0
$\begingroup$

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:

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

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

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

$$=0$$

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Mirko Jul 29 '15 at 15:03
  • $\begingroup$ $\int_0^1|f_m-f_n| = \int_{1/2-1/m}^{1/2}|f_m-f_n| \le (1/m)\cdot 1 = 1/m.$ $\endgroup$ – zhw. Jul 29 '15 at 15:04
0
$\begingroup$

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 $$|f_n (t)-f_m (t)|=-(f_n (t)-f_m (t))=f_m(t)-f_n (t). $$

For your second question, when you calculate your integrals you are forgetting the antiderivatives.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.