Does $\{nz^n\}_{n\in\mathbb N}$ converge uniformly in the open unit disc of $\mathbb{C}$?

Question

Let $E$ be the open unit disc about the origin in $\mathbb{C}$. Consider the sequence of functions $\{nz^n\}_{n\in\mathbb N}$ on $E$.

I'm trying to show that $\{nz^n\}_{n\in\mathbb N}$ converges pointwise on $E$, and then to investigate whether this convergence is uniform or not.

Attempt to show that $\{nz^n\}_{n\in\mathbb N}$ converges for every $z$ in $E$:

1. Let $z \in E$ so that $|z| < 1$.

2. I claim that $nz^n \rightarrow 0$ under such a constraint.

3. Let $\varepsilon > 0$.

4. Consider that

   $$ |nz^n - 0| = |nz^n| = n|z^n| = n |z|^n $$

   Now since $0 \le |z| < 1$, we have from real analysis that $|z|^n \rightarrow 0$. We also have from real analysis/calculus that the real sequence $|z|^n$ tends faster to $0$ than $n$ tends to $\infty$, in the sense that $n |z|^n \rightarrow 0$. (Equivalently, if instead $|z| > 1$ we have would that $\frac{n}{|z|^n} \rightarrow 0$).

5. It follows from the last step that -- for large enough $n$ -- we can make it so that $|nz^n - 0| < \varepsilon$, so that $nz^n \rightarrow 0$ as desired.

Is there an easy way to see the claim I make in (4) without handwaving? Also, does this sequence of functions on $E$ uniformly converge and if so why?

Answer 1

The sequence of functions $$f_n(z)=nz^n, \,\,\,n\in\mathbb N, $$ DOES ΝΟΤ converge uniformly to $0$ in the open unit disc $D$ as $$ f_n(z_n)=1, \quad\text{for}\,\,\, z_n=n^{-1/n}\in D, $$ even worse $$ f_n(z_n)=\frac{n}{2}, \quad\text{for}\,\,\, z_n=2^{-1/n}\in D. $$ In fact $$ \sup_{z\in D}\lvert\, f_n(z)\rvert=n, $$ and therefore $f_n$ DOES NOT converge uniformly in the open unit disc. Note that if $f_n:X\to\mathbb C$ converges uniformly to $f$, then for every $x_n\to x$, then $f_n(x_n)\to f(x)$.

On the other hand $f_n(z)=nz^n$ DOES converge uniformly to $0$ in any open disc $D_r$, $r<1$, since $$ \sup_{z\in D_r}\lvert f_n(z)\rvert=nr^r\to 0, $$ as $n\to \infty$. This can be proved using for example ratio test for sequences.

Answer 2

For 4., you can use the following:

Lemma. If $\delta>0$, then $\frac{(1+\delta)^n}{n}\rightarrow\infty$ or, equivalently, $n(1+\delta)^{-1}\rightarrow 0$.

Proof. Let $n\geq 2$. By the Binomial Theorem, $$\frac{(1+\delta)^n}{n}=\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\delta^n\geq\frac{1}{n}\binom{n}{2}\delta^2=\frac{n-1}{2}\delta^2\rightarrow\infty.\mathbf{QED}$$

To see that $nz^n$ does not converge uniformly in $E$, you could simply notice that $nE^n\supseteq E$ for every $n$. I'm going to explain this further:

By contradiction, suppose that $nz^n$ converges uniformly in $E$. Since the uniform limit (if it exists) is equal to the pointwise limit, we would have $nz^n\rightarrow 0$ uniformly in $E$. Let $\varepsilon$ be any number $0<\varepsilon<1$. Then for any $n\in\mathbb{N}$, $\varepsilon^{1/n}\in E$ but $|n(\varepsilon^{1/n})^n-0|\geq \varepsilon$, and that contradicts the uniform convergence (we had to show this for some $\varepsilon$, but this actually holds for any $\varepsilon$, which is obviously stronger).

Therefore, $nz^n$ does not converge uniformly in $E$.