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The function $f_n(x)=3n^3(x-1/n)^2$ for $x\in[0,2]$ is given, and I need to show whether the sequence of functions $(f_n)_{n\in\Bbb{N}}$ converges uniformly to $f(x)=\lim\limits_{n \to \infty}f_n(x)$, which I have determined to be $3x^2$. It's given that $(f_n)_{n\in\Bbb{N}}$ converges pointwisely, but I'm not sure whether that information is useful in this case, and I'm not sure how to proceed to determine whether it's uniformly convergent or not. Any help would be appreciated

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    $\begingroup$ Are you sure it is $2n^3(x-\frac{1}{n})^2$? It seems to approach $+\infty$ for any $x \in [0,2]$. Isn't it something like $3(x-\frac{1}{n})^2$? $\endgroup$ – LucaMac Sep 2 '18 at 10:33
  • $\begingroup$ Yes, it does say $3n^3(x-1/n)^2$, but another subtask did specify that it's only that for $0 \lt x \le n/2$ and that it's $0$ elsewhere, I didn't mention it here because in that subtask you had to find the integral for $f_n(x)$ and I didn't think that it could be relevant here $\endgroup$ – pavus Sep 2 '18 at 10:41
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    $\begingroup$ The sequence of functions doesn't converge pointwise, so it can't converge uniformly. $\endgroup$ – Vinícius Novelli Sep 2 '18 at 10:42
  • $\begingroup$ The task specifically says that it converges pointwisely... I am now seriously concerned about whether the task itself has mistakes $\endgroup$ – pavus Sep 2 '18 at 10:44
  • $\begingroup$ Also if we consider $0 < x \leq \frac{n}{2}$ it goes to $+\infty$, isn't it $\frac{2}{n}$ ? In this way its pointwise limit would be $0$. $\endgroup$ – LucaMac Sep 2 '18 at 10:44
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Ok! So, in the comments we discovered $f_n(x) = 3n^3 (x- \frac{1}{n})^2$ only if $0 < x \leq \frac{2}{n}$ and $f_n(x)=0$ otherwise.

Now this clearly converges pointwisely to $0$. (For any $x>0$ for $n$ big enough $x > \frac{2}{n}$)

But $$ ||f_n-0|| = \sup\limits_{x \in (0,\frac{2}{n}]} |f_n(x)| \geq f_n(\frac{2}{n}) = 3n$$ and thus the convergence is not uniform.

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  • $\begingroup$ Thank you very much, both for the answer and being patient in the comments! $\endgroup$ – pavus Sep 2 '18 at 11:00

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