$g_n(z)=z^n$ uniformly on $D={z: |z|<1}$?

Question

I am looking at this example:

$g_n(z)=z^n$ and domain $D={z: |z|<1}$

I see that every $g_n$ will converge to 0 for $n \rightarrow \infty$. Thus it converges.

Now, how can I show that it is or is not uniformly?

Answer 1

Try to compute the limit of $g_n\left(1-\frac1n\right)$ when $n\to\infty$. If the limit exists and is not zero, the convergence cannot be uniform on the disk $D$.

Once this is done, you might note that $1-\frac1n$ converges to $1$ which is not in the disk $D$ and try to imagine subsets of $D$ on which the convergence is uniform.

Edit: In view of the comments, let us recall the following facts:

• For every $c$, $\left(1-\frac{c}n\right)^n\to\mathrm e^{-c}$ when $n\to\infty$.
• For every $c$ and positive $a\gt1$, $\left(1-\frac{c}{n^a}\right)^n\to1$.
• For every $c\ne0$ and positive $a\lt1$, $\left(1-\frac{c}{n^a}\right)^n\to0$.
• Similar statements hold for the limit of $\left(1-x_n\right)^n$ for sequences $(x_n)$ such that $x_n\to0$, the limit depending on the asymptotics of $x_n$.