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Let $f_n:(-1,1) \rightarrow \mathbb R: x\mapsto x^n$. I want to show that the sequence $(f_n)_{n=0}^\infty$ can not converge uniformly. How can i prove that ?

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Thus, doing a proof by contradiction, i want to show that $$ \forall f \in \mathbb{R}^{(-1,1)} \exists \epsilon_0 > 0 \forall N \in \mathbb N \exists n \geq N \exists x_n \in (-1,1): |x^n - f(x)| \geq \epsilon_0$$ – Epsilon Nov 11 '12 at 0:23
The point wise limit is $0$. Calculate $\lim_{n\to\infty}\|f_n\|$. – Goku Nov 11 '12 at 0:23
up vote 3 down vote accepted

The pointwise limit is zero.

Fix $\varepsilon>0$. Fix $n$. Then $f_n((2\varepsilon)^{1/n})>\varepsilon$. So the convergence cannot be uniform (because that would mean that for $n$ big enough you can make your $f_n$ less than $\varepsilon$ at all points).

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Thanks. That's it. But you say "fix n". Shouldn't it be "let $n \in \mathbb N$ arbitrary" ? – Epsilon Nov 11 '12 at 0:30
Yes, you are right. But it's just terminology, and usually the way to prove that something occurs for "arbitrary $n$" is to fix one (arbitrary) and do something with it. – Martin Argerami Nov 11 '12 at 0:31

Note that $\lim_{n\to\infty}(1-\frac{1}{n})^n=\frac{1}{e}$,which is a contradiction of uniform convergence,since the pointwise limit is zero

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If it converges uniformly, then it has to converge uniformly to its pointwise limit, which is 0. You can show that it doesn't behave that way by using (the negation of) the definition of uniform convergence.

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So its sufficient to show that $\exists \epsilon > 0 \forall N \in \mathbb N \exists n \geq N \exists x \in (-1,1): |x^n| \geq \epsilon$ ? – Epsilon Nov 11 '12 at 0:27

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