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Prove that for any smooth function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ such that $\left|\frac{d\phi}{dx}\right|<1$ for all $x$ in $(0,\pi)$,

$\left(\int_0^\pi \cos(\phi(x)) \; dx\right)^2 + \left(\int_0^\pi \sin(\phi(x))\;dx\right)^2 > 4.$

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    @RobertIsrael You're right... apologies.2012-03-07

1 Answers 1

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Your inequality says $\left| \int_0^\pi e^{i\phi(x)}\ dx \right| > 2$. In fact if |\phi'| < 1 we have $|\phi(x) - \phi(\pi/2)| < |x - \pi/2|$ for $0 < x < \pi$, so $\left| \int_0^\pi e^{i\phi(x)}\ dx \right| \ge \text{Re}\ e^{-i \phi(\pi/2)} \int_0^\pi e^{i \phi(x)}\ dx = \int_0^\pi \cos(\phi(x)-\phi(\pi/2))\ dx > \int_0^\pi \cos(x - \pi/2)\ dx = 2$ Here I'm using the facts that $\cos$ is even and is decreasing on $[0,\pi]$. No smoothness of $\phi$ is necessary, just differentiability.

More generally, if you replaced |\phi'| < 1 by |\phi'| < k where $0 < k < 2$, the inequality would become $\left|\int_0^\pi e^{i\phi(x)}\ dx \right| > \frac{2}{k} \sin\left(\frac{k \pi}{2}\right)$

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    Well, one ingredient was checking the boundary case $\phi' = 1$ and seeing that you got equality there. That's useful because it means you can't use any estimates that would not be tight in that case. The fact that $|z| = \text{Re}( e^{i\theta} z)$ for suitable $\theta$ is pretty standard.2012-03-07