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That's really embarrassing, however I need to ask it. I could not prove that the normed space $\langle \mathscr{C} [0, 1], \| \cdot \|_1 \rangle $ is complete (as a metric space), where $\| f\|_1 = \int_0^1 |f(x)|dx$. I don't really know how to explicit a continuous functions as a limit of a Cauchy sequence. I tried to prove that the pointwise limit (of the Cauchy sequence) have just a finite number of discontinuities (I don't know if it's true) and then approximate it by a continuous function.

Thanks in advance.

user40276
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    It is not complete. Perhaps you can come up with a sequence of continuous functions that converges (in $L_1$-norm) to the characteristic funnction of $[0,1/2]$. And then show why that answers your question. – GEdgar Mar 21 '13 at 01:12
  • @GEdgar This question appeared in some lecture notes of one professor of mine. But your counter example seems plausible. – user40276 Mar 21 '13 at 01:21
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    I just did this today while studying for my prelim. Here's a nice sequence of functions in $C([0,1])$: $f_n(x)=\begin{cases}0&0\leq x\leq(1-n)/n\ nx & (1-n)/2\leq x\leq (1+n)/2\ 1 & (1+n)/2\leq x\leq 1\end{cases}$. It's easy to show that $(f_n)$ is Cauchy with respect to $|\cdot|1$, and hence this space is not complete. It's more tedious to show that it's not Cauchy with respect to $|\cdot|{L^2}$, to verify that it's not a counterexample to $C([0,1])$ being complete with respect to $|\cdot|_{L^2}$. – gmoss Mar 21 '13 at 01:28
  • @gmoss Thanks for the example. – user40276 Mar 21 '13 at 01:35
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    The completion of $(C(X),|\cdot|_1)$ is the $L_1(X)$ space (by definition). On the other hand, as I can recall, it is complete w.r.t. the $\max$ norm. – Berci Mar 21 '13 at 01:47

1 Answers1

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This space is not complete. Consider functions $$ f_n(t)=\begin{cases}0\qquad\qquad\qquad\qquad t\in\left(0,\frac{1}{2} -\frac{1}{2n}\right)\\\frac{1}{2}+n\left(t-\frac{1}{2}\right)\qquad t\in\left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right]\\1\qquad\qquad\qquad\qquad t\in\left(\frac{1}{2}+\frac{1}{2n},1\right)\end{cases} $$ It is easy to check that $\{f_n\}_{n=1}^\infty\subset \mathscr{C}([0,1])$. Moreover this is a Cauchy sequence. Indeed $$ \lim\limits_{N\to\infty}\sup\limits_{m,n>N}\Vert f_n-f_m\Vert_1 =\lim\limits_{N\to\infty}\sup\limits_{m,n>N}2^{-2}|n^{-1}-m^{-1}| \leq\lim\limits_{N\to\infty}2^{-1}N^{-1}=0 $$ Assume that there exist $f\in \mathscr{C}([0,1])$ such that $\lim\limits_{n\to\infty}\Vert f_n-f\Vert_1=0$.

Assume that there exist $t_0\in\left(\frac{1}{2},1\right)$ such that $f(t_0)\neq 1$. Since $f$ is continuous then there exists a neighborhood $U_\delta(t_0)$ such that $|f(t)-f(t_0)|\leq\frac{|1-f(t_0)|}{2}$ for all $t\in U_\delta(t_0)$. In particular, by reverse triangle inequality we have $$ \begin{align} |1-f(t)| &=|(1-f(t_0))-(f(t)-f(t_0))|\\ &\geq\left||1-f(t_0)|-|f(t)-f(t_0)|\right|\\ &\geq\frac{|1-f(t_0)|}{2}. \end{align} $$ Now take $\delta'=\min\left(\delta,1-t_0,t_0-\frac{1}{2}\right)$, so we can assume $U_{\delta'}(t_0)\subset \left(\frac{1}{2},1\right)\cap U_\delta(t_0)$. Pick any natural number $N>\frac{1}{2(t_0-\delta')-1}$, then for all $n>N$ we have $U_{\delta'}(t_0)\subset \left(\frac{1}{2}+\frac{1}{2n},1\right)$. Note that for all $n>N$ and $t\in U_{\delta'}(t_0)$ we have $f_n(t)=1$, so $$ \Vert f_n-f\Vert_1= \int\limits_{(0,1)}|f_n(t)-f(t)|dt\geq \int\limits_{(t_0-\delta',t_0+\delta')}|f_n(t)-f(t)|dt= $$ $$ \int\limits_{(t_0-\delta',t_0+\delta')}|1-f(t)|dt\geq \int\limits_{(t_0-\delta',t_0+\delta')}\frac{|1-f(t_0)|}{2}dt =\delta'|1-f(t_0)|>0 $$ and as the consequence $$ 0=\lim\limits_{n\to\infty}\Vert f_n-f\Vert_1\geq\delta'|1-f(t_0)|>0 $$ Contradiction, therefore $f(t_0)=1$ for all $t_0\in\left(\frac{1}{2},1\right)$. Similarly one can show that for all $t_0\in\left(0,\frac{1}{2}\right)$ we also have $f(t_0)=0$. Now note that $$ \lim\limits_{t\to 1/2-0}f(t)=0\qquad\qquad\lim\limits_{t\to 1/2+0}f(t)=1 $$ so $f$ is not continuous. Contradiction, hence $\mathscr{C}([0,1])$ is not complete.

