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Math Help - Inequality with absolute value of a complex integral

  1. #1
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    Inequality with absolute value of a complex integral

    I'm stuck trying to prove a step inside a lemma from Serre; given is

    0<a<b
    0<x

    To prove:

    |\int_{a}^{b}e^{-tx}e^{-tiy}dt|\leq\int_{a}^{b}e^{-tx}dt

    I've tried using Cauchy-Schwartz for integrals, but this step is too big (using Mathematica, it lead to a contradiction); something simpler must do the trick.
    Thanks in advance.
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  2. #2
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    Quote Originally Posted by bernardbb View Post
    I'm stuck trying to prove a step inside a lemma from Serre; given is

    0<a<b
    0<x

    To prove:

    |\int_{a}^{b}e^{-tx}e^{-tiy}dt|\leq\int_{a}^{b}e^{-tx}dt

    I've tried using Cauchy-Schwartz for integrals, but this step is too big (using Mathematica, it lead to a contradiction); something simpler must do the trick.
    Thanks in advance.
    \left| \int_{a}^{b}e^{-tx}e^{-tiy}dt\right| \leq\int_{a}^{b}\left| e^{-tx}e^{-tiy}\right| dx = \int_a^b e^{-tx}dt
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  3. #3
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    As far as I know, that only holds if f(x) is real, which it is not.
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  4. #4
    Moo
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    Quote Originally Posted by bernardbb View Post
    As far as I know, that only holds if f(x) is real, which it is not.
    It works with complex functions too...

    You can note that the absolute value and the modulus of a complex number are equivalent !

    |a+ib|=sqrt(aČ+bČ)
    modulus>>|a+i0|=sqrt(aČ)=|a|<<absolute value
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  5. #5
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    As far as I can tell, that proves it for complex values z with Im(z)=0, which doesn't get me any further.
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  6. #6
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    What Moo says is correct (as always). If f is a (complex-valued) function of a real variable t on the interval [a,b], let S_P(f) = \sum_{k=1}^n(x_k-x_{k-1})f(\xi_k) be a Riemann sum for f (where P is the partition given by the points x_0,\ldots,x_n and each \xi_k is a point in [x_{k-1},x_k]). Then the triangle inequality shows that |S_P(f)|\leqslant S_P(|f|). In the limit, as the mesh of the partition goes to 0, it follows that \left|\int_a^b\!\!f(t)\,dt\right|\leqslant\int_a^b  \!\!|f(t)|\,dt.
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  7. #7
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    Quote Originally Posted by Opalg View Post
    What Moo says is correct (as always).
    Actually Moo is not always correct, I got her making mistakes.

    Here is a way to prove that: \left| \int_a^b \bold{f}(t) dt \right| \leq \int_a^b |\bold{f}(t)| dt where \bold{f}:[a,b]\to \mathbb{R}^n is continous.

    Let \bold{u}\in \mathbb{R}^n be a unit vector.
    We see that,
    \left| \bold{u} \cdot \int_a^b \bold{f}(t) dt \right| = \left| \int_a^b \bold{f}(t) \cdot \bold{u} dt \right| \leq \int_a^b |\bold{f}(t)\cdot \bold{u}| dt \leq \int_a^b |\bold{f}(t)| dt

    Take \bold{u} pointing in the direction of \int_a^b \bold{f}(t)dt and so \bold{u}\cdot \int_a^b \bold{f}(t)dt = \int_a^b \bold{f}(t)dt

    To see the complex case just let n=2.
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  8. #8
    Moo
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    lol, of course I make mistakes !!!!

    And TPH does too
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