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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.

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

BCLC
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2 Answers2

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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 $\sup_{z\in A}|f(z)-f_n(z)|>n$.

Calvin Khor
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  • ok thanks. is my answer correct please? or if it's too long does it look correct please? – BCLC Oct 11 '21 at 15:33
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    I have no clue what you mean when you write $$ a <b \overset{\text {actually}} = c$$ because $b\neq c$. You refer to (2.1) but later make a bullet point that starts with 2.1 (not (2.1)). It looks like the broad strokes might be right on a glance. Similar issues with other post – Calvin Khor Oct 11 '21 at 23:30
  • thanks Calvin Khor. why are $b$ and $c$ not equal? $\frac{1}{\varepsilon = \frac12} - 1 = 1$ right? – BCLC Oct 13 '21 at 11:54
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    @JohnSmithKyon well OK, now I understand; didn't realise $\epsilon$ was fixed. But I have no ability to understand things like $a(=)b$ or $1/(a=b)-c=d$, if you don't explain further. IMO, it would have been better to just write $$\dots =\frac1{N+1} = \frac{\frac1\epsilon-1}{N+1},$$ or if you really want the brackets, $$\dots=\frac1{N+1} \left(= \frac{\frac1\epsilon-1}{N+1}\right),$$ so that I can still understand after deleting everything in the brackets. Or maybe even write out a full sentence, like: " $$...=\frac1{N+1},$$ which is actually equal to $\frac{\frac1\epsilon-1}{N+1}$." – Calvin Khor Oct 13 '21 at 13:08
  • ok thanks Calvin Khor. going to your answer, 1 - your choices for $\varepsilon$ and $n_{\varepsilon,N}$ what exactly? for $n_{\varepsilon,N}$, i guess your choice is the same as mine $n_{\varepsilon,N}=N+1$? – BCLC Oct 13 '21 at 13:57
  • 2 - how would you do this if we didn't conveniently have that $A$ intersects $\mathbb R$? Pick a point that has modulus of $\frac{1}{n}$? – BCLC Oct 13 '21 at 14:00
  • @JohnSmithKyon (1) for my answer, I have no interest in choosing $\epsilon$ or really any quantities directly found in the definition you wrote, which would make writing the answer a hassle. It simply suffices to find $z_n$ such that $|f_n(z_n) - f(z_n)|$ does not converge to zero. (2) yes, choose a sequence such that $z_n \to 0$, since $0$ is the problem point of the limit function (and not of any of the $f_n$s). My experience is that it doesn't matter what the $f_n$ are, this usually works – Calvin Khor Oct 13 '21 at 14:07
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    ok thanks Calvin Khor (btw Daily comment vote limit reached) – BCLC Oct 13 '21 at 19:52
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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 $|z_{\varepsilon,N}|^2 = r$.

details for uniform:

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

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

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

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

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

BCLC
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