
cauchysequenze
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 cauchysequenze
(field) norm: $\displaystyle u_2 = \sqrt{<u, u>} \ \ \forall u \in C[0,2]$
$\displaystyle f_n(x)=\begin{cases} 0, & 0\le x\le 1 \frac{1}{n} \\ n(x(1\frac{1}{n})), & 1\frac{1}{n}<x<1\\ 1, & 1\le x \le 2 \end{cases}$

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

Let m<n. We want to show that $\displaystyle \f_nf_m\_2\to0$ as $\displaystyle m,\,n\to\infty$, where $\displaystyle \f_nf_m\_2^2 = \int_0^2(f_n(x)f_m(x))^2\,dx$. 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.