In the first that inequality holds for any $z$ by uniform continuity. But since the LHS holds for any smaller choice of delta, it holds in the limit as $\delta \to 0$. So I concluded that the derivative was bounded, though now I see that your comment above concerned this very conclusion and I haven't used LVK's fact about that difference being entire.

    With the integral, by induction and the Gauss MVT I conclude that $f^{(n-1)}(0) = 0$ for all n. So by uniqueness, it must be identically zero on the unit disk, and therefore zero on the entire plane.

     – Kannaguchi O. Oct 01 '12 at 02:09
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    @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. –  Oct 01 '12 at 02:46

1 Answers1

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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 $f:z \to 0$

rewritten
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    And how exactly you are going to construct $z_j$ and guarantee exponential growth? – leshik Oct 01 '12 at 02:54
  • 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 01 '12 at 03:33
  • 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 01 '12 at 03:37