# Showing $(C[0,1], d_1)$ is not a complete metric space

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.

• Just compute $d_1(f_n, f_m)$ ... Commented May 31, 2012 at 21:54
• 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 $d_1(f_m,f_n) < \frac{2}{N}$. Commented May 31, 2012 at 22:05
• It happens to most of us... Commented May 31, 2012 at 22:15
• 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. Commented May 31, 2012 at 22:39
• @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. Commented May 31, 2012 at 23:08

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, how do you see that $|f_m(x) - f_n(x) | = | m(x - 1/2) - n(x- 1/2)| \leq 1$ ? Commented Jul 26, 2016 at 10:55
• @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? Commented Jul 26, 2016 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) Commented Feb 11, 2021 at 3:44