Check if $\left\{\frac{1}{z^2+\frac 1 n}\right\}_{n=1}^{\infty}$ is uniform (or at least pointwise) convergent.

Question

Check if $\{f_n\}_{n=1}^{\infty}$ is uniform (or at least pointwise) convergent, where $$f_n(z) = \frac{1}{z^2+\frac 1 n}, f_n: A \to \mathbb C,$$ $$A = \{z \in \mathbb C | |z| < 1, \Re(z)>0\}$$ $$\stackrel{\text{I believe}}{=} \{z \in \mathbb C | |z|^2 < 1, \Re(z)>0\}$$ $$\stackrel{aka}{=} \{(x,y) \in \mathbb R^2 | x^2+y^2<1, x>0\}$$ i.e. (I believe) the open right half of the unit disc.

• Note: $\frac{n}{nz^2+1} = \frac{1}{z^2+\frac 1 n}$, and I guess $\frac{n}{nz^2+1}$ is very different from $\frac{z}{nz^2+1}$ or $\frac{x}{nx^2+1}$ as in here: Does $f_n(x) = \frac{x}{1 + nx^2}$ converge uniformly for $x \in \mathbb{R}$?

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1. Is this correct for Pointwise? (I'll type details as an answer.)

• Related question: Complex non-/uniform convergence: About dependence of the index $N$ on modulus $|z|$ vs complex number $z$

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2. Is this correct for Uniform? --> See here: Check if $\{\frac{1}{z^2+\frac 1 n}\}_{n=1}^{\infty}$ is uniform convergent.

• Related question: Uniform convergence: Question about in/dependence of the indices on other variables

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3. Quick clarification: Just because the $N$ in proving pointwise convergence is true technically has a '$z$' in its formula, doesn't mean not uniformly convergent right?

I guess yes because in general for a uniformly convergent sequence it's possible we could still find an $N$ that depends on $z$ but another $N$ that doesn't depend on $z$, i.e. we didn't look hard enough or something. After all, the condition for uniform convergence is that we find some large enough index $N$ not that every damn index $N$ is gonna be independent.

Answer 1

Of course, pointwise convergence is very easy to check: fix $z\in A$, and then investigate the convergence of the real sequence $(f_n(z))_n$. This sequence indeed converges, towards the limit value $f_\infty (z) = \frac{1}{z^2}$.

Now suppose $f_n$ converges uniformly. Then its limit is necessarily $f_\infty$. However, we have:

$$ | f_n(z) - f_\infty(z)| = \frac{1}{z^2(nz^2+1)}$$

which CANNOT be bounded independently of $z$ since (for instance) :

$$| f_n(\frac{1}{\sqrt{n}}) - f_\infty(\frac{1}{\sqrt{n}})| = \frac{n}{2}$$

Which goes to infinity as $n$ goes to infinity. So the series is not uniformly convergent.

Answer 2

outline for pointwise:

I think yes, $\{f_n\}_{n=1}^{\infty}$ is pointwise convergent, namely to $f:A \to \mathbb C, f(z) = \frac{1}{z^2}$.

Proof: For any $(\varepsilon,z) \in (0,\infty) \times A$, choose $N_{(\varepsilon,z)} = \frac{1+\varepsilon|z|^2}{\varepsilon|z|^4}$. Then $|f_n(z)-f(z)| < \varepsilon$ whenever $n > N_{(\varepsilon,z)}$. QED

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details for pointwise:

Suppose $n > N$. We must show that

$$|\frac{-\frac 1 n}{(z^2 + \frac 1 n) z^2} = \frac{1}{z^2 + \frac 1 n} - \frac{1}{z^2}| < \varepsilon.$$

Step 1.

$$n > N $$ $$\iff n \varepsilon|z|^4 > 1+ \varepsilon|z|^2$$ $$\iff n \varepsilon|z|^4 - \varepsilon|z|^2 > 1$$ $$\iff \varepsilon(n |z|^2 - 1) > \frac{1}{|z|^2}$$ $$\implies \varepsilon|n |z|^2 - 1| > \frac{1}{|z|^2}$$ $$\iff \varepsilon > \frac{1}{|z|^2 |n |z|^2 - 1|}$$

Step 2.

$$\frac{1}{|z|^2 |n |z|^2 - 1|} \ge \frac{1}{|z|^2 |nz^2 +1|} \ \text{by reverse triangle inequality}$$

$$\frac{1}{|z^2| |nz^2 +1|}$$

$$\frac{1}{|(z^2)(nz^2 +1)|}$$

$$\frac{1}{|n(z^2)(z^2 + \frac{1}{n})|}$$

$$|\frac{1}{n(z^2)(z^2 + \frac{1}{n})}|$$

$$|\frac{\frac{1}{n}}{(z^2)(z^2 + \frac{1}{n})}|$$

$$= |\frac{-\frac 1 n}{(z^2 + \frac 1 n) z^2}|$$

Step 3.

$$\frac{-\frac 1 n}{(z^2 + \frac 1 n) z^2} = \frac{1}{z^2 + \frac 1 n} - \frac{1}{z^2}$$

Step 4. By Steps 1 and 2, we have $\varepsilon > |\frac{-\frac 1 n}{(z^2 + \frac 1 n) z^2}|$

Step 5. By Steps 3 and 4, we have $\varepsilon > |\frac{1}{z^2 + \frac 1 n} - \frac{1}{z^2}|$.