Let $f_n(x) = x^n.$
$f_n(x)$ converges uniformly on $[0,r]$ for $r<1.$
Proof: Since $r < 1,$ $r^n \rightarrow 0 $ as $n \rightarrow\infty.$ Then, for given $\varepsilon >0 $, we can pick $M\in N$ such that $r^n<\varepsilon$ for all $n\ge M.$ Also, for $x\in[0,r], x\le r$ Then, for $n\ge M, |x^M-0|\le|r^M| = r^M<\varepsilon.$
- The definition of uniform convergence is $\forall\varepsilon>0,\exists M$ s.t. if $n\ge M, |x^n-0|<\varepsilon$ for all $x\in S.$ But, in my text, it uses $|x^M-0|$ instead of $|x^n-0|.$ Is there any reason for that? Can I use $|x^n-0|?$
While $f_n(x)$ converges uniformly on the above domain, it does not converge uniformly on $[0,1).$
Proof: Suppose $f_n$ converges uniformly. Pick $\varepsilon = 1/2. $ Then, $\exists M$ such that for $n\ge M, |x^n-0| <1/2$ for all $x\in[0,1).$ It implies $x^n<1/2 \rightarrow x<(1/2)^{\frac1n}$, but $\exists x$ such that $(1/2)^{\frac1n}<x<1$. Therefore, it is a contradiction.
- However, I am wondering why this proof is not applied to the first case. For example, if $r= 0.\overline{9},$ the first case and the second case look the same. Could you elaborate on the difference between the first and the second case?
Thank you in advance.
High powers of a (positive) number less than $1$ approach $0$.
$0.\bar9=1$, on the other hand (as noted in the comments)...