I am looking at periodic functions $f$ on $[0,2\pi]$ satisfying these constraints:
- $f(0) = 1$.
- $f$ is zero-centered, i.e. its integral from 0 to $2\pi$ is 0.
- $f$ is even. (More specifically, it's positive definite.)
- $f$ has a continuous first derivative.
I suspect that there is some sense in which cosine is the the smoothest function satisfying these constraints, but I'm not sure how to appropriately define "smooth" here nor how to prove that cosine is the smoothest such function. Is there a known result along these lines?
Well, if I use $S\triangleq\int_0^{2\pi}f''(t)^2dt$ as my measure of smoothness (smaller number means smoother function), I've verified that cosine isn't the smoothest function. If $f$ has a continuous first derivative then it is square integrable over $[0,2\pi]$, and hence its Fourier series converges. I then find that
$$S = 8\pi\sum_{k=1}^\infty k^4 a_k^2$$
where $f(t)=2\sum_{k=1}^\infty a_k \cos(k t)$ is the Fourier series for $f$. Subject to the condition that $f(0)=1$, i.e. $\sum_{k=1}^\infty a_k = 1/2$, I find that $S$ is minimized when $a_k\propto 1/k^4$ for all $k$.