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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.

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    Just compute $d_1(f_n, f_m)$ ...2012-05-31
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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 $d_1(f_m,f_n) < \frac{2}{N}$.2012-05-31
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    @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.2012-05-31
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    It happens to most of us...2012-05-31
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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.2012-05-31
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    @MartianInvader Thanks for valuable comment. Do i need to write $f_n$ is converging to some discontinuous function?2012-05-31
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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.2012-05-31
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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-xtdt2012-06-01
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    @JonasMeyer Thank you very much. Now i am reading those problems.2012-06-01

1 Answers 1

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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 $|f_m(x)-f_n(x)| \leq 1$ when $x \in (\frac{1}{2}-\frac{1}{N}, \frac{1}{2}+\frac{1}{N})$.

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

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    @copper.hat, how do you see that $|f_m(x) - f_n(x) | = | m(x - 1/2) - n(x- 1/2)| \leq 1$ ?2016-07-26
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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?2016-07-26