Complex inner product proof

Question

I have just solved this problem in the real inner product space with $\langle \cdot , \cdot \rangle$ as the inner product.

Show that in a real inner product space we have:

$\langle x,y \rangle = \frac{1}{4} \left(||x+y||^2 - ||x-y||^2\right)$

Not so bad.

The second part of the question is confusing for me.

In a complex inner product space we have

$\Re \langle x,y \rangle = \frac{1}{4} \left(||x+y||^2 - ||x-y||^2\right)$

and

$\Im \langle x,y \rangle = \frac{1}{4} \left(||x+iy||^2 - ||x-iy||^2\right)$

Anyone have any idea how to solve this?

Answer 1

Note that we have

$$\begin{align} ||x\pm y||^2&=\langle x\pm y,x\pm y\rangle\\\\ &=||x||^2+||y||^2\pm \left(\langle x,y\rangle+\langle y,x\rangle\right)\\\\ &=||x||^2+||y||^2\pm 2\text{Re} \{\langle x,y\rangle\}\\\\ \end{align}$$

Therefore, it is easy to see that

$$||x+ y||^2-||x-y||^2=4\text{Re}\{\langle x,y\rangle\} \tag 1$$

whereupon dividing by $4$ on both sides of $(1)$ yields the coveted result

$$\text{Re}\{\langle x,y\rangle\}=\frac14\left(||x+ y||^2-||x-y||^2\right)$$

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Similarly, we have

$$\begin{align} ||x\pm i y||^2&=\langle x\pm i y,x\pm i y\rangle\\\\ &=||x||^2+||y||^2\mp i\left(\langle x,y\rangle-\langle y,x\rangle\right)\\\\ &=||x||^2+||y||^2\pm 2\text{Im} \{\langle x,y\rangle\}\\\\ \end{align}$$

Therefore, it is easy to see that

$$||x+ iy||^2-||x-iy||^2=4\text{Im}\{\langle x,y\rangle\} \tag 2$$

whereupon dividing by $4$ on both sides of $(2)$ yields the coveted result

$$\text{Im}\{\langle x,y\rangle\}=\frac14\left(||x+ iy||^2-||x-iy||^2\right)$$