Q:

Fourier series: Prove the integral is less than $2\pi$.

Let $f\in R(\mathbb{T})$ which is $C^1$. Also, $\int_0^{2\pi} \left|f'(t)\right|^2 dt < 2\pi$
  Prove:

$\sum_{n\ne 0} \left|\hat{f}(n)\right|^2 < 1$.
$\exists c\in\mathbb{C}: \int_0^{2\pi} \left|f(t)-c\right|^2 dt < 2\pi$.

Proving #1 was easy. I just used the identity: $\hat{f'}(n) = in\hat{f}(n)$.
I had troubles proving #2. I got an hint to choose $c = \int_0^{2\pi} f(x) \, dx$ (Not sure how could I think about this myself though).

A:

You can write $f(t)=\hat {f}(0) + \sum _{n \neq 0}\hat{f}(n)e^{i nt}$. 
Then by Parseval's formula : $$\frac{1}{2\pi}\int |f(x)-\hat{f}(0)|^2dx = \sum _{n\neq 0}|\hat{f}(n)|^2< 1 .$$

