Prove that $\langle x, y \rangle = \overline{\langle y, x \rangle}$

Question

Let $X$ be a normed linear space over the field $\mathbb C$ with the norm $\|\cdot \|$. Let $x,y \in X$. Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$. Prove that $\langle x, y \rangle = \overline{\langle y, x \rangle}.$

My attempt:

\begin{align*} \overline{\langle y, x \rangle} &= \overline{\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert y +i^k x\Vert^2} \\ &= \overline{\frac{1}{4}(\Vert y +x\Vert^2)+i\Vert y +i x\Vert^2-\Vert y - x\Vert^2-i\Vert y -ix\Vert^2)} \\ &= \frac{1}{4}(\overline{\Vert y +x\Vert^2)}-i\overline{\Vert y +i x\Vert^2}-\overline{\Vert y - x\Vert^2}+i\overline{\Vert y -ix\Vert^2}). \end{align*}

Can I take the complex conjugate inside the norm?

Answer 1

Since the norm is a positive real number by definition/construction, it isn't affected by the complex conjugation. Then, it is only a matter of rearranging the terms. At the end, we have : $$ \begin{align} \overline{\langle y, x \rangle} &= \overline{\frac{1}{4} \sum_{k =0}^{3} i^{k} \|y + i^kx\|^2} \\ &= \frac{1}{4} \sum_{k =0}^{3} (-i)^{k} \|y + i^kx\|^2 \\ &= \frac{1}{4} \sum_{k =0}^{3} (-i)^{k} \|x + (-i)^ky\|^2 \\ &= \frac{1}{4} \left(\|x+y\|^2 - i\|x-iy\|^2 - \|x-y\|^2 + i\|x+iy\|^2\right) \\ &= \frac{1}{4} \left(\|x+y\|^2 + i\|x+iy\|^2 - \|x-y\|^2 - i\|x-iy\|^2\right) \\ &= \frac{1}{4} \sum_{k =0}^{3} i^{k} \|x + i^ky\|^2 \\ &= \langle x,y \rangle \end{align} $$ Formally, the rearrangement corresponds to the change of variable $k' := 3 - k$.

Answer 2

You cannot bring the complex conjugate inside the norm as $\|x+iy\|$ is not necessarily equal $\|x-iy\|.$ Instead I would use $$i^k\|x+i^ky\|^2= i^k\|i^{-k}x+y\|^2=\overline{i^{-k}\|y+i^{-k}x\|^2}$$ On summing up the terms you get the conclusion.