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Suppose a metric $d$ is defined on the space of entire functions as follows:

$$d(f, g)=\sum_{n=1}^{\infty} \min \left\{\frac{1}{2^{n}}, \max _{|z| \leq n}|f(z)-g(z)|\right\}$$ Is the operator of differentiation (the operator sending $f$ to $f^\prime$) continuous on this metric space of functions? Explain why or why not.

Here is my try with Cauchy's Integral Formula.
For $|z|<n$, $$\begin{aligned}|f^\prime(z)-g^\prime(z)|&= |\frac{1}{2\pi i}\int_{|w|=n}\frac{f(w)-g(w)}{(w-z)^2} dw|\\ &\leq \frac{1}{2\pi} \max_{|w|\leq n}|f(w)-g(w)| \int_{|w|=n} \frac{1}{(w-z)^2} dw. \end{aligned}$$

Since $z$ can be very closed to the boundary, $|w-z|$ can be very small. I don't know how to bound the integral in the inequality.

Any idea to solve this question?

Ariel So
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  • So, if attempting to prove that differentiation is continuous is running into difficulty, have you tried the opposite, proving that it is not continuous? In, particular, guided by what has gone wrong with your attempt to prove that it is? – Paul Sinclair Jan 07 '20 at 14:05
  • it might help to note that this metric exactly gives the topology of pointwise convergence – Lukas Rollier Jan 08 '20 at 19:18
  • @LukasRollier: No, this is a metric which gives locally uniform convergence. – Lukas Geyer Jan 09 '20 at 21:22
  • Hint: You can bound $|f'(z)-g'(z)|$ for $|z| \le n$ in terms of the bound of $|f(z)-g(z)|$ for $|z| \le 2n$. – Lukas Geyer Jan 09 '20 at 21:24

1 Answers1

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For convenience I will use the notation $M_n = \max\limits_{|z|=n} |f(z)-g(z)|$ and $M_n' = \max\limits_{|z|=n} |f'(z)-g'(z)|$. The trick is to integrate over a larger circle. I.e., for $|z| \le n$ you have $$ \begin{split} |f'(z)-g'(z)| &= \left| \frac{1}{2\pi i} \int_{|w|=2n} \frac{f(w)-g(w)}{(w-z)^2} \, dw \right| \le \frac{M_{2n}}{2\pi} \int_{|w|=2n} \frac{|dw|}{|w-z|^2} \\ & \le \frac{M_{2n}}{2\pi} \cdot \frac{4\pi n}{n^2} = \frac{2M_{2n}}n \le 2 M_{2n}. \end{split} $$ Now if $d(f,g) < 2^{-2N}$ for some positive integer $N$, then by the definition of $d$ we have $M_{2n} < 2^{-2N}$ as long as $n \le N$, which implies that $|f'(z)-g'(z)| \le 2^{-2N+1}$ for $1 \le n \le N$, and so $M_{n}' \le 2^{-2N+1}$ for $1 \le n \le N$. This shows that $$ d(f',g') \le \sum_{n=1}^N M_n' + \sum_{n=N+1}^\infty \frac{1}{2^n} \le N 2^{-2N+1} + 2^{-N} $$ and since this goes to $0$ as $N \to \infty$, it shows that $f \mapsto f'$ is continuous with respect to this metric.

Lukas Geyer
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