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I know that $(C[0, 2], d_{\infty})$ is a complete metric space, being $C[0, 2]$ the set of continuous functions in the closed interval $[0, 2]$ and $d_\infty$ the distance metric induced by the infinite norm, i.e., $d_\infty\{f(x), g(x)\} = ||f(x) - g(x)||_\infty = \sup \{|f(x) - g(x)|\}$ with $x \in C[0, 2]$.

Given the following sequence of functions:

$f_n(x) = \begin{cases} 0 & \mbox{if } 0 \leq x \leq 1 - \frac{1}{n}, \\ nx + 1 - n & \mbox{if } 1 - \frac{1}{n} < x \leq 1, \\ 1 & \mbox{if } 1 < x \leq 2, \end{cases}$

I would like know if, in this metric space, this sequence is a Cauchy sequence, and if so, to which limit it converges.

The origin of this question is that, using a norm-2 induced metric, this sequence converges to the non continuous step function. This shows that $(C[a, b], d_{2})$ is not complete. But I when I tried to calculate the limit of this sequence for $d_\infty$ I found myself lost.

I would really appreciate any hint.

Thanks in advance!

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up vote 2 down vote accepted

Draw a picture of the difference between $f_n$, $f_m$. This should give a hint.

More explicitly: Suppose $n>m$, and let $x = 1-\frac{1}{n}$. Then $f_n(x)=0$, and $f_m(x) = 1-\frac{m}{n}$. Thus $d_{\infty}(f_n,f_m) \geq 1-\frac{m}{n}$, for all $n>m$. In particular, we can choose $n=2m$, and get $d_{\infty}(f_{2m},f_m) \geq \frac{1}{2}$, so the sequence $f_n$ cannot be Cauchy.

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