Show there exist a constant $c\in \Bbb{C}$ such that $\int_{0}^{1}|{f-c}|^2<{1\over 36}$

Question

Let $f:\Bbb{R}\to \Bbb{C}$ be a $1$-periodic function, $f\in C^1$ and $\int_{0}^{1}|f'|^2\le 1$.

a. Show $\sum_{k\ne 0}|{\hat{f}(n)}|^2\le {1\over 4\pi^2}$ (I did it already, and that question is also available somewhere on this website).

b. Show there exist a constant $c\in \Bbb{C}$ such that $\int_{0}^{1}|{f-c}|^2<{1\over 36}$.

This is where I am stuck; I know that $\int_{0}^{1}|f|^2=\sum_{k\in \Bbb{Z}}|{\hat{f}(n)}|^2\le {1\over 4\pi^2}<{1\over 36}$ and I also know that $\int_{0}^{1}|{f-c}|^2\le \int_{0}^{1}|f|^2$ for any constant (that, actually, I am not sure of. It is the complex after all.). If the last inequality actually holds, then it works for any $c\in \Bbb{C}$. If it doesn't- why doesn't it, and how do I find such a $c$?

Answer 1

Let $m=\int_{0}^{1}f(x)\,dx$ be the mean value of $f$ over $[0,1]$. By considering the Fourier series of $f$ we have:

$$f(x) = m +\sum_{n\geq 1}\left( s_n \sin(2\pi n x) + c_n \cos(2\pi n x)\right)$$ and: $$ \| f'(x) \|^2 = 2\pi^2\sum_{n\geq 1} n^2 (s_n^2+c_n^2)\leq 1 $$ hence: $$ \| f(x) - m\|^2 = \frac{1}{2}\sum_{n\geq 1}(s_n^2+c_n^2)\leq\frac{1}{2}\sum_{n\geq 1}n^2(s_n^2+c_n^2)\leq\frac{1}{4\pi^2}<\frac{1}{36} $$ so a good choice for your constant $c$ is just $c=m=\int_{0}^{1}f(x)\,dx$.

This is just Wirtinger's inequality.

Answer 2

If you know that $\sum_{k\ne 0}|{\hat{f}(n)}|^2\le {1\over 4\pi^2}$, then you only need to consider $|\hat{f}(0)|$ and you can then apply Parseval's theorem linking the sum of the Fourier Coefficients with the Integral of the function squared. So if you can modify $f$ by adding a constant to make $\hat{f}(0) = 0$, you should be good to go...