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Let $f_n(x)=\frac{x^2}{x^2+(1-nx)^2}$ with $x \in [0,1]$ and $n \geq 1$. Prove $f_n$ converges (pointwise) to the zero function but has no uniformly convergent subsequence.

I've tried directly using the definition of convergence: Given any $\epsilon \geq 0$ we have that (after taking away the absolute value because every $f_n$ is non-negative and using the method of completing the square): $$\frac{x^2}{x^2+(1-nx)^2} < \epsilon \iff x^2n^2 - 2nx - (x^2(\frac{1}{\epsilon}-1)-1)>0 \iff x^2(n^2-\frac{2}{x}n + \frac{1}{x}^2)-1-(x^2(\frac{1}{\epsilon}-1))>0$$ And from there i end up getting $$n> \sqrt{\frac{1}{\epsilon}-1} + \frac{1}{x}$$ So, whenever this last inequality holds $f_m(x)$ will be within $\epsilon$ from zero, provided that $m \geq n$, and this proves that the sequence is pointwise converging to the zero function. Now, if i had to prove that the sequence is not uniformly convergent it would be enough to say that the function $n(x)=\sqrt{\frac{1}{\epsilon}-1} + \frac{1}{x}$ is not bounded above so there cannot exist an $m \in \mathbb{N}$ that "works" for every $x$ but i can't realize how can i extend this to any subsequence.

I've been looking on the internet and i found Arzelà-Ascoli theorem, i don't know if it is useful to prove this but i don't think we are supposed to use it since we haven't seen it in class. So how could i extend my reasoning to finish the proof? Thanks

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Since $f_n\left(\frac1n\right)=1$ for all $n$, for any subsequence $\{f_{n_k}\}$ we have $$\sup_{x\in[0,1]}f_{n_k}(x)=1, $$ and so $f_{n_k}$ does not converge uniformly.

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  • $\begingroup$ Thanks, that helps. I'll try to "translate" it into an $\epsilon -N$ proof. $\endgroup$
    – la flaca
    Apr 3 '16 at 19:23

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