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Could someone please use the following theorem to show how a pointwise convergent function is sometimes uniform convergent using the following theorem through the means of an example.

Example: $f_n:[0,1] \rightarrow \mathbb{R}$ defined by $f_n(x)=n^2x(1-x)^n$. This function is pointwise convergent to the zero function but is not uniformly convergent on $[0,1]$.

Thanks!

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Hint. Take $f_n(x)=n^ax(1-x)^n$ and $E=[0,1]$ for $a\geq 0$. Then $f_n$ is pointwise convergent to $f=0$. Show that
$$M_n=\sup_{x\in E}|f_n(x)-f(x)|=f_n\left(\frac{1}{n+1}\right)=\frac{n^a}{n+1}\left(1+\frac{1}{n}\right)^{-n}$$ which goes to $0$ iff $a<1$ (why?). For such $a$, the sequence $(f_n)_n$ is uniformly convergent to $f$ in $[0,1]$.

P.S. Note that $f_n$ is non negative in $[0,1]$ and $f_n'(x)=n^a(1-x)^{n-1}(1-(n+1)x)$. Hence $f_n$ attains its maximum value at $1/(n+1)$.

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  • $\begingroup$ Hi, thanks for that! Could you please explain the steps behind your reasoning. The answer to the example is that the f_n is not uniformly convergent. Thank you for your help. It is much appreciated $\endgroup$ – USERMATHS May 7 '17 at 13:07
  • $\begingroup$ My sequence is different from yours... By the way I 'm not sure what you are asking for. Could you restate your question please? $\endgroup$ – Robert Z May 7 '17 at 13:10
  • $\begingroup$ Thank you. Yes, the sequence in my question is pointwise convergent to the zero function. I was hoping for some help which shows the steps as to how we can show that the sequence is not uniformly convergent using the theorem 6.7 attached above. Please could you show the steps involved in this process. I'd be really grateful. Thank you very much. :) $\endgroup$ – USERMATHS May 7 '17 at 13:13
  • $\begingroup$ See my P.S. Any doubt? In case tell me what is unclear. $\endgroup$ – Robert Z May 7 '17 at 13:31

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