# $f(x)$ is non-negative and $\int_a^bf(x)dx = 1$, show that $[\int_a^bf(x)\cos{kx}dx]^2 + [\int_a^bf(x)\sin{kx}dx]^2 \leq 1$

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.

• 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.
– Did
Commented Mar 15, 2015 at 11:08

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

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.

• Any "elementary" solutions?? :) Commented Mar 15, 2015 at 8:33

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):

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