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:
$$ \left| \int_C^ \ f(z) dz \right| \leq \int_C^\ |f(z)| |dz| $$
On
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.
On
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 $$ \left\lvert \int_a^b f(x)dx \right\rvert = \left\lvert \int_a^b f_+(x)dx - \int_a^b f_-(x)dx\right\rvert \leq\\ \left\lvert \int_a^b f_+(x)dx \right\rvert + \left\lvert\int_a^b f_-(x)dx\right\rvert =\\ \int_a^b f_+(x)dx + \int_a^b f_-(x)dx = \\ \int_a^b (f_+(x) + f_-(x))dx = \int_a^b\lvert f(x)\rvert dx\\ $$
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.