Question

Define fn : [0,1] --> R by the

equationfn(x) = xn. Show that the
sequence(fn(x)) converges for each x belongs to [0,1],
but thatthe sequence (fn) does not converge uniformly.

Answer 1

The sequence of functions
`\{x^(n)\}_{n\in \mathbb{N}}` converges to the function

`f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}`.

Explanation:

The limit
`\lim_(n\to \infty )c^(n)` exists and converges to zero whenever
`\lvert c \rvert <1`. But, if
`c=1` the sequence
`\{c^(n)\}` is constant and all its terms are equal to
`1`, then converges to
`1`. Using this result, consider the sequence of functions
`\{f_(n)\}` defined on the interval
`[0,1]` by
`f_(n)(x)=x^(n)`. Then, for all
`0\leq x<1` we have that
`\lim_(n\to \infty)x^(n)=0`. Now, if
`x=1`, then
`\lim_(n\to \infty )x^(n)=1`. Therefore, the limit function of the sequence of functions is

`f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}`.

To show that the convergence is not uniform consider
`0<\varepsilon<1`. For any
`n>1` choose
`x\in (0,1)` such that
`\varepsilon^(1/n)<x<1`. Then

`\varepsilon <x^(n)=\lvert f(x)-f_(n)(x)\rvert`

This implies that the convergence is not uniform.