Let $f\in C^1$, $2\pi$-periodic, and let's assume $\int_{-\pi}^\pi |f'|^2 \le 1$. Prove:
- $$\sum_{n\in\mathbb{Z}} |\hat f(n)|^2 \le \frac{1}{2\pi}$$
- There's a $c\in\mathbb{C}$ such that: $$\int_{-\pi}^\pi |f-c|^2 \lt 1$$
Now, I already proved #1 by using Parsevals' Theorem. I know that the answer for #2 is $c = \hat f(0)$.
consider the function $g = f-c$. Then,
$$\hat g(n) = \hat {f-c}(n) = \hat f(n) - \hat c(n)$$
Now, for $n\ne 0$, $\hat g(n) = \hat f(n)$ but for $n=0$, $\hat g(n) = 0$.
If $\hat f(0) \ne 0$.
$$\int_{-\pi}^\pi |f-c|^2 = 2\pi \|f-c\|^2 = 2\pi \sum_{n\in\mathbb{Z}}|\hat{f-c}(n)|^2 \lt \sum_{0 \ne n\in\mathbb{Z}}|\hat{f-c}(n)|^2 \le \frac{1}{2\pi}2\pi = 1$$
BUT! can we tell for sure that $\hat f(0) \ne 0$? How?