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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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.

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

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