Answer to Question #122939 in Real Analysis for Ruksan

Question

Show that {fn}, where fn : [0, 1] → R is the map fn(x) = x
n, is not uniformly convergent.

Answer 1

Given that , $(f_n)$ be a sequence of function.

Where $f_n:[0,1]\rightarrow\R$ defined by $f_n(x)=x^n$ .

Clearly, if $x=1$ then the sequence $(f_n(1))$ converges to 1 .

If $0\leq x <1$ then the sequence $(f_n(x))$ converges to 0 as we known that $lim_{n\to \infty} x^n=0$ if $0\leq x<1$ .

Let $f(x)=\begin{cases} 1 & \text{if} \ x=1 \\ 0 & \text{if} \ 0\leq x <1 \end{cases}$

Thus the sequence of function $(f_n(x))$ convergent to $f$ on the set $[0,1]$ .

If $n_k=k \ and \ x_k=(\frac{1}{2})^{ \frac{1}{k}}$ then

$| f_{n_k}(x_k)-f(x_k)|=|\frac{1}{2}-0|=\frac{1}{2}$ .

Therefore the sequence $(f_k)$ doesn't converge uniformly on $[0,1] \ to \ f$ .