When $\|f_n\|_{L^2}=1$ where $f_n(x)= f(x-n)+(-1)^n f(x+n)$?

Question

Let $f: \mathbb R \to \mathbb C $ be a function and define $f_n(x)= f(x-n)+(-1)^n f(x+n).$

Can we expect to choose $f\in L^2(\mathbb R)$ such that $\|f_n\|^2_{L^2}=1$ for all $n\in\mathbb Z$?

Side thought: We know that $L^2(\mathbb R)$ is complex Hilbert space with inner product $\langle f, g \rangle = \int f \bar{g}$ and using properties of inner product we have $$\|f_n\|_{L^2}^2= 2\|f\|^2_{L^2}+ 2 \text{Re} \langle f(x-n), (-1)^nf(x+n)\rangle $$

Answer 1

Idea for negative answer: for $\epsilon>0$ exists $M>0$ s.t. $\|f\chi_{[-M,M]}\|_{L^2} > 1 - \epsilon.$ ($\chi_{\cdots} =$ characteristic function). For $n$ large enough, $$\|f_n\|_{L^2}^2\approx\|f(\cdot - n)\|_{L^2}^2 + \|f(\cdot + n)\|_{L^2}^2 = 2\|f\|_{L^2}^2.$$