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Discuss the following sequences limit and whether their convergence is uniform in the region $\alpha\le |z|\le\beta$ with finite $\alpha , \beta \gt 0$.

a. $f_n (z) = {1 \over nz^2}$

So...

$$ \Big| {1 \over nz^2}\Big| = {1 \over n|z|^2} \le {1 \over n\alpha^2} $$

Then let

$$ N = {1 \over \epsilon\alpha^2} $$ Thus $$ \Big| {1 \over nz^2}\Big| \lt \epsilon, \ for \ n \gt N $$

and converges uniformly.

d. $f_n (z) = {1 \over 1+(nz)^2}$

This is the one I'm not so sure about, part of me wants to kind of just do like last time.

$$\Bigg|{1 \over 1+(nz)^2}\Bigg| \le \Big| {1 \over nz^2}\Big| = {1 \over n|z|^2} \le {1 \over n\alpha^2} $$

and then say I have uniform convergence once again. But then I'm just not sure what that singularity ($z={i \over n}$) is meaning for this sequence. Basically I'm confused if that affects uniform convergence. Does it not in this case because my region is bounded by $\alpha$?. And since it is, then I only have a finite amount of functions in my sequence that actually have a singularity? So in other words does uniform convergence depend on all functions in my sequence or just infinitely many (the tail end of sequence).

Sorry I'm not sure if I'm making any sense so some elucidation on this subject would be MUCH APPRECIATED

Thanks

levon r.
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1 Answers1

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The first part is fine.

Concerning the second example: It seems that you used the "inequality" $$|1+w^2| \geq |w^2|$$ (where $w=n \cdot z$ in your case) to obtain the first inequality. But this estimate is not correct for all $w \in \mathbb{C}$. Take for example $w=\imath$, then $$|1+\imath^2| = 0 \not \geq |\imath^2|=1$$ This means that your proof doesn't work.

Hint Use the Triangle inequality instead, i.e. $$|1+w^2| \geq ||w|^2-1| \geq |w|^2-1$$

The singularities aren't a problem at all. Since $\alpha>0$ you can choose $N \in \mathbb{N}$ large enough such that $\left|\frac{\imath}{n}\right| < \alpha$ for all $n \geq N$. Thus, the singularities are outside the given domain for large $n$. And, convergence is always about the tail of the sequence, right from the definition! (For all $\varepsilon>0$ there exists $N$ large enough such that ....)

saz
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