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Math Help - Integral Inequality

  1. #1
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    Integral Inequality

    Hi,

    I have the following inequality
    \int_0^t \! \, F(\tau) d\tau \leq M
    where F(\tau)>0 is a continuous differentiable function and M is a constant.

    My question, what is the upper bound for
    \int_0^t \! \, F^n(\tau) d\tau \leq ?
    where n is positive integer larger then 2.
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  2. #2
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    Beijing, China
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    There is no upper bound. Choose a number c such that 0<c<t, let f(x) = M/c when x < c and 0 otherwise. f is not continuous differentiable but one is always able to find a differetiable function F as close as possible to f. Now \int_0^t f(x)dx = M, and \int_0^t f^n(x)dx=c(\frac{M}{c})^n=\frac{M^n}{c^{n-1}}. Choose a small c you can get an arbitary big result.
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  3. #3
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    Thank you for your reply. How if t>1, if I choose a sufficiently large c, 1<c<t, then the bound will get smaller at t increase.
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