# Problem in showing that a sequence is a Cauchy sequence on a space with the integral metric.

I'm having difficulty following what is going on and understanding parts in the following example. It is quite similar to an example I posted before (Changing of the limits of integration with the integral metric.), but then, I haven't seen many examples of these problems before.

Given the sequence of functions for $n\ge2$,

$$f_n(x)=\begin{cases} 0 & \text{for 0\le x\le\frac12-\frac1n} \\ n(x-\frac12)+1 & \text{for \frac12-\frac1n\lt x\lt\frac12} \\ 1 &\text{for x\ge\frac12} \end{cases}$$

in the space of continuous functions on the interval $[0,1]$, $C([0,1])$, with the integral metric $d_{int}$,

$$d_{int}(f,g)=\int_0^1 |f(x)-g(x)| \,dx\\$$

In moving to show that this is a Cauchy sequence we consider $d_{int}(f_n,f_m)$ with $n\ge m$:

$$d_{int}(f_n,f_m)=\int_0^1 |f_n(x)-f_m(x)| \,dx\\$$

$$=\int_{1/2-1/m}^{1/2} f_m(x) \,dx-\int_{1/2-1/n}^{1/2} f_n(x) \,dx\\$$

$$=\frac12\left(\frac1m-\frac1n\right)$$

Here are the points I am struggling with:

1. How is it that one can get rid of the modulus sign whilst still maintaining equality throughout the consideration? All tricks for dealing with the modulus that I have seen so far deal with the inequalities $"\ge"$. What could I use that doesn't involve such inequalities?

2. How is it that I get the final result from the second part of the consideration (i.e. from the statement at the second equals sign)? This is one of the things that I tried so far, although I am sure I have missed something very minuscule and rudimentary:

$$\int_{1/2-1/n}^{1/2} f_m(x) \,dx-\int_{1/2-1/n}^{1/2} f_n(x) \,dx\\$$

$$=f_m(1/2)-f_m(1/2-1/m)-(f_n(1/2)-f_n(1/2-1/n))$$

$$=1-0-(1-0)$$

$$=0$$

• when you split into two integrals, the lower bound of the first should be 1/2-1/m, not 1/2-1/n ... or it is even more complicated than that but this is a hint. The comment below by @zhw is even better, using a suitable estimate to avoid exact messy computation – Mirko Jul 29 '15 at 15:03
• $\int_0^1|f_m-f_n| = \int_{1/2-1/m}^{1/2}|f_m-f_n| \le (1/m)\cdot 1 = 1/m.$ – zhw. Jul 29 '15 at 15:04

The absolute value of $|a|$ is $a$ or $-a$, depending on whether $a$ is nonegative or not.
In your case, the condition $n\geq m$ implies $f_m (t)\geq f_n (t)$, so $$|f_n (t)-f_m (t)|=-(f_n (t)-f_m (t))=f_m(t)-f_n (t).$$