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I am trying to show and disprove uniform convergence in the following example:

$D=\{z\in \C | |z| < 1 \} \ and \ f_{n}:D\rightarrow \C : f_{n}(z)=\frac{1}{1+z^{n}} $

Proposition 1: $f_{n}$ converges uniformly in all $B(0)$ with $0

Proof 1: With $f_{n}(z)= (1+z^n)^{-1}$ put

f'_{n}(z)=-nz^{n-1}(1+z^n)^{-2}=0 \Rightarrow z=0

(I wanted to use that \lim sup |f_{n} - f | = 0 but I fail at finding the supremum of $|f_{n}-f|$) How does one find the supremum?

So I try finding an estimate instead: $|f_{n} - f| = |\frac{1}{1+z^{n}}-1| = |\frac{-z^{n}}{1+z^{n}}| \le \frac{r^{n}}{|1-r^{n}|} =: \epsilon$

Is this done correctly?

Proposition 2: $f_{n}$ does not converge uniformly in D

Proof 2: How can one show that something does not converge uniformly??

Thanks for suggestions.

  • 1
    For the first proposition, why do you compute the derivative? You can write $|1-r^n|=1-r^n\geq 1-r$ so $\sup_{z\in B(0,r)}|f_n(z)-f(z)|\leq \frac{r^n}{1-r}$.2019-04-29

1 Answers 1

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If $|z|<1$, then $z^n \to 0$ as $n \to \infty$, so $f_n$ converges pointwise to the constant function $1$.

On the other hand $\sup_{z\in D} |f_n| = +\infty$. (To see this, for a fixed $n$, take a sequence of points in $D$ converging to a $n$:th root of $-1$.)

On $D_r = \{ z : |z| < r \}$, (if $r < 1$) $ \left| \frac{1}{1+z^n} - 1 \right| = \left| \frac{z^n}{1+z^n} \right| < \frac{r^n}{1-r^n} \to 0\quad\text{as $n\to\infty$}.$

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    I actually didn't notice the $B(0)$ notation, but from your original post, it looks like you mean the same thing as what I wrote as $D_r$, i.e. the open disc, centered at the origin with radius $r$.2012-02-12