Define fn : [0,1] --> R by theequationfn(x) = xn. Show that thesequence(fn(x)) converges for each x belongs to [0,1],but thatthe sequence (fn) does not converge uniformly.

Accepted Solution

Answer:The sequence of functions [tex]\{x^{n}\}_{n\in \mathbb{N}}[/tex] converges to the function [tex]f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}[/tex].Step-by-step explanation:The limit [tex]\lim_{n\to \infty }c^{n}[/tex] exists and converges to zero whenever [tex]\lvert c \rvert <1 [/tex]. But, if [tex]c=1[/tex] the sequence [tex]\{c^{n}\}[/tex] is constant and all its terms are equal to [tex]1[/tex], then converges to [tex]1[/tex]. Using this result, consider the sequence of functions [tex]\{f_{n}\}[/tex] defined on the interval [tex][0,1][/tex] by [tex]f_{n}(x)=x^{n}[/tex]. Then, for all [tex]0\leq x<1[/tex] we have that [tex]\lim_{n\to \infty}x^{n}=0[/tex]. Now, if [tex]x=1[/tex], then [tex]\lim_{n\to \infty }x^{n}=1[/tex]. Therefore, the limit function of the sequence of functions is [tex]f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}[/tex].To show that the convergence is not uniform consider [tex]0<\varepsilon<1[/tex]. For any [tex]n>1[/tex] choose [tex]x\in (0,1)[/tex]  such that [tex]\varepsilon^{1/n}<x<1[/tex]. Then[tex]\varepsilon <x^{n}=\lvert f(x)-f_{n}(x)\rvert[/tex]This implies that the convergence is not uniform.