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Check convergence of

$f_{n}(x)=x^{n}-x^{2n}$ where $x\in(0,1)$

Please verify my answer, I'm not sure I'm doing it correctly. Thanks in advance!

Steve
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1 Answers1

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Convergence is not uniform. Let $f(x)=x^n-x^{2n}=x^n(1-x^n)$, then $f'(x)=nx^{n-1}(1-2x^n)$. Maximum of $f$ on $]0,1[$ is found at $x=(\frac{1}{2})^\frac{1}{n}$, and this maximum is $1\over 4$.

However, there is obviously pointwise convergence on $]0,1[$.

nouk
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  • why does maxima being at given point make it non uniform convergence? Would you bother to explain? –  Mar 26 '13 at 23:02
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    Of course. Uniform convergence means $sup |f_n-f| \rightarrow 0$. How could that happen if $f$ is the null function and for all $\epsilon>0$, there is some $x$ such that $f(x)>\epsilon$ (actually, $f(x)={1\over4}$) ? I comes just from definition of uniform convergence. – nouk Mar 27 '13 at 07:16
  • There is a typo in the preceding comment, it's $f_n(x)>\epsilon$ – nouk Mar 27 '13 at 11:11
  • There is a further typo, it should be $f_n(x_n)\gt\epsilon$. – Did Mar 29 '13 at 07:23