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I'm stuck in the next Analysis problem:

Let $(f_n)_{n\in \mathbb{N}}$ with $f:[0,1]\rightarrow \mathbb{R}$, $f_n(x)=x^n$ Prove that the sequence is not uniformly Cauchy.

Attempts:

So first of all the negation of the definition of being Cauchy is:

$\exists \varepsilon >0$ such that for all $N\in \mathbb{N}$ with $m,n\geq N$ we have $|f_n(x)-f_m(x)|\geq \varepsilon$

So to work this out, there are two obvious inequalities

$|f_n(x)-f_m(x)|=|x^n-x^m|\geq |x^n|-|x^m|$

$|f_n(x)-f_m(x)|=|x|^m |x^{n-m}-1|$

But I can't manage to find a lower bound for the difference $|f_n-f_m|$. Should I try for specific values of $\varepsilon$?

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  • $\begingroup$ Try a proof by contradiction. Suppose the sequence of functions $(f_{n})_{_{n\geq 1}}=(x^{n})_{_{n\geq 1}}$ is uniformly Cauchy. There is a theorem that this is equivalent uniform convergence. What can you now say about your given sequence of functions? $\endgroup$ – Procore Jun 3 '17 at 5:40
  • $\begingroup$ Oh! Right, the sequence of functions is NOT uniformly convergent. I get it! Thanks :) $\endgroup$ – User117E29 Jun 3 '17 at 5:48
  • $\begingroup$ You got it! My pleasure. Also, if you want to prove it directly, the negation of the definition of a uniformly Cauchy sequence of real-valued functions defined on a (fixed) set $E\subseteq\mathbb{R}$, say $(g_{n})_{_{n\geq 1}}$, is there exists an $\varepsilon>0$ such that for all $N\in\mathbb{N}$ there exists an $m\geq N$ and an $n\geq N$ such that $\big|g_{n}(x)-g_{m}(x)\big|\geq\varepsilon$ at some $x\in E$ - I mainly wanted to mention to remember the dependence on the domain of definition in addition to the index. $\endgroup$ – Procore Jun 3 '17 at 5:54
  • $\begingroup$ Also, an easy way to show a contradiction that the given sequence of functions doesn't converge uniformly, fix $\varepsilon=\frac{1}{2}$ (or any real number strictly less than 1), so that an $N\in\mathbb{N}$ must exist such that $x^{N}<\frac{1}{2}$ for all $x\in[0,1)$ (note that $x^{n}$ does converge to $0$ on $[0,1)$ as $n\rightarrow+\infty$). So, for any sequence $(x_{k})_{_{k\geq 1}}$ such that $\lim\limits_{k\rightarrow+\infty}x_{k}=1$ we have $x^{N}_{k}<\frac{1}{2}$. However, by continuity, $1=1^{N}=\lim\limits_{k\rightarrow+\infty}x_{k}^{N}\leq\frac{1}{2}$, a clear contradiction. $\endgroup$ – Procore Jun 3 '17 at 6:12
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If you know what is the limit of $(1-1/m)^m$, try with $n=2m$, $x=1-1/m$ and your second idea, ie you have $|f_n(x)-f_m(x)|=(1-\frac{1}{m})^m (1-(1-\frac{1}{m})^m)$, and choose $\varepsilon$ sufficiently small.

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  • $\begingroup$ Oh I see! Do people just naturally know about using this particular m's and n's? Because I think I would never manage to get those specific ones! :) $\endgroup$ – User117E29 Jun 3 '17 at 5:51
  • $\begingroup$ This sequence , or $v_n=(1+\frac{1}{n})^n$ is a classic counter-example for a frequent error, namely that if $x_n>0$ has limit $1$, and if $y_n\to +\infty$, then $x_n^{y_n}\to 1$. It is frequently used. $\endgroup$ – Kelenner Jun 3 '17 at 5:55

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