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Suppose $f(x)$ is non-negative and integrable on $[a, b]$, and that $ \int_a^bf(x)dx = 1 $. Prove that $$ [\int_a^bf(x)\cos{kx}dx]^2 + [\int_a^bf(x)\sin{kx}dx]^2 \leq 1.$$ Thanks!

There is a hint that the problem has something to do with the Cauchy-Schwarz Inequality and has a simple elementary solution.

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    $\begingroup$ Use Cauchy-Schwarz indeed: $$\int f(x)\cos(kx)dx=\int \sqrt{f(x)}\cdot\sqrt{f(x)}\cos(kx)dx\leqslant\int f(x)dx\cdot\int f(x)\cos^2(kx)dx=\int f(x)\cos^2(kx)dx.$$ Repeat with the sine, use that $\cos^2+\sin^2=1$ and once again that $f$ integrates to $1$, QED. $\endgroup$
    – Did
    Mar 15, 2015 at 11:08

4 Answers 4

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$$ \left[\int_a^bf(x)\cos kx dx\right]^2 + \left[\int_a^bf(x)\sin kx dx\right]^2= $$ $$ \int_a^b\int_a^bf(x)\cos kx f(y)\cos kydxdy + \int_a^b\int_a^bf(x)\sin kx f(y)\sin kydxdy = $$ $$ \int_a^b\int_a^bf(x)f(y)cos(kx-ky)dxdy\le $$ $$ \int_a^b\int_a^bf(x)f(y)dxdy=\left[\int_a^bf(x)dx\right]^2=1. $$

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The measure $d\mu (x)=f(x) dx$ is a probability measure by assumption. Hence, Jensens inequality yields

$$ (\int \sin(x) f(x)\, dx)^2 = (\int \sin(x) \,d\mu)^2 \leq \int (\sin(x))^2 \,d\mu =\int (\sin(x))^2 f(x)\, dx. $$

The same estimate holds for $\cos$. Adding these estimates finishes the proof.

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    $\begingroup$ Any "elementary" solutions?? :) $\endgroup$
    – leafpile
    Mar 15, 2015 at 8:33
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By Cauchy Schwarz,

$ [\int_a^bf(x)\cos{kx}dx]^2 + [\int_a^bf(x)\sin{kx}dx]^2$ $\begin{align}&=[\int_a^b\sqrt{f(x)}\sqrt{f(x)}\cos{kx}dx]^2 + [\int_a^b\sqrt{f(x)}\sqrt{f(x)}\sin{kx}dx]^2 \\&\leq \int_a^bf(x)dx\int_a^bf(x)\cos(kx)^2dx+\int_a^bf(x)dx\int_a^bf(x)\sin(kx)^2dx \\&=\int_a^bf(x)dx\\&=1 \end{align}$

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Yet another solution (props to this guy):

$$ [\int_a^bf(x)\cos{kx}dx]^2 + [\int_a^bf(x)\sin{kx}dx]^2=\left | \int_{a}^{b} f(x)e^{ikx}dx\right |^{2}\leq \left(\int_{a}^{b}\left | f(x)e^{ikx} \right |dx\right)^{2}=1$$

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