cauchy-sequenze

**Rapha** · November 10th 2008, 07:29 AM

Hello

It is $<u,v> = \int^2_0 \overline{u}(x) v(x) dx \ \ \forall u,v \in C[0,2]$ (defines a dot product)

Show that $f_n$ is a cauchy-sequenze
(field) norm: $||u||_2 = \sqrt{<u, u>} \ \ \forall u \in C[0,2]$

$f_n(x)=\begin{cases} 0, & 0\le x\le 1- \frac{1}{n} \\ n(x-(1-\frac{1}{n})), & 1-\frac{1}{n}<x<1\\ 1, & 1\le x \le 2 \end{cases}$

--------------
I think $\lim_{n \to \infty} <f_{n+1}, f_n> = 0$ is okay, but it is

$\lim_{n \to \infty} <f_{n+1}, f_n> = \infty$

any help would be really apprectiated

Rapha

**Opalg** · November 10th 2008, 08:34 AM

Let m<n. We want to show that $\|f_n-f_m\|_2\to0$ as $m,\,n\to\infty$ , where $\|f_n-f_m\|_2^2 = \int_0^2(f_n(x)-f_m(x))^2\,dx$ . But $f_m(x)$ and $f_n(x)$ coincide on the intervals [0,1-(1/m)] and [1,2]. So we need only worry about the integral on the interval from 1-(1/m) to 1. Even on this interval, the integral is a bit messy to evaluate exactly, so I suggest using a little trick.

It's easy to see that if m<n then $f_m(x)\geqslant f_n(x)$ . Therefore $0\leqslant f_m(x) - f_n(x) \leqslant f_m(x)\leqslant1$ , and the integral that we want to evaluate is less than $\int_{1-(1/m)}^11\,dx$ . That integral is (very!) easy to evaluate, and it gets small as m gets large.

Thread: cauchy-sequenze

LinkBack

Thread Tools

Display

cauchy-sequenze

Similar Math Help Forum Discussions

[SOLVED] Subsequence of a Cauchy Sequence is Cauchy

cauchy

not cauchy

cauchy

Cauchy

Search Tags