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Math Help - Estimate an integral

  1. #1
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    Estimate an integral

    Hello everyone.

    It is a,b,q \in \mathbb{R}, a<b and q \ge 1

    Furthermore let c \in ]a,b[ \mbox{ and  } j_0 \in \mathbb{N} : c-(1/j_0) > a.

    f_j:[a,b]\to [0,1] is for every j \ge j_0 a continious function, e.g.

    f_j=\begin{cases} 0, & \mbox{iff } a\le x \le c-\frac{1}{j+1} \\ 1, & \mbox{iff } c<x<b \\ ((j+1)x-(j+1)c+1)^{1/q} & \mbox{iff } c-\frac{1}{j+1}<x\le c \end{cases}

    for all i \ge j \ge j_0 it is

    ||f_j-f_i||^q_q = \int^b_a(f_j-f_i)^q

    I don't get the following lines:

    = \int^{\displaystyle c-\frac{1}{i+1}}_{\displaystyle  c-\frac{1}{j+1}} \ f_j^q+\int^c_{\displaystyle c-\frac{1}{i+1}}\ (f_j-f_i)^q

    I don't unterstand the next line either:

    \le \frac{1}{j+1}+\frac{2^q}{i+1}

    Can anyone explain it to me, please?

    Kind regards
    Rapha
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  2. #2
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    Thanks
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    You have a definition, I assume, of a norm
    <br /> <br />
||f_j-f_i||^q_q = \int^b_a(f_j-f_i)^q<br />

    that has rules on how to evaluate it -
    f = 0 for x between a and c-...
    1 for x between, and so on.

    Now the part you didn't understand is trying to do the integral to evaluate the norm. Note the norm integral is from a to b. For x just a little larger than a the f's are zero, so there is no need to do an integral. For x just a little less than c, the f is the really complicated expression, so the integral is just written in terms of the f's. For x in the little space between 1/j+1 and 1/i+1, one of the f's is zero so you can ignore it.

    I suspect the next line is actually doing the integral.
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  3. #3
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    Quote Originally Posted by qmech View Post
    You have a definition, I assume, of a norm
    <br /> <br />
||f_j-f_i||^q_q = \int^b_a(f_j-f_i)^q<br />

    that has rules on how to evaluate it -
    f = 0 for x between a and c-...
    1 for x between, and so on.

    Now the part you didn't understand is trying to do the integral to evaluate the norm. Note the norm integral is from a to b. For x just a little larger than a the f's are zero, so there is no need to do an integral. For x just a little less than c, the f is the really complicated expression, so the integral is just written in terms of the f's. For x in the little space between 1/j+1 and 1/i+1, one of the f's is zero so you can ignore it.

    I suspect the next line is actually doing the integral.
    That makes sense.
    Thank you very much, qmech!
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