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I was solving this problem:

For the following {$f_n$} sequence, determine the pointwise limit of {$f_n$} (if it exists) on the invterval, and indicate if {$f_n$} converges uniformly towards this function.$$f_n(x)=\sqrt[n]{x}, on [0,1]$$

I ended up with the pointwise limit being $\lim_{n\to\infty}x^{1/n}=1$ if $0<x\leq 1$ and $\lim_{n\to\infty}x^{1/n}=0$ if $x=0$. My problem is to determine if it does or doesn't converge uniformly, I saw a question about it on this website, but I didn't understand it. The answer was that it doesn't converge uniformly, can you explain me why? Thanks.

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  • $\begingroup$ read through the definition of uniform convergence $\endgroup$ – Sean Nemetz Feb 9 at 23:34
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    $\begingroup$ What is the limit of $f_n(x)$ as $n\rightarrow \infty$? Is it continuous on [0,1]? $\endgroup$ – Theo C. Feb 9 at 23:36
  • $\begingroup$ @TheoC. it's not helpful, because that's (continuity of limiting function) not the definition of uniform convergence (but merely a consequence). $\endgroup$ – enedil Feb 9 at 23:37
  • $\begingroup$ The limit is $0$ if $x=0$ and $1$ if $0<x\leq 1$, so it isn't continous on $[0,1]$. $\endgroup$ – davidllerenav Feb 9 at 23:38
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    $\begingroup$ Show we can't find $N$ that does not depend on $x$ such that $|f_n(x) - f(x) | < \epsilon$ for $n > N$ and for ALL $x \in (0,1]$. $\endgroup$ – RRL Feb 9 at 23:51
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Hint:

$$|1 -f_n(2^{-n})| = \frac{1}{2} \not\to 0$$

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  • $\begingroup$ I don't understand what you did there. Where does that $f_n(2^{-n})$ comes from? $\endgroup$ – davidllerenav Feb 10 at 0:13
  • $\begingroup$ The limit is $f(x) = 1$ for $0 < x \leqslant 1$ and $f(0) = 0$. For uniform convergence we must have for any $\epsilon > 0$, $N(\epsilon)$ such that $|f_n(x) - f(x)|< \epsilon$ for all $n > N$ and for all $x \in (0,1]$. Take $\epsilon = 1/4$ say. So if $x \neq 0$ we have $|f_n(x) - f(x)| = 1 - x^{1/n}$ But for any $n$ no matter how large if we look at $x = 2^{-n}$ we get $1 - (2^{-n})^{1/n} = 1/2$ and this is not less than $1/4$. $\endgroup$ – RRL Feb 10 at 0:21
  • $\begingroup$ I see. You choose $x=2^{-n}$ for convenience, right? In the case that $x=0$, it woudl be $|0-x^{1/n}| right? $\endgroup$ – davidllerenav Feb 10 at 0:31
  • $\begingroup$ Yes it was a convenient choice. We don't need to worry about $0$ where convergence is automatic since $f_n(0) = f(0) = 0$. It is points near $0$ in $(0,1]$ that cause problems. If $f_n$ converges uniformly on $[0,1]$ then it converges uniformly on $(0,1]$. So if it fails to converge uniformly on $(0,1]$ that is enough to show. $\endgroup$ – RRL Feb 10 at 0:34
  • $\begingroup$ Ok, thanks. What other $x$ could I choose? $\endgroup$ – davidllerenav Feb 10 at 1:22

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