# Uniform convergence of $x^n$ on $(-1,1)$.

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 ?

• Thus, doing a proof by contradiction, i want to show that $$\forall f \in \mathbb{R}^{(-1,1)} \exists \epsilon_0 > 0 \forall N \in \mathbb N \exists n \geq N \exists x_n \in (-1,1): |x^n - f(x)| \geq \epsilon_0$$ – user42761 Nov 11 '12 at 0:23
• The point wise limit is $0$. Calculate $\lim_{n\to\infty}\|f_n\|$. – user44687 Nov 11 '12 at 0:23

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).
• Thanks. That's it. But you say "fix n". Shouldn't it be "let $n \in \mathbb N$ arbitrary" ? – user42761 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
• So its sufficient to show that $\exists \epsilon > 0 \forall N \in \mathbb N \exists n \geq N \exists x \in (-1,1): |x^n| \geq \epsilon$ ? – user42761 Nov 11 '12 at 0:27