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Suppose $f_n(x) = x^n$ for $x\in [0,1]$. Does $f_n$ converge uniformly on $[0,1]$ and $\left[0,\frac12 \right]$?

My attempt :

$\lim_{n \to \infty} f_n(x)= F(x)$

Where $F(x)=0$ if $x=0$

$F(x)= \frac{1}{1-x}$ if $0

I said it's not uniformly convergent since $F(x)$ is not continuous at the point $0$ $\lim_{x \to 0} F(x) \neq F(0)$

Since if $f_n$ was to be uniformly continuous $\lim_{n \to \infty}f_n$ would have been continuous, I conclude $f_n$ is not uniformly convergent.

What did I do wrong here? Answer sheet says that it's not uniformly convergent on $[0,1]$ but says it is uniformly convergent on $[0,\frac{1}{2}]$ But my approach applies to $[0,\frac{1}{2}]$ and proves again.

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    $x^n\to0$ for $x\in[0,1)$. Are you thinking of $f_n(x)=1+x+\cdots+x^n$? – 2017-02-24
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    Perhaps did you mean $F(x)$ is not continuous at $1$? – 2017-02-24

2 Answers 2

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The pointwise limit function $F(x)$ is $0$ for $x \in [0,1)$ and $1$ for $x=1$. So F is actually continuous at $x=0$, but not at $x=1$. Hence convergence is not uniform on $[0,1]$.

On the other hand, for any $\epsilon > 0$, there is a natural number N such that for all $x\in[0,1/2]$ and $n\ge N$, $x^n <\epsilon$. So $f_n \to 0$ uniformly on $[0,1/2]$.

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For $x\in [0,x_0]$ for $00$

$$|x^n|\le |x_0|^n<\epsilon$$

whenever $n>\frac{\log(\epsilon)}{\log(x_0)}$. Hence the convergence of $x_n$ is uniform on any compact subset of $[0,1]$.

However, the convergence is not uniform on $[0,1]$ since for $\epsilon=1/4$ and any $N$ we can find $n>N$ (say $n\ge 2$) and find $x$ (say $x=1-1/n$) such that

$$x^n=(1-1/n)^n\ge 1/4$$

This is precisely the negation of uniform convergence.

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    Please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – 2017-04-12