Does $f_{n}(z) = \frac{1}{1+n^{2}|z-e^{in}|}$ converge pointwise or uniformly?

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?

• 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 – Pedro Aug 12 at 13:43

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.
• 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? – BaroqueFreak Aug 15 at 2:25