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Prove that the sequence $\left((nx)/(1+4n^2x^2)\right)_{n\in\mathbb N}$ is not uniformly convergent on $(-a,a)$, where $a > 0$

My attempt:

$\lim_{n\to\infty}(nx)/(1+4n^2x^2) = 0 = f(x)$

Now, compute$$M_n = \sup_{-a<x<a}|f_n(x) - f(x)| = \sup_{-a<x<a}\left|(nx)/(1+4n^2x^2) - 0\right| = \sup_{-a<x<a}(nx)/(1+4n^2x^2)$$

Finally, compute $\lim_{n\to\infty} M_n$

The theorem says that if $\lim_{n\to\infty}$ $M_n$ = $0$, then $\bigl(f_n(x)\bigr)_{n\in\mathbb N}$ is uniformly convergent. And the question is to prove that it is not. Any help please? What is the supremum of $f_n(x)$, i.e what is $M_n$?

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  • $\begingroup$ What do you get for your supremum? $\endgroup$ – George Dewhirst May 9 at 10:50
  • $\begingroup$ @GeorgeDewhirst I dont know what is the supremum, I edited the question, can you review it? $\endgroup$ – dodo bc May 9 at 10:52
  • $\begingroup$ Sure basically you need to differentiate the function and set equal to zero, and find the turning points. See if they lie in $[-a,a]$. I suspect that for $n$ large enough they will. The turning point is likely to be a maximum. Supremum means maximise the function on the given interval. Once you have the maximum, it might be clear that this max as a function of $n$ won't tend to $0$ $\endgroup$ – George Dewhirst May 9 at 10:58
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You can compute $\sup f_n$ by the standard method of computing $f_n'$ and determining where it is equal to $0$. It turns out that $f_n'(x)=0\iff x=\pm\frac1{2n}$ and that $f_n\left(\pm\frac1{2n}\right)=\pm\frac14$. So, $\sup\lvert f_n\rvert=\frac14$, and therefore your sequence does not converge uniformly to the null function.

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  • $\begingroup$ Yes, that's what i was trying to do. I just got stuk at +-1/2. Thank you! $\endgroup$ – dodo bc May 9 at 11:00
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We have $f_n(\frac{1}{n})=\frac{1}{5}$ for all $n$. Hence $M_n \ge \frac{1}{5}$ for all $n$.

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