$\sum_{n \geq 1} |c_n|^2 \leq \frac {1}{8}$ Let $f: \mathbb{R} \to \mathbb{R}$ be a $2\pi$ -periodic Riemann integrable function such that $1 \leq f(x) \leq 2$ for all $ x \in \mathbb{R}$. Let $f(x)$ ~ $\sum_{n \in \mathbb{Z}} c_n e^{inx}$ be the Fourier expansion of $f$. Prove that:
A) $\sum_{n \geq 1} |c_n|^2 \leq \frac {1}{8}$
B) There exists a continuous $2\pi$-periodic function $g: \mathbb{R} \to \mathbb{R}$ with Fourier expansion $g(x)$ ~ $\sum_{n \geq 1} \frac {|c_n|}{n} \cos {nx}$.
I know that the Fourier coefficients are defined as $c_n=\frac {1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}dx$. I am also familiar with the Dirichlet kernel as an alternate way of finding partial sums. This is a test review problem so a full solution would be appreciated. 
 A: For A), notice that $f(x) - c_0 \sim \sum_{n \neq 0} c_n e^{inx}$. Then by the Parseval's identity and the identity $c_{-n} = \overline{c_n}$, we have
$$ 2 \sum_{n \geq 1} |c_n|^2 = \sum_{n \neq 0} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |f(x) - c_0|^2 \, dx. $$
So it suffices to show that the LHS is $\leq \frac{1}{4}$. To this end, we write
\begin{align*}
\frac{1}{2\pi} \int_{-\pi}^{\pi} |f(x) - c_0|^2 \, dx
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(\left(f(x) - \tfrac{3}{2}\right) - \left(c_0 - \tfrac{3}{2}\right) \right)^2 \, dx \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(f(x) - \tfrac{3}{2}\right)^2 \, dx - \left(c_0 - \tfrac{3}{2}\right)^2 \\
&\leq \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(f(x) - \tfrac{3}{2}\right)^2 \, dx.
\end{align*}
Since $1 \leq f(x) \leq 2$, it follows that $|f(x) - \frac{3}{2}| \leq \frac{1}{2}$. Plugging this inequality gives the desired bound.
For B), by Cauchy-Schwarz inequality we have
$$\sum_{n\geq 1} \frac{|c_n|}{n} \leq \left( \sum_{n\geq 1} |c_n|^2 \right)^{1/2}\left( \sum_{n\geq 1} \frac{1}{n^2} \right)^{1/2} < \infty. $$
Therefore by the Weierstrass M-test, the series defining $g(x)$ converges uniformly and so $g$ is continuous.
