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 $E$:
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Let $z \in E$ so that $|z| < 1$.
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I claim that $nz^n \rightarrow 0$ under such a constraint.
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Let $\varepsilon > 0$.
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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 $\frac{n}{|z|^n} \rightarrow 0$).
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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?
Best Answer
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.