So I have a vectorspace with the norm $||\cdot||$. And I know the parallelogram law is valid because $(V, <\cdot , \cdot >)$ is a Hilbert space. So I know: $$ ||x+y||^2+ ||x-y||^2=2(||x||^2+||y||^2) $$.
My task is to show that: $$||x_n - x_m||^2 = 2 \left[ ||x_n - x||^2 + ||x-x_m||^2 -2 \left |\left | x- \frac{x_n + x_m}{2} \right |\right | ^2 \right ]$$.
What I did was: $$ ||x_n - x_m||^2 = ||(x_n -x) +(x - x_m)||^2 = 2 \left[ ||x_n - x||^2 + ||x-x_m||^2 \right ] - ||x_n -2x +x_m ||^2 = 2 \left[ ||x_n - x||^2 + ||x-x_m||^2 -\left |\left | x- \frac{x_n + x_m}{2} \right | \right| ^2 \right ]. $$ But I am missing the coefficient $2$ on the last norm! And I really don't know how to get it. Thank you for your help!
What you did is correct but when you will take 2 out of the square it will become 4, as follows-$$2[||x_n-x||^2+||x-x_m||^2]-[2||x-\frac{x_n+x_m}{2}||]^2$$ $$=2[||x_n-x||^2+||x-x_m||^2]-4||x-\frac{x_n+x_m}{2}||^2$$ from where the desired result follows. Hope it is helpful