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Show that the sequence of functions $\left\{ x^n \right\}$ converges uniformly on $[0,k],k<1$, but non-uniformly on $[0,1]$.

For $x\in[0,1)$ $$\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}x^n=0$$ and for $x=1$ $$\lim_{n\to \infty}f_n(x)=1.$$

Therefore, the limit function is $$f(x)=\cases{ 0 &$\quad 0\leq x <1$\\ 1 &$\quad x=1$}$$ Let $\epsilon >0$ be arbitrary. Then, for each $x\in (0,1)$ $$|f_n(x)-f(x)|<\epsilon\implies x^n<\epsilon \implies n\ln x<\ln\epsilon \implies n>\frac{\ln\epsilon}{\ln x}$$
For $x=0,1$ $$|f_n(x)-f(x)|=0<\epsilon,\quad\forall n\geq1 $$ Choose a natural number $N(\epsilon,x)=\left[\frac{\ln\epsilon}{\ln x}\right]+1$ Here $[\cdot]$ is the box function.

Note that $N(\epsilon,x)\to +\infty$ as $x\to 1^-$. So, $N(\epsilon,x)$ is not bounded on $[0,1]$.

Hence, the convergence of $\left\{ x^n \right\}$ is not uniform on $[0,1]$.

I faced problems in showing uniform convergence of $\left\{ x^n \right\}$ on $[0,k]$, $k<1$.

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You can use $\epsilon$ or $\varepsilon$; using $\in$ is rather unusual; compare $\epsilon$ or $\varepsilon$; and $\in$. –  Martin Sleziak Aug 7 '12 at 14:59
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up vote 2 down vote accepted

Given $\varepsilon>0$ take $n_0$ such that $k^{n_0} <\varepsilon$. As $x<k<1$, we have \begin{equation} |x^n-0|=|x^n|<|k^{n}|<|k^{n_0}|<\varepsilon \quad \forall x \in[0,k], \forall n\ge n_0. \end{equation}

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Hint: given $\epsilon > 0$, find $n$ such that $k^n < \epsilon$.

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