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Math Help - cauchy-sequenze

  1. #1
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    cauchy-sequenze

    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
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  2. #2
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    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.
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