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Can anyone help me to prove that these integral inequalities hold?

Here $x$ is a real value:

$ \left| \int_a^b\ f(x) dx \right| \leq \int_a^b\ |f(x)| dx $

Here $z$ is a complex value:

$ \left| \int_C^ \ f(z) dz \right| \leq \int_C^\ |f(z)| |dz| $

  • 2
    @user71815 Thanks, I know. My goal with this comment was to help the OP to precise the setting of their question (but, as you can see, the OP was not interested in that).2013-06-12

4 Answers 4

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Notice by the triangle inequality

$ \Big|\sum_{i=1}^n f(\xi_i)\Big|\Delta x_i\ \leq \sum_{i=1}^n |f(\xi_i)|\Delta x_i $ Now when applying limits as $\Delta x_i$ go to $0$, then we obtain the first part of your question.

3

Let's denote $ f_+(x) = \max\{0, f(x)\}\\ f_-(x) = \max\{0, -f(x)\} $ We have $ f(x) = f_+(x) - f_-(x)\\ \lvert f(x) \rvert = f_+(x) + f_-(x)\\ \int_a^b f_{\pm} (x) dx \geq 0 $ and so $ \left\lvert \int_a^b f(x)dx \right\rvert = \left\lvert \int_a^b f_+(x)dx - \int_a^b f_-(x)dx\right\rvert \leq\\ \left\lvert \int_a^b f_+(x)dx \right\rvert + \left\lvert\int_a^b f_-(x)dx\right\rvert =\\ \int_a^b f_+(x)dx + \int_a^b f_-(x)dx = \\ \int_a^b (f_+(x) + f_-(x))dx = \int_a^b\lvert f(x)\rvert dx\\ $

2

Hint for the 2nd integration inequality:

Set $\overline{w}:=\int f(z) dz$ and have a look at $\int \overline{w} f(z) dz$. What happens to the lhs? Write the rhs as a double integral.

Hope this helps.


Complete proof of the 2nd inequality:

We know $\displaystyle \int f(z) dz = \int \text{Re } f(x) dx + i \int \text{Im } f(x) dx$

Set $\overline{w}:=\int f(z) dz$. Then we get $|w|=|\overline{w}|=\left| \int f(x) dx \right|$ and further more

$\overline{w} \int f(x) dx= \left( \int Re f(x) dx + i \int Im f(x) dx \right)\left( \int Re f(x) dx + i \int Im f(x) dx \right)=\left| \int f(x) \right|^2$ and so $\displaystyle \overline{w} \int f(x) dx \in \mathbb{R}^+$. We finally have

$\left| \int f(x) \right|^2 = \overline{w} \int f(x) dx = Re \left( \int \overline{w} f(x) dx \right) = \int Re(\overline{w}f(x))dx \leq\\ \int |\overline{w}f(x)|dx= \int |\overline{w}||f(x)|dx= |\overline{w}| \int |f(x)| dx =\left| \int f(x) dx \right| \int |f(x)| dx \iff\\ \left| \int f(x) dx \right| \leq \int |f(x)| dx$

I hope I didn't mixed up some absolute values.

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    You should spell out that '$|dz|$' stuff..2012-10-18
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  1. Proving the first inequality

Suppose, WLOG, that

$\int_{a}^{b}f(x)dx \neq 0$

Define

$ \lambda = \frac{\left| \int_{a}^{b}f(x)dx \right|}{\int_{a}^{b}f(x)dx} $


Remark 1: $\lambda \in \{\pm 1,\pm i\}$.

Remark 2: Integrals are homogeneous, i.e.,

$\forall a\in \mathbb{C} :\int_{a}^{b}a f(x)dx = a \int_{a}^{b}f(x)dx$

Remark 3: $\forall a\in \mathbb{C}: |Re(a)|\leq |a|$


So, we have

$\mathbb{R^+} \ni \left|\int_{a}^{b}f(x)dx\right| = \lambda \int_{a}^{b}f(x)dx = \int_{a}^{b} \lambda f(x)dx = \int_{a}^{b}Re\{\lambda f(x)\}dx \leq \int_{a}^{b}|Re\{\lambda f(x)\}|dx \leq \int_{a}^{b}|\lambda f(x)|dx \leq |\lambda|\int_{a}^{b} |f(x)|dx$

Finally,

$ \left|\int_{a}^{b}f(x)dx\right| \leq \int_{a}^{b}\left|f(x)\right| dx $

  1. Proving the second inequality:

$\left| \int_{\gamma}f(z)dz \right| = \left| \int_{a}^{b}f(\gamma(t)) \gamma ' (t) dt \right| \leq \int_{a}^{b} \left| f(\gamma(t)) \gamma ' (t)\right| dt = \int_{a}^{b} \left| f(\gamma(t))\right| \left| \gamma ' (t)\right| dt = \int_{\gamma}\left|f(z)\right||dz|$