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I am completely stuck on this problem: $C[0,1] = \{f: f\text{ is continuous function on } [0,1] \}$ with metric $d_1$ defined as follows:

$d_1(f,g) = \int_{0}^{1} |f(x) - g(x)|dx $.

Let the sequence $\{f_n\}_{n =1}^{\infty}\subseteq C[0,1]$ be defined as follows:

$ f_n(x) = \left\{ \begin{array}{l l} -1 & \quad \text{ $x\in [0, 1/2 - 1/n]$}\\ n(x - 1/2) & \quad \text{$x\in [1/2 - 1/n, 1/2 +1/n]$}\\ 1 & \quad \text{ $x\in [1/2 +1/n, 1]$}\\ \end{array} \right. $

Then $f_{n}$ is cauchy in $(C[0,1], d_1)$ but not convergent in $d_1$.

I have proved that $f_{n}$ is not convergent in $(C[0,1])$ since it is converging to discontinuous function given as follows:

$ f_n(x) = \left\{ \begin{array}{l l} -1 & \quad \text{ $x\in [0, 1/2 )$}\\ 0 & \quad \text{$x = 1/2$}\\ 1 & \quad \text{ $x\in (1/2 , 1]$}\\ \end{array} \right. $

I am finding it difficult to prove that $f_{n}$ is Cauchy in $(C[0,1], d_1)$. I need help to solve this problem.

Edit: I am sorry i have to show $f_n$ is cauchy

Thanks for helping me.

Srijan
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    Just compute $d_1(f_n, f_m)$ ... – martini May 31 '12 at 21:54
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    The sequence is Cauchy with $d_1$. That is why you are having problems showing it is not Cauchy. If $m,n > N$, then it is straightforward to get the estimate $d_1(f_m,f_n) < \frac{2}{N}$. – copper.hat May 31 '12 at 22:05
  • @copper.hat I must be stupid enough wasted one hour to show not cauchy but at the time when i wrote title , i got the intution of something wrong here. Now i am trying. – Srijan May 31 '12 at 22:13
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    It happens to most of us... – copper.hat May 31 '12 at 22:15
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    Be careful when you say "$f_n$ is not convergent since it is converging to discontinuous function". Remember, "convergent" means something different here than pointwise or uniform convergence - for example, the functions $f_n(x) = n^{-x}$ converge to the 0 function, even though $f_n(0) = 1$ for all $n$! So be sure you're using correct reasoning to show that the sequence isn't convergent in the given metric space. – MartianInvader May 31 '12 at 22:39
  • @MartianInvader Thanks for valuable comment. Do i need to write $f_n$ is converging to some discontinuous function? – Srijan May 31 '12 at 22:53
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    @srijan In this metric space, there are no discontinuous functions, so "converging to some discontinuous function" doesn't make sense. You need to prove that there's no continuous function to which the $f_n$ converge, ie there's no continuous function $f$ such that $\int_0^1 \lvert f(x) - f_n(x) \rvert dx$ approaches zero. Just showing that the functions converge pointwise to a discontinuous function isn't enough to prove this, as my above example shows. – MartianInvader May 31 '12 at 23:08
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    Related questions: http://math.stackexchange.com/questions/21878/examples-of-function-sequences-in-c0-1-that-are-cauchy-but-not-convergent and http://math.stackexchange.com/questions/97171/cauchy-sequence-in-x-on-0-1-with-norm-int-01-xtdt – Jonas Meyer Jun 01 '12 at 04:03
  • @JonasMeyer Thank you very much. Now i am reading those problems. – Srijan Jun 01 '12 at 04:32

1 Answers1

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Suppose $m,n > N$. Then $f_m(x) = f_n(x) = -1 $ when $x \in [0, \frac{1}{2}-\frac{1}{N}]$. Similarly, $f_m(x) = f_n(x) = +1 $ when $x \in [\frac{1}{2}+\frac{1}{N},1]$. And $|f_m(x)-f_n(x)| < 1$ when $x \in (\frac{1}{2}-\frac{1}{N}, \frac{1}{2}+\frac{1}{N})$.

Hence $d_1(f_m,f_n) = \int_{0}^{1} |f_m(x) - f_n(x)|dx = \int_{\frac{1}{2}-\frac{1}{N}}^{\frac{1}{2}+\frac{1}{N}} |f_m(x) - f_n(x)|dx < \frac{2}{N}$.

copper.hat
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  • @copper.hat, how do you see that $|f_m(x) - f_n(x) | = | m(x - 1/2) - n(x- 1/2)| \leq 1$ ? – Kamil Jul 26 '16 at 10:55
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    @Kamil: Take $x \ge {1 \over 2}$, then we have $f_m(x),f_n(x) \in [0,1]$, hence $|f_m(x)-f_n(x)| \le 1$. Same reasoning for $x<{1 \over 2}$ with $[-1,0]$. Does this answer your question? – copper.hat Jul 26 '16 at 14:36
  • Sir, isn't $|f_m(x)-f_n(x)|<1$ when $x \in (\frac{1}{2}-\frac{1}{N}, \frac{1}{2}+\frac{1}{N})$ (I mean strictly less than. Because endpoints are not included) – Akash Patalwanshi Feb 11 '21 at 03:44
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    @AkashPatalwanshi Yes, I have updated the answer. – copper.hat Feb 11 '21 at 05:20