1
$\begingroup$

Kreyszig - Introduction to Functional Analysis - Page 135 Exercise 4.

If an inner product space $X$ is real, show that the condition $\|x\|=\|y\|$ implies that $\langle x+y,x-y\rangle=0$.

Okay so $\|x\|=\|y\|\implies \|x^2\| = \|y\|^2\implies \def\l{\langle}\def\r{\rangle} \l x,x\r=\l y,y\r$

I tried playing around from here, but couldn't get it. Below is just a bunch of steps that might be in the right direction, not sure:

$$\begin{align*} \l x,x \r - \l y,y\r &=0\\ \l x,x\r+\l y,y\r &= 2\l y,y\r\\ \quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad\,2\l y,y \r &= \l x+y,x+y\r\,\,\,\text{ (by parallelogram equality) }\\ \l x,x\r+ \l y,y\r + 2\l x,y\r &= 2\l y,y\r\\ \l x,x\r + 2\l x,y\r &= \l y,y\r\\ \l x,x\r + \l x,y\r + \l x-y, y\r &= 0\\ \l x,x \r + \l y,x\r + \l x-y,y\r &= 0\\ \l x+y,x\r + \l x-y,y\r &=0 \end{align*}$$

I felt like I was getting close, but not sure from this point. Ideas?

| cite | improve this question | | | | |
$\endgroup$
3
$\begingroup$

Real inner products are bilinear (that is, linear in each argument) and symmetric. Therefore $$\begin{align*} \langle x+y,x-y\rangle&=\langle x+y,x\rangle-\langle x+y,y\rangle\\ &=\langle x,x\rangle+\langle y,x\rangle-\langle x,y\rangle-\langle y,y\rangle\\ &=\langle x,x\rangle +\langle x,y\rangle-\langle x,y\rangle -\langle y,y\rangle\\ &=\langle x,x\rangle-\langle y,y\rangle\\ &=\|x\|^2-\|y\|^2 \end{align*}$$ equals $0$ if and only if $\|x\|^2=\|y\|^2$, which is the case if and only if $\|x\|=\|y\|$.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thanks for the answer, and for the format edit in my post. $\endgroup$ – Functional Analysis Sep 5 '15 at 4:55
3
$\begingroup$

$$ \langle x+y,x-y\rangle = \langle x,x\rangle + \langle y,x\rangle - \langle x,y\rangle - \langle y,y\rangle = \langle y,x\rangle - \langle x,y\rangle $$ Now since you are over the reals, $\langle y,x\rangle = \langle x,y\rangle$

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.