I've been thinking about this problem for a while now with no success:
Let $f:\mathbb{R}\to \mathbb{R}$ in $PC$ (piecewise continuous) and $1$-periodic. Using the function: $$g(x)=f(x)+f\big(x+\frac{1}{2}\big)$$ prove:
$$ \sum_{n=-\infty}^\infty \big|\hat{f}(2n) \big|^2=\frac{1}{2}\int_0^1\bigg(f(x)^2+f(x)f\big(x+\frac{1}{2}\big)\bigg) $$
where $\hat{f}(n)$ is the $n$-th Fourier coefficient.
I've been messing around with Parseval's equality, since the function is in $L^2$, and exploring both the LHS and the RHS, as well as by integrating $g$ to see if there's any term that looks like it might yield the equality, but nothing came of this. I cannot see any other way of approaching this without using Parseval - so I'm convinced it has to be related, but I just can't see the way (I might be mistaken on this, of course).
My closest attempt yielded (unless I have a mistake) that the RHS equals $\frac{1}{4}\sum_{m=-\infty}^{\infty}|\hat{g}(m)|^2$, but I still couldn't find a way for it to equal the LHS.
What am I missing?
Your attempt is correct. It suffices to notice that $$\hat g(m)=\begin{cases} 2\hat f(m), & m \text{ is even,}\\ 0, & m \text{ is odd}.\end{cases}$$ Indeed, using the definition of Fourier coefficients, \begin{align*} \hat g(m)&=\hat f(m)+\int_0^1 f(x+1/2) e^{-2\pi imx}\,dx\\ &=\hat f(m)+e^{\pi im}\int_0^1 f(x+1/2) e^{-2\pi im(x+1/2)}\,dx\\ &=\hat f(m)+(-1)^m\hat f(m),\qquad m\in\mathbb Z. \end{align*} Here we use a result for periodic fucntions: if $h(x)=h(x+T)$ and $h$ is integrable on $[0,T]$, then $$\int_0^Th(x)\,dx=\int_a^{a+T}h(x)\,dx,\qquad a\in\mathbb R.$$ See here for a proof of the above result.