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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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    $\begingroup$ Just compute $d_1(f_n, f_m)$ ... $\endgroup$
    – martini
    May 31 '12 at 21:54
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    $\begingroup$ 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 $d_1(f_m,f_n) < \frac{2}{N}$. $\endgroup$
    – copper.hat
    May 31 '12 at 22:05
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    $\begingroup$ It happens to most of us... $\endgroup$
    – copper.hat
    May 31 '12 at 22:15
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    $\begingroup$ 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. $\endgroup$ May 31 '12 at 22:39
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    $\begingroup$ Related questions: math.stackexchange.com/questions/21878/… and math.stackexchange.com/questions/97171/… $\endgroup$ Jun 1 '12 at 4:03
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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 $d_1(f_m,f_n) = \int_{0}^{1} |f_m(x) - f_n(x)|dx = \int_{\frac{1}{2}-\frac{1}{N}}^{\frac{1}{2}+\frac{1}{N}} |f_m(x) - f_n(x)|dx < \frac{2}{N}$.

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  • $\begingroup$ @copper.hat, how do you see that $|f_m(x) - f_n(x) | = | m(x - 1/2) - n(x- 1/2)| \leq 1$ ? $\endgroup$
    – Kamil
    Jul 26 '16 at 10:55
  • $\begingroup$ @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? $\endgroup$
    – copper.hat
    Jul 26 '16 at 14:36
  • $\begingroup$ 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) $\endgroup$ Feb 11 '21 at 3:44
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    $\begingroup$ @AkashPatalwanshi Yes, I have updated the answer. $\endgroup$
    – copper.hat
    Feb 11 '21 at 5:20

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