# $f_{n}=\frac{1}{1+z^{n}}$ uniform convergence

$\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\s}{\sigma}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\F}{\mathbb{F}}$

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 $0<r<1$

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.

-
For the first proposition, why do you compute the derivative? You can write $|1-r^n|=1-r^n\geq 1-r$ so $\sup_{z\in B(0,r)}|f_n(z)-f(z)|\leq \frac{r^n}{1-r}$. – Davide Giraudo Feb 12 '12 at 20:51
Thanks......... – VVV Feb 12 '12 at 21:42

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 $r < 1$) $$\left| \frac{1}{1+z^n} - 1 \right| = \left| \frac{z^n}{1+z^n} \right| < \frac{r^n}{1-r^n} \to 0\quad\text{as n\to\infty}.$$
Thank you. I understand the pointwise convergence, and also can see the $sup|f_{n}|$.I am confused because the estimate chain I want use to prove that all B(0) with 0<r<1 are uniform convergent you used to show that (D_{r}) is not uniform convergent with it. I am not sure what is right and wrong and what is the difference between $D_{|z|<1}$ and $B_{|r|<1}(0)$. – VVV Feb 12 '12 at 21:54
The sequence converges uniformly on each $D_r$ for $r < 1$, but not on $D_1$. This kind of behaviour is pretty common, but takes some time getting used to. – mrf Feb 12 '12 at 22:46
B(0) is the ball with center at 0 and radius $r 0<r<1$; $D_{1}$ is also a ball with radius one but not center 0. And thats the only difference with them? Thanks. – VVV Feb 12 '12 at 23:05
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 $r$. – mrf Feb 12 '12 at 23:09