Check if $\{\frac{1}{z^2+\frac 1 n}\}_{n=1}^{\infty}$ is uniform 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? --> See here: Check if $\{\frac{1}{z^2+\frac 1 n}\}_{n=1}^{\infty}$ is uniform (or at least pointwise) convergent.

• 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? (Again, I'll type details as an answer.)

• 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

Simpler to just try say $z_n=1/n\in A$ (for e.g. $n>2)$, since things are going real bad near zero. $$f(1/n)-f_n(1/n)=\frac1{1/n^2}-\frac1{1/n^2+1/n}=n^2-\frac{n^2}{1+n}>n^2-\frac {n^2}{0+n}=n(n-1)> n. $$ Hence $\sup_{z\in A}|f(z)-f_n(z)|>n$.

Answer 2

outline for uniform:

I think no, $\{f_n\}_{n=1}^{\infty}$ is not uniformly convergent (to its pointwise convergent limit).

Proof: Choose $\varepsilon = \frac12$. Let $N>0$. We must find $n_{\varepsilon,N} > N$ and $z_{\varepsilon,N} \in A$ s.t. $|f_n(z)-f(z)| \ge \varepsilon = \frac12$. Choose $n_{\varepsilon,N} = N+1$ and any $z_{\varepsilon,N}$ s.t. $|z_{\varepsilon,N}|^2 < \frac{1}{N+1} (\stackrel{\text{actually}}{=}) \frac{\frac 1 \varepsilon - 1}{N+1} $ (see (2.1)). QED

• 2.1. Such $z_{\varepsilon,N}$ exists because: For $0 < \frac{1}{N+1} < 1$ (where $\frac{1}{N+1} < 1$ because $N>0$), we can think of $\frac{1}{N+1}$ as a (positive) squared radius smaller than 1: Then pick an even smaller squared radius $0 < r < \frac{1}{N+1} (< 1)$. Then pick any $z_{\varepsilon,N} \in A$ whose radius squared is $|z_{\varepsilon,N}|^2 = r$.

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details for uniform:

$$|\frac{1}{z^2+\frac 1 n} -\frac 1 {z^2}| $$ $$=|\frac{-\frac{1}{n}}{(z^2+\frac 1 n)(z^2)}| $$ $$=|\frac{\frac{1}{n}}{(z^2+\frac 1 n)(z^2)}| $$ $$=|\frac{1}{n(z^2+\frac 1 n)(z^2)}| $$ $$=|\frac{1}{(nz^2+1)(z^2)}| $$ $$=|\frac{1}{nz^2+1}| |\frac{1}{z^2}| $$ $$= |\frac{1}{z^2}| |\frac{1}{nz^2+1}|$$ $$= \frac{1}{|z|^2} \frac{1}{|nz^2+1|}$$ $$ \ge \frac{1}{|nz^2+1|} \ \text{because} \ z \in A$$ $$ \ge \frac{1}{n|z|^2+1} \ \text{by triangle inequality}$$ $$= \frac{1}{(N+1)|z|^2+1} \ \text{by choice of n}$$ $$> \frac 12 \ \text{by choice of z because...}$$

$$\frac{1}{(N+1)|z|^2+1} > \frac 12$$

$$\iff 2 > (N+1)|z|^2+1$$

$$\iff 1> (N+1)|z|^2$$

$$\iff \frac{1}{N+1} > |z|^2$$