# How the sup norm works to make a normed space complete

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.

• $x^n$ is not a Cauchy sequence with respect to this norm is it? – Yanko Dec 31 '18 at 22:02
• @Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{\left\lVert \cdot \right\rVert}_{sup}) )$ ? – user249018 Dec 31 '18 at 22:51
• 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. – DanielWainfleet Jan 1 '19 at 4:49
• In $C[0,1]$ the set $\{f_n: n\in \Bbb N\}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace. – DanielWainfleet Jan 1 '19 at 4:56

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.
• 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 ? – user249018 Dec 31 '18 at 22:10
• Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with. – Sri-Amirthan Theivendran Dec 31 '18 at 22:18
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 $$\|f_{n^2}-f_n\|_\sup \ge (f_{n^2}-f_n)\left(\sqrt[n]{1-\frac1n}\right)= \left(1-\frac1n\right)^n - \left(1-\frac1n\right) \xrightarrow{n\to\infty} \frac1e - 1 \ne 0$$