3
$\begingroup$

I'm beginning my way through Coddington's Intro to ODE's and I'm a little thrown off in the preliminary section in a proof regarding complexed valued functions. ( I should note that I've taken a course in ODES, but my background in Complex anything-besides-the-basics is subpar. )

Particularly the book shows that:

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

The proof is pretty short, so I guess I'll just map it out until the me-thrown-off point.

First, let

$ F \, = \, \int_a^b f(x) \, dx \quad $

and

$ u = \cos(\theta) + i\sin(\theta). $

Then, let $ F \, = \, \left| F \,\right| u $ where $ F \neq 0 $.

Since $u\overline{u} = 1$,

$ \left| F \right| = \,\overline{u} F \, = \;\overline{u} \int_a^b f(x) \, dx = \;...$

and the step that loses me:

$ ...\; = \, Re\left[ \; \overline{u} \int_a^b f(x) \, dx \; \right] \, = \; ... $

Maybe my unfamiliarity with complex-valued functions is making me miss something obvious, but I'm stumped. For instance, I've tried expanding $ \overline{u} \int_a^b f(x) \, dx $ to:

$ ( \cos(\theta) - i \sin(\theta) ) \int_a^b f(x) \, dx $ $ = \; \cos(\theta) \int_a^b f(x) \, dx - i sin(\theta) \int_a^b f(x) \, dx $ $ = \; \cos(\theta) \left( \int_a^b \left ( Re \, f \, \right) (x) \, dx + i \int_a^b \left ( Im \, f \, \right) (x) \, dx \right)- i sin(\theta) \left( \int_a^b \left ( Re \, f \, \right) (x) \, dx + i \int_a^b \left ( Im \, f \, \right) (x) \, dx \right) $ $ = \; \cos(\theta) \int_a^b \left ( Re \, f \, \right) (x) \, dx + i \cos(\theta) \int_a^b \left ( Im \, f \, \right) (x) \, dx - i sin(\theta) \int_a^b \left ( Re \, f \, \right) (x) \, dx + sin(\theta) \int_a^b \left ( Im \, f \, \right) (x) \, dx $ $ = \; \overline{u} \int_a^b \left( Re \, f \right) (x) \, dx + \left( i \cos(\theta) + \sin(\theta)\right) \int_a^b \left ( Im \, f \, \right) (x) \, dx $

I felt as if I was on the right track but I hit a wall at this point, and it began to feel like I was convoluting something simple. Anyway the proof finishes with:

$ ... \; = \; \int_a^b Re\left[ \overline{u} f(x) \right] \, dx \; \leq \int_a^b \left| f(x) \right| \, dx $

Any help is appreciated :)

  • 0
    Related: https://math.stackexchange.com/questions/20900222018-11-27

2 Answers 2

4

By definition, the quantity $|F|$ is real. Since $|F| = \overline{u}\int_a^bf(x)\,dx,$ it follows that $Re\left[\overline{u}\int_a^b f(x)\,dx\right] = Re\,|F| = |F| = \overline{u}\int_a^b f(x)\,dx.$

  • 0
    So $ Re[ ... ] $ is shown to reiterate that is it real, so that the next step logically follows. Hmm I think I got it. Thanks!2012-05-31
1

The real part is inserted not for emphasis, but so you can use the bound $\Re (z) \leq |z|$, vis-à-vis:

$|\int_a^bf(x)\,dx| = \overline{u}\int_a^bf(x)\,dx = \int_a^b \overline{u} f(x)\,dx = \Re (\int_a^b \overline{u} f(x)\,dx)$

(Sorry, I don't know how to continue the chain properly...)

$\Re (\int_a^b \overline{u} f(x)\,dx) = \int_a^b \Re (\overline{u} f(x))\,dx \leq \int_a^b |\overline{u} f(x)|\,dx = \int_a^b |f(x)|\,dx.$

  • 0
    Happens to all of us.2012-05-31