Function that converge uniformly

Does the sequence $x, ..., x^n$ of functions converge uniformly on the interval $[0,k]$ for $k\in(0,1)$? If it does, prove it. How about on the interval $[0,1]?$

Can you help me complete the proof?

For the interval $[0,k]$ I got it does and for $[0,1]$ it doesn't. So let $\{f_n\} = \{x^n\}$, and suppose $f^n \to f$. Then I must show that for $\epsilon>0$, there exists an $N$ such that $d(f,f^n)<\epsilon$ whenever $n>N$ for all $x.$

For $k\in (0,1),$ it is clear to see that $x^n\to 0$ as $n$ approaches infinity. Then I must show $|x^n|<\epsilon$ whenever $n$ is greater than some $N$.

On $[0,k]$, $x^n$ attains its max at $x=k$ so $x^n<k^n$. Then note that $k^n$ decreases with increasing $n$, so we choose $N$ such that $k^N<\epsilon$.

$\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.

Let $x\in[0,1)$ and $t\in[0,x]$. Suppose $n\in\mathbb{N}$. Then $$|t^n| \le x^n.$$ Since $x^n\to 0$ as $n \to\infty$, the series converges uniformly to $0$ on $[0,x]$.
Furthermore, note that the $x^n$ converge pointwise to $0$ on $(0,1)$ and to $1$ at $x = 1$. Since the target functions is not continuous, the convergence on $[0,1]$ cannot be uniform.
• Thanks and for when $x\in [0,1]$ how can I complete that? – user104235 Nov 3 '13 at 2:00
• This works for any nonnegative $x$ chosen that is less than $1$. – ncmathsadist Nov 3 '13 at 2:09
you can use that the funtions $f_n(x)=x^n$ are continuous at $x=1$, then you can say that for any $\epsilon>0$ and any $n\in\mathbb{N}$ there exists $\delta>0$ such that $f_n(x)>\epsilon$ for all $x\in(1-\delta,1)$. Then the sequence can not converge uniformly to 0.
• How can I show that there exists a $\delta$ such that $f_n(x)>\epsilon$? – user104235 Nov 3 '13 at 2:01
• using the continuity at 1, you can get $|f_n(x)-1|<1-\epsilon$ (take $\epsilon$ less than 1) for $x\in(1-\delta,1)$. This implies $f_n(x)>\epsilon$. – fernanfio Nov 3 '13 at 2:36