Why is $\|x+y\|^2=4 \| (x+y)/2 \|^2$ in Normed space?

Question

Why is $\|x+y\|^2=4 \| (x+y)/2 \|^2$ in normed space?

This could be an application of parallelogram law, but I don't see how.

Answer 1

By linearity of the norm, $\forall a \geq 0,||ax|| = a ||x||$.

$\|z\|^2 = \frac{a}{a} \|z\|^2 = a \|\frac{z}{a^2}\|^2$

Apply it for $z = x+y$ and $a = 4$.

Answer 2

$$\|x+y\|^2= \|2 \cdot \frac{x+y}{2}\|^2 = \left(|2|\|\frac{x+y}{2}\|\right)^2 = 4\|\frac{x+y}{2}\|^2$$

using $\|2z\|= 2\|z\|$ for all $z$