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$F$ is nonnegative function on real line such that integral of $F$ on real line is $1$.

Show that integral of $F(x)\exp(-ix)$ on real line is strictly less than $1$.

By plugging abs value inside the integral, we can easily get integral is less than or equal to $1$, but i think it will be need more to prove inequality is strict!!

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Is there any reason to assume the integral of $F(x)\exp(-ix)$ is real? Or are you taking the magnitude of the answer? – Jason DeVito Oct 16 '12 at 5:06
oh yes. Exactly as you said – Detectives Oct 16 '12 at 5:15
up vote 2 down vote accepted

I assume you mean you want $\left| \int_{\mathbb R} F(x) e^{-ix}\; dx\right| < 1$. Take $\theta$ so that $\left| \int_{\mathbb R} F(x) e^{-ix}\; dx\right| = \text{Re} \;e^{i\theta} \int_{\mathbb R} F(x) e^{-ix}\; dx$. Now $\text{Re} \; e^{i\theta} \int_{\mathbb R} F(x) e^{-ix}\; dx = \int_{\mathbb R} F(x) \cos(x-\theta)\; dx$. But if that is $1$, then $\int_{\mathbb R} F(x) (1 - \cos(x-\theta))\; dx = 0$, which implies that $F(x) = 0$ almost everywhere and thus that $\int_{\mathbb R} F(x)\; dx = 0$, contradiction.

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Wow.. What a Surprising idea this is.. THANK YOU. – Detectives Oct 16 '12 at 5:29

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