A well-known theorem of elementary geometry states that the sum of the squares of the sides of a parallelogram equals the sum of the squares of its diagonals. Prove that this so-called para elogram law is true for the norm in R:
First I was not sure whether the following is correct:
$||f-g||^2=(f,f)-2(f,g)+(g,g)$
If it is then:
$2||f||^2+2||g||^2=||f+g||^2+||f-g||^2$
I will first show from RHS -> LHS
$||f+g||^2+||f-g||^2$
$=(f+g,f+g)+(f-g,f-g)$
$=(f,f)+2(f,g)+(g,g)+(f,f)-2(f,g)+(g,g)$
$=2(f,f)+2(g,g)$
$=2||f||^2+2||g||^2$
Now from LHS -> RHS
$2||f||^2+2||g||^2$
$=2(f,f)+2(g,g)$
$=(f,f)+(f,f)+(g,g)+(g,g)+0$
$=(f,f)+(f,f)+(g,g)+(g,g)+2(f,g)-2(f,g)$
$=(f,f)+2(f,g)+(g,g)+(f,f)-2(f,g)+(g,g)$
$=(f+g,f+g)+(f-g,f-g)$
$=||f+g||^2+||f-g||^2$
You can add $0$ to the last line, the write that $0$ as $2(f,g)-2(f,g)$, then use those to complete the squares. And your first formula is correct.