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Let $(x_n)_{n\in N}$ be the sequence of continuous bounded functions $x_n=x^n$ on the unit interval $[0,1]$ and $C([0,1])$ be the space of continuous bounded real-valued functions on $[0,1].$ It is well known that this Vector space is complete in respect to the sup norm. But we also know that $(x_n)=x^n$ is converging towards a discontinuous function which is thus not element of $C[0,1]$.

I am wondering in what sense does the sup norm resolves this problem, i.e. how the limit of the sequence $x_n=x^n$ become an element of $C([0,1], {\left\lVert \cdot \right\rVert}_{sup}) \,\,?$ Many thanks.

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    $\begingroup$ $x^n$ is not a Cauchy sequence with respect to this norm is it? $\endgroup$ – Yanko Dec 31 '18 at 22:02
  • $\begingroup$ @Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{\left\lVert \cdot \right\rVert}_{sup}) )$ ? $\endgroup$ – user249018 Dec 31 '18 at 22:51
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    $\begingroup$ Instead of speaking of "the limit of $x^n,\, $" bear in mind that "the function $x^n\,$" is actually the set $\{(x,x^n):x\in [0,1[\}, $ and there may be more than one kind of limit of a sequence of infinite sets. A point-wise limit is not the same thing as a uniform limit. Convergence in the $\sup$ norm means uniform convergence only..... $x^n$ does not converge uniformly. $\endgroup$ – DanielWainfleet Jan 1 '19 at 4:49
  • $\begingroup$ In $C[0,1]$ the set $\{f_n: n\in \Bbb N\}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace. $\endgroup$ – DanielWainfleet Jan 1 '19 at 4:56
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The pointwise limit of the sequence is not become an element of $C([0,1])$. This poses no contradiction to the fact that $C([0,1])$ is complete. Indeed, the sequence of functions given by $f_{n}(x)=x^n$ is not uniformly Cauchy.

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  • $\begingroup$ Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ? $\endgroup$ – user249018 Dec 31 '18 at 22:10
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    $\begingroup$ Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with. $\endgroup$ – Sri-Amirthan Theivendran Dec 31 '18 at 22:18
  • $\begingroup$ So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits? $\endgroup$ – user249018 Dec 31 '18 at 22:28
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Your sequence $(f_n)_n$ is not Cauchy with respect to $\|\cdot\|_{\sup}$, not even on $[0,1)$.

Indeed, for any $n \in\mathbb{N}$ we have $$\|f_{n^2}-f_n\|_\sup \ge (f_{n^2}-f_n)\left(\sqrt[n]{1-\frac1n}\right)= \left(1-\frac1n\right)^n - \left(1-\frac1n\right) \xrightarrow{n\to\infty} \frac1e - 1 \ne 0$$

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