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I've been searching for a function series $ f_n:[0,\infty] -> \mathbb{R} $ such that $ (f_n)_{n\geq1} $ uniformly converges to $f$, but $(f^2_n)_{n\geq1}$ does not uniformly converges to $f^2$.

I've tried it with many functions. Does anyone have a hint?

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Hint: $$f_n(x) = x-\frac1n{}$$

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  • $\begingroup$ I've got (2xn-1)/n^2 should be greater then or equals ɛ​. Is that correct? $\endgroup$ – Maddude Jul 17 '17 at 11:10
  • $\begingroup$ That's right. Now, for your given $N\in \Bbb N $, choose $x$ large enough that $\frac{2xN - 1}{N^2} > \epsilon$, and you have disproven uniform convergence. (Some boks require you to find an $x$ such that $\frac{2x(N+1) - 1}{(N+1)^2} > \epsilon$, but they're both very doable.) $\endgroup$ – Arthur Jul 17 '17 at 11:16
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    $\begingroup$ Well, no, it's not a full solution, as there are a lot of details to type out. But if you're wondering where the idea came form , then that's precicely because $f^2 - f_n^2$ is, in general, roughly $2f$ times larger than $f-f_n$ (as you discovered). You just need an example where that allows you to make the difference arbitrarily large for any $n$, and that means having an $f$ that gets arbitrarily large. Second, $f-\frac1n$ is the standard way of making a non-trivial sequence that converges uniformly to $f$. $\endgroup$ – Arthur Jul 17 '17 at 11:22
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    $\begingroup$ Yes, exactly. When disproving uniform convergence of a sequence $f_n\to f$, you're free to pick an $\epsilon$ (you've chosen $1$, which is an excellent choice in my opinion, as long as it works with whatever example your working on; some times, when you're not free to choose the function yourself, $\frac12$ or $\frac14$ might be needed), then you're given an arbitrary $N$, and then you get to choose an $x$. If the given quintuple $f,f_n,\epsilon,N,x$ violates the inequality in the definition of uniform convergence, then you've shown that the sequence is not uniformly convergent. $\endgroup$ – Arthur Jul 17 '17 at 11:41
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    $\begingroup$ The order, and who chooses what (whether you choose or you're given an arbitrary value) is important in order to determine what you're actually showing: Different ways give different results, including proving uniform convergence (arbitrary $\epsilon$, you choose $N$, arbitrary $x$), proving pointwise convergence (arbitrary $\epsilon$, arbitrary $x$, you choose $N$), or disproving pointwise convergence (you choose $\epsilon$, you choose $x$, arbitrary $N$). Note specifically that switching between proving and disproving only swaps which numbers are arbitrary and which ones you get to choose. $\endgroup$ – Arthur Jul 17 '17 at 11:46

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