Calculate $4||x+y||2+||x−y||2 $

Question

Let's take $x, y \in X$ where $X$ is vector space with norm inducted by inner product.

Additionally $||3x+y||= 2$, $||x+3y|| = 2$.

I want to calculate $4||x+y||^2+||x-y||^2$ $\\$

My work so far

Because $X$ has a norm inducted by inner product then for $x,y \in X$ we have $$||x+y||^2 + ||x-y||^2 = 2(||x||^2+||y||^2)$$

So:

$4||x+y||^2 + ||x-y||^2 = 3||x+y||^2 + (||x+y||^2 + ||x-y||^2) = 3||x+y||^2 + 2(||x||^2 + ||y||^2)$

And I'm not sure why to do next... I'm not sure how can I was trying to rewrite this expression to somehow include information that $||3x+y|| = ||x+3y|| = 2$ but I wasn't able to. Could you please give me a hint how it can be calculated ?

Answer 1

You are on the right track with the parallelogram law. Apply it to $u=3x+y$ and $v=x+3y$: $$ 16 = 2 (\Vert 3x+y\Vert^2 + \Vert x+3y\Vert^2 ) = \Vert 4x+ 4y \Vert^2 + \Vert 2x - 2y \Vert^2 \\ = 16 \Vert x+y \Vert^2 + 4 \Vert x-y \Vert^2 = 4 (4\Vert x+y \Vert^2 + \Vert x-y \Vert^2) $$

Answer 2

Use the fact that $$\|x \pm y\|^2 = \|x\|^2 + \|y\|^2 \pm 2\Re\langle x, y \rangle.$$