MHB Absolute Value of Complex Integral

[kalish1]

Let $[a,b]$ be a closed real interval. Let $f:[a,b] \to \mathbb{C}$ be a continuous complex-valued function. Then $$\bigg|\int_{b}^{a} f(t)dt \ \bigg| \leq \int_{b}^{a} \bigg|f(t)\bigg| dt,$$ where the first integral is a complex integral, and the second integral is a definite real integral.

There's a neat "rotational" proof of this in D'Angelo's An Introduction to Complex Analysis and Geometry.

Question:** Can this fact also be proven using the Cauchy-Schwarz Inequality? If so, some help would be nice.

Thank you...

 

[chisigma]

Let $[a,b]$ be a closed real interval. Let $f:[a,b] \to \mathbb{C}$ be a continuous complex-valued function. Then $$\bigg|\int_{b}^{a} f(t)dt \ \bigg| \leq \int_{b}^{a} \bigg|f(t)\bigg| dt,$$ where the first integral is a complex integral, and the second integral is a definite real integral.

There's a neat "rotational" proof of this in D'Angelo's An Introduction to Complex Analysis and Geometry.

Question:** Can this fact also be proven using the Cauchy-Schwarz Inequality? If so, some help would be nice.

Thank you...

Remembering the definition of Riemann Integral...

http://mathhelpboards.com/analysis-50/riemann-integral-two-questions-14927.html

$\displaystyle \int _{a}^{b} f(x)\ dx = \lim_{\text{max} \Delta_{k} \rightarrow 0} \sum_{k=1}^{n} f(x_{k})\ \Delta_{k}\ (1)$

... and the so called Triangle Inequality...

Triangle Inequality -- from Wolfram MathWorld

$\displaystyle | \sum_{k=1}^{n} a_{k}| \le \sum_{k=1}^{n} |a_{k}|\ (2)$

... You easily met the goal...

Kind regards

$\chi$ $\sigma$