Parallelogram law in L1 space - Did not understand counter-example

Question

Parallelogram law in $L_1$ space In this topic they gave a counter-example to the title.
My problem is, their counter-example is wrong!
if you find the norm of: $||f-g||$ by their norm, you will receive 0! Thus the equality is True! They dont get $2 \ne 1$, they get $1=1$. Am I right? please check the norm there, I will add my calculation even to prove it! $$\left(\int _0^1\left(f\left(x\right)+g\left(x\right)\right)dx\right)^2=(\int _0^{\frac{1}{2}}f\left(x\right)dx-\int _{\frac{1}{2}}^1g\left(x\right)dx)^2=(\int _0^{\frac{1}{2}}1dx-\int _{\frac{1}{2}}^11dx)^2=\left(\frac{1}{2}-\frac{1}{2}\right)^2=\left(0\right)^2=0$$
and as you can see from my calculation, it is not equal to $1$!!. Please tell me what am I doing wrong \ are they wrong. Thanks.

Answer 1

$\|f-g\|$ is not $\int (f(x)-g(x))dx$. It is $\int |f(x)-g(x)|dx$ which is equal to $1$.