# Thread: Inequality with absolute value of a complex integral

1. ## 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:

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

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.

2. Originally Posted by bernardbb
I'm stuck trying to prove a step inside a lemma from Serre; given is

0<a<b
0<x

To prove:

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

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.
$\displaystyle \left| \int_{a}^{b}e^{-tx}e^{-tiy}dt\right| \leq\int_{a}^{b}\left| e^{-tx}e^{-tiy}\right| dx = \int_a^b e^{-tx}dt$

3. As far as I know, that only holds if f(x) is real, which it is not.

4. Originally Posted by bernardbb
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

5. As far as I can tell, that proves it for complex values z with Im(z)=0, which doesn't get me any further.

6. 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 $\displaystyle S_P(f) = \sum_{k=1}^n(x_k-x_{k-1})f(\xi_k)$ be a Riemann sum for f (where P is the partition given by the points $\displaystyle x_0,\ldots,x_n$ and each $\displaystyle \xi_k$ is a point in $\displaystyle [x_{k-1},x_k]$). Then the triangle inequality shows that $\displaystyle |S_P(f)|\leqslant S_P(|f|)$. In the limit, as the mesh of the partition goes to 0, it follows that $\displaystyle \left|\int_a^b\!\!f(t)\,dt\right|\leqslant\int_a^b \!\!|f(t)|\,dt$.

7. Originally Posted by Opalg
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: $\displaystyle \left| \int_a^b \bold{f}(t) dt \right| \leq \int_a^b |\bold{f}(t)| dt$ where $\displaystyle \bold{f}:[a,b]\to \mathbb{R}^n$ is continous.

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

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

To see the complex case just let $\displaystyle n=2$.

8. lol, of course I make mistakes !!!!

And TPH does too