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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 :)

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  • $\begingroup$ Maybe this is not really answering your question (So I put it in comments, but if you let $b$ be a function of $x$, and differentiate both sides, then you get: $$\operatorname{sgn}\int_a^b f(x) \mbox{d}x\mbox{ }f(x)\leq|f(x)|$$, which is obvious $\endgroup$ Jun 9, 2013 at 3:07
  • $\begingroup$ Related: math.stackexchange.com/questions/2090022 $\endgroup$
    – Watson
    Nov 27, 2018 at 14:55

2 Answers 2

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

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  • $\begingroup$ So $ Re[ ... ] $ is shown to reiterate that is it real, so that the next step logically follows. Hmm I think I got it. Thanks! $\endgroup$
    – Chester
    May 31, 2012 at 0:40
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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...)

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

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  • $\begingroup$ Ah ok. Now it makes perfect sense. The world is momentarily right again. Thank you. I guess I got hung up on trying to explicity show that $$ \overline{u} \int_a^b f(x) \, dx $$ only contains real parts when it wasn't really necessary since the derivation starts with $$ \left| \int_a^b f(x) \, dx \right| $$ D'oh! $\endgroup$
    – Chester
    May 31, 2012 at 11:00
  • $\begingroup$ Happens to all of us. $\endgroup$
    – copper.hat
    May 31, 2012 at 15:17

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