Answer to Question #310714 in Real Analysis for Dhruv bartwal

Question

Let fn(x)= x^n is not uniformly continuous on [0,1] but is uniformly continuous on [0,k]

Answer 1

Solution

${f_n}\left( x \right) = {x^n}$ and $[0,1]$

For $x=1$

${f_n}\left( 1 \right) = {1^n} = 1$

Therefore,

$\mathop {\lim }\limits_{n \to \infty } {f_n}\left( 1 \right) = \mathop {\lim }\limits_{n \to \infty } \left( 1 \right) = 1\$

Now for $0 \le x < 1$

$\begin{array}{c} \mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {x^n}\\ \mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = 0 & & x < 1 \end{array}$

Therefore, 

$\mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \left\{ \begin{array}{l} 0 & 0 \le x < 1\\ 1 & x = 1 \end{array} \right.\$

This shows that the ${f_n}\left( x \right) = {x^n}$ is piecewise convergent over the interval $[0,1]$ . 

Now for the interval $[0,k]$

We have

$\begin{array}{c} \mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {x^n}\\ \mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \left\{ \begin{array}{l} 0 & x < 1\\ \infty & x > 1 \end{array} \right. \end{array}\$

Therefore, for any value of $k>1$ , the ${f_n}\left( x \right) = {x^n}$ is divergent. 

Hence

${f_n}\left( x \right) = {x^n}$ is piecewise convergent over the interval $[0,1]$  whereas not uniformlay continuously on the interval $[0, k]$