Let $f\in L^2([0,1)^2).$ Define $|F(x,y)|^2:= \frac{|f(x,y)|^2} {|f(x,y)|^2 + |f(x-(1/2),y )|^2}$ for all $x, y \in [0,1].$ Assume that $f$ is nice so everything make sense.
My question is: How to compute the $L^2([0,1)^2)-$norm of $F$? Is $F$ independent of choice of $f$? Is it constant?
Note that $\|F\|_{L^2}^2= \int_0^1 \int_0^1 \frac{|f(x,y)|^2} {|f(x,y)|^2 + |f(x-(1/2),y )|^2} dx dy$
Edit: We may realized $[0,1)^2=\mathbb T^2$ (torus). So $f$ is periodic function in each variable with period 1.
I guess that $F$ is properly defined as $$F(x,y)=\frac{f(x,y)}{\sqrt{\big|f(x,y)\big|^2+\Big|f\left(x-\frac12,y\right)\Big|^2}}\text{ for }x,y\in\mathbb{T}:=\mathbb{R}/\mathbb{Z}\,.$$ Let $T:\mathcal{H}\to\mathcal{H}$, where $\mathcal{H}:=\mathcal{L}^2\big(\mathbb{T}^2\big)$, be the operator $$(Tg)(x,y)=g\left(x-\frac12,y\right)\text{ for every }x,y\in\mathbb{T}\,.$$ Prove that $T$ is a unitary operator (or an isometric isomorphism) on $\mathcal{H}$ (i.e., $\|Tg\|_{\mathcal{H}}=\|g\|_{\mathcal{H}}$ for all $g\in\mathcal{H}$).
Now, $$\|F\|_{\mathcal{H}}^2+\|TF\|_{\mathcal{H}}^2=\int_0^1\,\int_0^1\,\frac{\big|f(x,y)\big|^2+\Big|f\left(x-\frac12,y\right)\Big|^2}{\big|f(x,y)\big|^2+\Big|f\left(x-\frac12,y\right)\Big|^2}\,\text{d}x\,\text{d}y\,.$$ Thus, $$2\,\|F\|_{\mathcal{H}}^2=\int_0^1\,\int_0^1\,1\,\text{d}x\,\text{d}y=1\,.$$