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I want to check whether the sequence below converges pointwise or uniformly

$$f_{n}(z):\{z\in\mathbb{C}:|z| = 1\}\to\mathbb{R}$$ $$\qquad\qquad f_{n}(z) = \frac{1}{1+n^{2}|z-e^{in}|}$$

I have tried using the triangle inequality $||z|-|w||\leq|z-w|$ but that gives me $f_{n}(z)\leq 1$.

Intuitively, I don't even know whether the sequence is meant to converge pointwise because while $n^{2}\to\infty$, $e^{in}$ is dense on the unit circle, so it seems to me that $|z-e^{in}|$ should be close to $0$ infinitely many times.

How can I prove/disprove this sequence converges pointwise and/or uniformly?

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    $\begingroup$ I don't think it converges pointwise. Here is a reason why: fix $z=\cos(t_0)+i\sin(t_0)$, the quantity $n^2\vert z-e^{in}\vert=n^2 [(t_0-n)\mod 2\pi]^2$ doesn't seem to converge to any particular value or diverge to $\infty$. Of course this may be (very) wrong but I would at least try this line of reasoning $\endgroup$ – Pedro Aug 12 at 13:43
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Put $\Bbb T=\{z\in\Bbb C: |z|=1\}$. Suppose to the contrary that for each point $z\in\Bbb T$ the sequence $\{f_n(z)\}$ converges to $f(z)$. Then $f(z)=0$, because, since the set $\{e^{in}:n\in\Bbb N\}$ is dense in $\Bbb T$, for any natural $N$ there exists $n>N$ such that $|z-e^{in}|>1$, so $|f_n(z)|<1/n^2$.

Now in order to answer the question we can try to apply a general topological tool which is Baire Theorem. For this we need to provide a cover of $\Bbb T$ by a countable many sets. Namely, for each natural $N$ put $$\Bbb T_N=\{z\in\Bbb T: |f_n(z)|\le 1/2\mbox{ for each }n\ge N\}.$$ The pointwise convergence of the sequence $\{f_n\}$ implies that $\Bbb T=\bigcup_{N\in\Bbb N}\Bbb T_N$. Since each function $f_n$ is continuous (being a composition of continuous functions; for instance, since a function $g_n(z):\Bbb T\to\Bbb R$, $z\mapsto 1+n^{2}|z-e^{in}|$ is continuous and positive, a function $f_n$ is continuous being a composition $jg_n$ of the function $g_n$ and a continuous function $j:\{x\in\Bbb R:x>0\}\to \Bbb R$, $x\mapsto x^{-1}$ ), a set $\{z\in\Bbb T: |f_n(z)|\le 1/2\} $ is closed being a preimage $f_n^{-1}([-1/2,1/2])$ of a closed set $[-1/2,1/2]$. Thus the set $\Bbb T_N$ is closed being an intersection of a family of closed sets. By Baire theorem, there exists $N\in\Bbb N$ such that $\Bbb T_N$ has non-empty interior in $\Bbb N$. There exists a number $n>N$ such that $e^{in}\in\Bbb T_N$. Then $f_{n}(e^{in})=1>1/2$, a contradiction.

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    $\begingroup$ Good answer, could I get some further hints on these questions - 1. What's the easiest way to see $f_{n}$ is continuous? 2. Why is $\mathbb{T}_{N}$ closed? 3. I'm not familiar with the Baire Theorem - what is the intuition behind it and why is it used here? $\endgroup$ – BaroqueFreak Aug 15 at 2:25
  • $\begingroup$ @BaroqueFreak I updated my answer. Sorry for the delay. $\endgroup$ – Alex Ravsky yesterday

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