# Cauchy-Schwarz Inequality proof (for semi-inner-product A-module).

I am reading a proof of the following Cauchy-Schwarz Inequality and I don't understand one part of the proof:

Theorem: Let $A$ be a $C^*$-algebra and let $E$ be a semi-inner-product $A$-module. Then $$\langle x,y \rangle ^* \langle x,y \rangle \leq \| \langle x,x \rangle \| \langle y,y \rangle \text{ for all } x,y\in E$$

The proof starts of by saying, without loss of generality we can assume $\| \langle x,x \rangle \| = 1$ But I don't understand why we can assume this.

The rest of the proof is just some calculations which I understand: for $a\in A$, $x,y\in E$ we have $$0 \leq \langle xa-y,xa-y \rangle = a^* \langle x,x \rangle a - \langle y,x \rangle a - a^* \langle x,y \rangle + \langle y,y \rangle \leq a^*a - \langle y,x \rangle a - a^* \langle x,y \rangle + \langle y,y \rangle$$ and by letting $a=\langle x,y \rangle$ we get the desired result.

I just need to understand why we can assume $\| \langle x,x \rangle \| = 1$. My guess is that if it does not equal 1 then we can replace it with an equivalent norm that equals 1, but I don't know if this is the correct reason.

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just take a scalar multiple of $x$ to make the norm one. – J. Loreaux Jun 13 '12 at 23:48

For any fixed $x$ and $y$ and $t > 0$, if one replaces $x$ with $tx$, both sides of the inequality you want to prove scale by $t^2$. So for all $t > 0$, the inequality holds for a given $x$ and $y$ if and only if it holds for $tx$ and $y$. Since $$\|\langle tx, tx\rangle\| = t^2 \|\langle x,x\rangle\|,$$ if $\langle x,x\rangle$ is nonzero, we can (by taking $t = \|\langle x,x\rangle\|^{-1/2}$) indeed reduce to the case that $\|\langle x, x\rangle\| = 1$.
It remains to handle the case that $\langle x,x\rangle = 0$, which can happen for nonzero $x$ (I think that's the point of the "semi" in "semi-inner-product space"). In this case, I don't see any obvious reason why $\langle x,y\rangle$ should be zero, but if one echoes the argument you gave for the $\|\langle x,x\rangle\| = 1$ case, one sees that for all $a$ in $A$ one has $$0 \leq \langle xa - y, xa - y\rangle = -\langle y,x\rangle a - a^* \langle x,y\rangle + \langle y,y\rangle.$$ For any $t > 0$, by putting $a = \frac{t}{2} \langle x,y\rangle$ in the above one deduces $$t \langle x,y\rangle^* \langle x,y\rangle \leq \langle y,y\rangle.$$ One gets a contradiction if $\langle x,y\rangle$ is nonzero by taking $t$ large enough (if $0 \leq a \leq b$ in a $C^*$ algebra, then $\|a\| \leq \|b\|$), so it must be that $\langle x, y \rangle = 0$, which is what we want in this case.
There is probably a simpler way to handle the $\langle x,x\rangle = 0$ case that I'm not seeing at the moment, but anyway, I think the above works.
An easier way to deal with the degenerate case is to exchange the roles of $x$ and $y$. – Martin Argerami Jun 14 '12 at 8:07