Relation between Fourier components of a positive function

Question

Here's a problem that has recently come up in my physics research:

Let f be a function on [0, 2 $\pi$], which yields positive real numbers.

Let the integral of $\int_0^{2\pi}f(x)= 1$. (Just for the sake of normalization).

Let the integral of $\int_0^{2\pi}f(x)\sin(x)= r$ (some real number between -1 and 1.)

Question: Max($\int_0^{2\pi}f(x)\cos(x))=$? The maximum is to be taken across all functions satisfying the above criteria.

Answer 1

Let's put this problem back into physics territory from whence it came. Think of $f(t)$ as density of unit circle at the point with polar angle $t$. Then the constraints are

1. Density is nonnegative
2. Total mass $m$ is $1$
3. The center of mass has $y$-coordinate equal to $r$ (it's $\frac{1}{m}\int_0^{2\pi}f(t)\sin t\,dt$ since $\sin t$ is the $y$-coordinate of the point with polar angle $t$).

And you want to maximize the $x$-coordinate of the center of mass. Now the solution is obvious:

The best you can do is put all the mass into the point $(\sqrt{1-r^2},r)$, i.e., let $f$ be the $\delta$-function at point $t_0$ with $(\cos t_0,\sin t_0)=(\sqrt{1-r^2},r)$. If the $\delta$-function is not allowed, approximate it by tall skinny rectangles centered at $t_0$, for which $\int_0^{2\pi}f(t)\cos t\,dt $ will approach the (unattainable) supremum $\sqrt{1-r^2}$.

Formal proof

Let $t_0$ be as above. For all $t\in [0,2\pi]$ we have $$\cos t \cos t_0+\sin t\sin t_0 = \cos (t-t_0)\le 1$$ Multiply by $f(t)$ and integrate: $$\int_{0}^{2\pi} f(t) (\cos t \cos t_0+\sin t\sin t_0)\,dt \le 1$$ hence $$ \int_{0}^{2\pi} f(t) \cos t\,dt \le \frac{1}{\cos t_0}\left( 1 - \sin t_0 \int_{0}^{2\pi} f(t) \sin t\,dt\right) =\frac{1-r^2}{\sqrt{1-r^2}}=\sqrt{1-r^2} $$