cauchy-sequenze

• Nov 10th 2008, 07:29 AM
Rapha
cauchy-sequenze
Hello

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

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

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

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

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

any help would be really apprectiated

Rapha
• Nov 10th 2008, 08:34 AM
Opalg
Let m<n. We want to show that $\displaystyle \|f_n-f_m\|_2\to0$ as $\displaystyle m,\,n\to\infty$, where $\displaystyle \|f_n-f_m\|_2^2 = \int_0^2(f_n(x)-f_m(x))^2\,dx$. But $\displaystyle f_m(x)$ and $\displaystyle f_n(x)$ 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 $\displaystyle f_m(x)\geqslant f_n(x)$. Therefore $\displaystyle 0\leqslant f_m(x) - f_n(x) \leqslant f_m(x)\leqslant1$, and the integral that we want to evaluate is less than $\displaystyle \int_{1-(1/m)}^11\,dx$. That integral is (very!) easy to evaluate, and it gets small as m gets large.