5
$\begingroup$

I think I have this problem almost done. I am taking $C([0,1])$ to be the set of all continuous function $f\colon[0,1] \to \mathbb{R}$. I have already shown that $\displaystyle\|f\| = \max_{x\in [0,1]} |f(x)|$ is indeed a norm on $C([0,1])$ and makes it a normed space.

My next step was to show that if $f_n$ is a Cauchy sequence in $C([0,1])$ then for all $x \in [0,1]$ $f_n (x)$ is a Cauchy sequence in $\mathbb{R}$. That wasn't too hard.

Now I'm stuck trying to show the pointwise convergence of each sequence, that is $f(x) = \lim_{n \to \infty} f_n (x)$, by showing that $\displaystyle\sup_{x \in [0,1]} |f(x) - f_n (x)| \to 0 ,(n \to \infty)$.

I don't think it would be to difficult to show that $f \in C([0,1])$ after that and then I would have shown that $C([0,1])$ is a complete metric space. Any help is greatly appreciated.

$\endgroup$
  • $\begingroup$ I think you should define $f(x)$ as the limit of the cauchy sequence $f_n(x)$ in $\mathbb{R}$. Then you should show that the convergence towards this $f$ is uniform. Or simpler: What is $f$? How do you define it? $\endgroup$ – Quickbeam2k1 Jan 14 '14 at 17:10
3
$\begingroup$

Assume $f_n$ is a Cauchy sequence in $C([0,1])$. You said you showed that $f_n(x)$ is a Cauchy sequence of real numbers. The standard norm is complete on $\mathbb R$, so for any $x \in [0,1]$ we know $f_n(x)$ converges to some real number $\alpha_x$.

Now, define a function $f$ on $[0,1]$ by $f(x)=\alpha_x$. We need to show that $f \in C([0,1])$ and that $\|f_n-f\| \to 0$ as $n \to \infty$. That will conclude the proof of completeness.

Let $\varepsilon >0$. We have, if $n, m$ are large enough, $$ |f(x) - f_n (x)|=\lim_{m \to \infty} |f_m(x)-f_n(x)| < \varepsilon $$ and then $\|f_n-f\|=\sup_{x \in [0,1]}|f(x) - f_n (x)| < \epsilon$. Since $\varepsilon>0$ was arbitrary, $\|f_n-f\| \to 0$.

Finally, $f$ is a uniform limit of continuous function, then it's continuous, so we are done.

$\endgroup$
  • $\begingroup$ may you, please, elaborate some more on the statement that $$|f(x) - f_n (x)|=\lim_{m \to \infty} |f_m(x)-f_n(x)| < \varepsilon$$ ? I am having a hard time understanding the equality. Thank you $\endgroup$ – karhas Jan 22 '16 at 21:16
  • $\begingroup$ @karhas it's just using the fact that $f_m \to f$, as $m \to \infty$. The absolute value is a continuous function so we can pass the limit under it, and $f_n$ does not depend on $m$. $\endgroup$ – Silvia Ghinassi Jan 22 '16 at 21:18
  • $\begingroup$ $f=\lim f_m$? with respect to which metric? isn't this what we want to prove? $\endgroup$ – karhas Jan 22 '16 at 21:23
  • 1
    $\begingroup$ @karhas Pointwise. Which is already true, we need to show it's true with respect to the metric. $\endgroup$ – Silvia Ghinassi Jan 22 '16 at 21:24
  • 1
    $\begingroup$ @karhas By compactness of $[0,1]$ you can choose an $N$ that works for all $x$. $\endgroup$ – Silvia Ghinassi Jan 25 '16 at 3:30
1
$\begingroup$

Hint: It is sufficient to show that $C([0,1])$ is a closed set in $B([0,1])$ where $B[(0,1)]$ consists of all limited functions and then proof that $B([0,1])$ (with the maximum norm) is complete.

$\endgroup$
  • $\begingroup$ +1, though this is less of a hint and more of an alternative approach. $\endgroup$ – Cameron Buie Aug 27 '15 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.