Norbert
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    Good complete answer to the problem. This same question has been asked many times, but this answer seems to be the only one that handles the subtlety to show that if the sequence has a limit, that limit has to be equal to the pointwise limit in this case (which is not continuous, hence the contradiction). IMO this question should kept and the linked question is the one that should be closed as duplicate instead. – PatrickR Mar 08 '20 at 07:02
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    I think there is a typo. Is $f$ identically zero on $(0,\frac{1}{2})$ instead? – Boar Apr 18 '20 at 08:34
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    @Steve, indeed. Thank you. – Norbert Apr 18 '20 at 08:38
  • @Norbert Nice anwser, I am trying to understand the mysterious calculations you've done, so I opened a question about that. Just letting you know in case you want to answer it, since you would be the most appropriate person. https://math.stackexchange.com/questions/4360066/how-do-i-make-sense-of-the-anwser-given-in-this-post-to-prove-the-non-completen – some_math_guy Jan 18 '22 at 17:42
  • @J.C.VegaO, thank you for proofreading my sloppy written post. There was no need to write your questions as a separate post. We can discuss it here in the comments. I've fixed all typos and mistakes that you've pointed out. Feel free to ask questions regarding the latest edit of my post. – Norbert Jan 18 '22 at 19:30
  • Thank you,I think in the last integrant you're missing the absolute value, since you're using $|1-f(t)|\geq\frac{|1-f(t_0)|}{2}$ I guess – some_math_guy Jan 18 '22 at 19:36
  • $\int\limits_{(t_0-\delta',t_0+\delta')}\frac{1-f(t_0)}{2}dt\geq \delta'(1-f(t_0))$ I thing this should be an equality – some_math_guy Jan 18 '22 at 19:40
  • If the objective is to have $U_{\delta'}(t_0)=(t_0-\delta',t_0+\delta')\subset \left(\frac{1}{2},1\right)$, then I would impose $t_0-\delta'>\frac{1}{2}$ and $t_0+\delta'<1$ . That is $\delta'<t_0-\frac{1}{2}$ and $\delta'<1-t_0 $, so I would take $\delta'=\min\left(1-t_0,t_0-\frac{1}{2}\right)$ . So why $\delta'=\min\left(\delta,t_0-\frac{1}{2}\right)$? Shouldn't it be $\delta'=\min\left(1-t_0,t_0-\frac{1}{2}\right)$ or $\delta'=\min\left(\delta,1-t_0,t_0-\frac{1}{2}\right)$ (not sure how to put the $\delta$ inside) – some_math_guy Jan 18 '22 at 19:45
  • I am not sure how you get $ \left||1-f(t_0)|-|f(t)-f(t_0)|\right| =\frac{|1-f(t_0)|}{2}$ in the third line of the reverse triangle inequality – some_math_guy Jan 18 '22 at 19:58
  • I am not sure why you use $ \lim\limits_{N\to\infty}\sup\limits_{m,n>N}$ to prove $f_n$ is Cauchy. I have only seen people take the limit for $n,m \to 0$ of$ ||f_n -f_m||_1$ – some_math_guy Jan 18 '22 at 20:01
  • @J.C.VegaO I want to ensure that $U_{\delta'}(t_0)\subset \left(\frac{1}{2},1\right)\cap U_\delta(t_0)$, hence the choice of $\delta'$ as $\min\left(\delta,1-t_0,t_0-\frac{1}{2}\right)$. – Norbert Jan 19 '22 at 10:15
  • @J.C.VegaO fixed the third line of the reverse triangle inequality – Norbert Jan 19 '22 at 10:17
  • @J.C.VegaO I use $\lim_{N\to\infty}\sup_{n,m>N}$ because this is a more rigorous way to take the limit as $m$ and $n$ tend to infinity – Norbert Jan 19 '22 at 10:18
  • Yes, but in the reverse triangle inequality, how do you do this $\left||1-f(t_0)|-|f(t)-f(t_0)|\right| \geq\frac{|1-f(t_0)|}{2}.$ – some_math_guy Jan 19 '22 at 11:38
  • Note that $|f(t)-f(t_0)|\leq \frac{|1-f(t_0)|}{2}$, so $-|f(t)-f(t_0)|\geq -\frac{|1-f(t_0)|}{2}$, hence $|1-f(t_0)|-|f(t)-f(t_0)|\geq |1-f(t_0)|-\frac{|1-f(t_0)|}{2}=\frac{|1-f(t_0)|}{2}$. Finally $||1-f(t_0)|-|f(t)-f(t_0)||=|\frac{|1-f(t_0)|}{2}|$. This is so trivial! – Norbert Jan 19 '22 at 22:29