Is $\langle x,y \rangle=\overline{\langle y,x \rangle}$?

Question

can anyone help me?

Be $(X,\|.\|)$ a normed space that satisfies the parallelogram law. Let's define $\Phi : X\times X \rightarrow \mathbb{C}$ by:

$\Phi (x,y)=\dfrac{1}{4}[ \|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2]$

Prove

$\Phi(x,y)=\overline{\Phi(y,x)}$

I resolved

$\overline{\Phi(x,y)}=\dfrac{1}{4}[ \|x+y\|^2-\|x-y\|^2-i\|x+iy\|^2+i\|x-iy\|^2]$

Answer 1

A Hermitian inner product must have the conjugate symmetry $$ \langle x,y\rangle=\overline{\langle y,x\rangle}\,. $$ For $\Phi(x,y)=\langle x,y\rangle$ this is not hard to show by noticing that $$ \|x+iy\|^2=\|i(-ix+y)\|^2=|i|^2\|(-ix+y)\|^2=\|y-ix\|^2\,. $$ Likewise, $$ \|x-iy\|^2=\|-y-ix\|^2=\|y+ix\|^2\,. $$ Then \begin{align} \Phi (x,y)&=\dfrac{1}{4}[ \|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2]\\ &=\dfrac{1}{4}[ \|x+y\|^2-\|x-y\|^2+i\|y-ix\|^2-i\|y+ix\|^2] \end{align} from which $\Phi(x,y)=\overline{\Phi(y,x)}$ follows.