1. ## Inner Product Space

Hi guys, I'm trying to show that for $\displaystyle x,y\in S$ where S is an inner product space that

$\displaystyle ||x+y||\cdot||x-y||\leq ||x||^2+||y||^2$

This looks simple but I just can't crack it at all

2. I haven't worked it out, but it may be easier to manipulate the square of a norm. That is, it would be enough to show

$\displaystyle \|x+y\|^2\cdot \|x-y\|^2\leq (\|x\|^2+\|y\|^2)^2$

This way, you could write $\displaystyle \|x+y\|^2=\langle x+y,x+y\rangle$ and $\displaystyle \|x-y\|^2=\langle x-y,x-y\rangle$. The inner products are much nicer to play with.

3. Originally Posted by kenndrylen
Hi guys, I'm trying to show that for $\displaystyle x,y\in S$ where S is an inner product space that

$\displaystyle ||x+y||\cdot||x-y||\leq ||x||^2+||y||^2$

This looks simple but I just can't crack it at all
Denote by $\displaystyle \langle,\rangle$ the corresponding inner product, so:

$\displaystyle ||x+y||^2\cdot ||x-y||^2=\langle x+y, x+y\rangle\cdot\langle x-y,x-y\rangle=$

$\displaystyle \left(||x||^2+||y||^2+2Re\langle x,y\rangle\right)\left(||x||^2+||y||^2-2Re\langle x,y\rangle\right)=$

$\displaystyle ||x||^4+||y||^4-4(Re\langle x,y\rangle )^2+2||x||^2||y||^2\leq ||x||^4+||y||^4+2||x||^2||y||^2$ , and we're done

Tonio

4. Thanks so much, both of you