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Let $F: [\alpha,\beta] \to \mathbb C$ be a continuous function, $F(t) = u(t)+iv(t)$. Define the integral of $F$ over $[\alpha,\beta]$ to be \begin{align*} \int_\alpha^\beta F(t) dt = \int_\alpha^\beta u(t) dt + i \int_\alpha^\beta v(t) dt. \end{align*}

I was trying to prove the following:

Is $F$ continuous on $[\alpha,\beta]$, then \begin{align*} \left|\int_\alpha^\beta F(t) dt\right| \le \int_\alpha^\beta |F(t)| dt. \end{align*}

My (incomplete) solution: \begin{align*} \left|\int_\alpha^\beta F(t) dt\right| &= \left|\int_\alpha^\beta u(t)dt + i \int_\alpha^\beta v(t)dt\right| \le \left|\int_\alpha^\beta u(t)dt\right| + \left|\int_\alpha^\beta v(t)dt\right| \\ &\le \int_\alpha^\beta |u(t)|dt + \int_\alpha^\beta |v(t)|dt = \int_\alpha^\beta |u(t)| + |v(t)|dt. \end{align*}

The problem: My estimation is already too far. How can I correct this?

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A nice solution goes by picking a real $\theta$ such that $\displaystyle e^{i\theta}\int_\alpha^\beta F(t)\,dt=\left|\int_\alpha^\beta F(t)\,dt\right|$. We have $$ \left|\int_\alpha^\beta F(t)\,dt\right|=e^{i\theta}\int_\alpha^\beta F(t)\,dt=\int_\alpha^\beta e^{i\theta} F(t)\,dt. $$ This shows that the integral of the imaginary part of $e^{i\theta} F(t)$ is $0$, and the integral of the real part is the absolute value. But $\Re z\le|\Re z|\le|z|$, and we are done.

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  • $\begingroup$ Yes! This proof is in Baby Rudin, isn't it? I can't recall where I've seen it. $\endgroup$ – abnry Dec 5 '13 at 21:52
  • $\begingroup$ It may well be the standard approach. It also appears in Berenstein-Gay. $\endgroup$ – Andrés E. Caicedo Dec 5 '13 at 21:53
  • $\begingroup$ Thanks for the nice solution. How do we know that we can find such a $\theta$? $\endgroup$ – numerion Dec 5 '13 at 22:02
  • $\begingroup$ If $|z|=1$, then $z$ is on the unit circle, so it is $e^\theta=\cos\theta+\sin\theta$ for some $\theta$. Now, given any $t\ne0$, $z=t/|t|$ has norm $1$. In particular, this is the case if $t$ is the integral you are considering. If the integral is $0$, any $\theta$ works. $\endgroup$ – Andrés E. Caicedo Dec 5 '13 at 22:11
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As you have noticed, you end up with an inferior estimate carrying a factor of $2$ when you apply the real triangle inequality to each component in the complex case.

Hint: if $z$ is any complex number (i.e. the integral you are trying to estimate, which is a complex number), then $\alpha z=|z|$, where $\alpha$ is a complex number of norm $1$ (it is not important to know how $\alpha$ is obtained to apply this fact, though you have probably seen it in complex analysis).

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