# Proving two integral inequalities

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:

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

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What is $|dz|$? – Did Oct 26 '12 at 12:13
@Did For a contour integral around a path $\gamma(t)$, $|dz|$ is a shorthand for $|\gamma'(t)| dt$ – user71815 Jun 12 '13 at 14:37
@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). – Did Jun 12 '13 at 15:32

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.

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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 $$\left\lvert \int_a^b f(x)dx \right\rvert = \left\lvert \int_a^b f_+(x)dx - \int_a^b f_-(x)dx\right\rvert \leq\\ \left\lvert \int_a^b f_+(x)dx \right\rvert + \left\lvert\int_a^b f_-(x)dx\right\rvert =\\ \int_a^b f_+(x)dx + \int_a^b f_-(x)dx = \\ \int_a^b (f_+(x) + f_-(x))dx = \int_a^b\lvert f(x)\rvert dx\\$$

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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.. – Berci Oct 18 '12 at 10:31
Thank you for everything. – wlantakumi Oct 18 '12 at 10:49