# Absolute value integral inequality proof step

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 :)

• 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 – Abhimanyu Pallavi Sudhir Jun 9 '13 at 3:07
• – Watson Nov 27 '18 at 14:55

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

• So $Re[ ... ]$ is shown to reiterate that is it real, so that the next step logically follows. Hmm I think I got it. Thanks! – Chester May 31 '12 at 0:40

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...)

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

• 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! – Chester May 31 '12 at 11:00
• Happens to all of us. – copper.hat May 31 '12 at 15:17