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In my notes, there is this theorem:

A sequence in $R^n$ converges (to a limit in $R^n$) iff it is Cauchy.

I understand that this theorem applies to all complete metric spaces, not just to $R^n$.

Now, here is my question:

The sequence of continuous functions $(f_n)$ in the space of continuous functions $C[0,1]$ equipped with the metric

$$d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$$

also converges to a limit in $C[0,1]$ iff it is Cauchy since $C[0,1]$ is complete.

But, what the convergence here explicitly is? Is it pointwise convergence or uniform convergence?

I understand that uniform convergence implies pointwise convergence with the same limit. Is it possible in the case above that there is a sequence of functions that converges pointwise and Cauchy in $C[0,1]$ but does not converge uniformly?

So is it more appropriate is it to re-write the Theorem as a sequence of continuous function in converges uniformly in $C$ iff it is Cauchy?

Thank you for the clarification.

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  • $\begingroup$ Do you know when a sequence converges uniformly? $\endgroup$ Feb 5 '16 at 6:04
  • $\begingroup$ @MhenniBenghorbal. I am not sure, I thought we should prove it to see whether it converges or not. Is there a way? $\endgroup$
    – user71346
    Feb 5 '16 at 6:26
  • $\begingroup$ I think you need to take it step by step. First you need to crasp what it means that a sequence converges uniformly. Then you will see that the definition of complete spaces does not require for Cauchy sequences to converge uniformly. However in your case since you are using sup norm then Cauchy sequences will converge uniformly. $\endgroup$ Feb 5 '16 at 6:55
  • $\begingroup$ See here for uniform convergence. $\endgroup$ Feb 5 '16 at 7:04
  • $\begingroup$ @MhenniBenghorbal. Thanks. But why is it that if I use the supremum norm, Cauchy sequences have to converge uniformly? $\endgroup$
    – user71346
    Feb 5 '16 at 9:21
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In the space of continuous function if you take any sequence $\{f_n\}$ with respect to any metric $d(f,g)$, if it is uniformly bounded and equi-continuous then you can say it converges uniformly by Arzela-Ascoli Theorem.

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Definitely it is meant uniform convegence since a uniform limit of a sequence of continuous functions is again continuous while a pointwise limit of continuous functions is not continuous in general. The convergence is always related to your topological structure of your space, and in your case, the topological structure is just a topology induced by metric. This metric is the supremum norm of $f-g.$

For example $f_n(x)=x^n$ on $[0,1]$ converges pointwise to $\chi_{\{1\}}$ which is not continuous.

I think that in some classical spaces you can "omit" the norm since the norm is by the definition known and other maybe still important norms does not play a great role in that given case. For example:

$l^p$ $(1\leq p<\infty)$ is a Banach spaces equipped by the norm $\|\cdot\|_p$. Since every sequence in $l^p$ is also bounded, you can also define the supremum norm on $l^p$. However, this norm does not imply completeness of the space $l^p$. Although it is worthwhile noticing that $l^p$ is dense in $c_0$ in the supremum norm.

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For your first question the answer is positive. It is not very hard to prove the completeness of $C[0,1]$ (or actually any $C(K)$ where $K$ is compact), you should first try doing it on your own.

For the second question, under the usual metric (which is just what you provided in your post as $d(f,g)$), if a sequence $\{f_n\}$ converges in $C[0,1]$ then it is almost trivial to see that $\{f_n\}$ converges uniformly on $[0,1]$. You can surely find without difficult that this convergence is a rephrase of the definition of uniform convergence on $[0,1]$.

There sure exist continuous functional sequences (like $\{x^n\}$) defined on $[0,1]$ that converge pointwise but not uniformly. However, in these cases they do not converge in $C[0,1]$.

If you need any further help please feel free to leave a comment under my post.

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