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.
$$ \left[\int_a^bf(x)\cos kx dx\right]^2 + \left[\int_a^bf(x)\sin kx dx\right]^2= $$ $$ \int_a^b\int_a^bf(x)\cos kx f(y)\cos kydxdy + \int_a^b\int_a^bf(x)\sin kx f(y)\sin kydxdy = $$ $$ \int_a^b\int_a^bf(x)f(y)cos(kx-ky)dxdy\le $$ $$ \int_a^b\int_a^bf(x)f(y)dxdy=\left[\int_a^bf(x)dx\right]^2=1. $$
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.
By Cauchy Schwarz,
$ [\int_a^bf(x)\cos{kx}dx]^2 + [\int_a^bf(x)\sin{kx}dx]^2$ $\begin{align}&=[\int_a^b\sqrt{f(x)}\sqrt{f(x)}\cos{kx}dx]^2 + [\int_a^b\sqrt{f(x)}\sqrt{f(x)}\sin{kx}dx]^2 \\&\leq \int_a^bf(x)dx\int_a^bf(x)\cos(kx)^2dx+\int_a^bf(x)dx\int_a^bf(x)\sin(kx)^2dx \\&=\int_a^bf(x)dx\\&=1 \end{align}$
Yet another solution (props to this guy):
$$ [\int_a^bf(x)\cos{kx}dx]^2 + [\int_a^bf(x)\sin{kx}dx]^2=\left | \int_{a}^{b} f(x)e^{ikx}dx\right |^{2}\leq \left(\int_{a}^{b}\left | f(x)e^{ikx} \right |dx\right)^{2}=1$$