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Going to as little details as possible, here is a statement from Wald's text on QFT in curved spacetimes(I am not quoting the book)

He considers two vector spaces ${\cal S}$ and ${\cal H}$.

Note - For more details about ${\cal S}$ and ${\cal H}$, read this box. I believe for most part of the question, the following details are irrelevant, but I will provide them nonetheless. Otherwise, skip below.

He starts by considering the solution space of a classical system ${\cal S}$ with symplectic structure $\Omega$. This has a natural vector space structure. He complexifies it to ${\cal S}^{\mathbb C}$ and extends $\Omega$ to ${\cal S}^{\mathbb C}$ by complex linearity on each variable. He then defines the map $(\cdot, \cdot): {\cal S}^{\mathbb C} \times {\cal S}^{\mathbb C} \to {\mathbb C}$ on ${\cal S}^{\mathbb C}$ as $$ (y_1, y_2) = - i \Omega( \overline{y_1}, y_2) $$ This satisfies all the properties of an inner product except positive-definiteness. He then considers the subspace ${\cal H}$ of ${\cal S}^{\mathbb C}$ on which the inner product above is positive-definite. (There are of course many such choices of ${\cal H}$. Any one of them will do.)

He then shows that there is a one-one onto map $K: {\cal S} \to {\cal H}$. He shows that one can define a real inner product $\mu: {\cal S} \times {\cal S} \to {\mathbb R}$ on ${\cal S}$. He then goes on to show that one can use this to define a complex inner product on $\cal H$ as $$ \left( K y_1, K y_2 \right)_{\cal H} = \mu(y_1, y_2) - \frac{i}{2} \Omega(y_1, y_2)~\forall~y_1, y_2 \in {\cal S} $$ where $\Omega: {\cal S} \times {\cal S} \to {\mathbb R}$ is an antisymmetric function on ${\cal S}$, i.e. $\Omega(y_1, y_2) = - \Omega(y_2, y_1)$. He then uses the Cauchy-Schwarz Inequality for ${\cal H}$. This reads $$ \left( K y_1, K y_1 \right)_{\cal H} \left( K y_2, K y_2 \right)_{\cal H} \geq \left| \left( K y_1, K y_2 \right)_{\cal H} \right|^2 \geq \left| \text{Im} \left( K y_1, K y_2 \right)_{\cal H} \right|^2 $$ Expanding it out, he writes $$ \mu(y_1, y_1) \mu(y_2, y_2) \geq \mu(y_1, y_2)^2 + \frac{1}{4} \Omega(y_1, y_2)^2 \geq \frac{1}{4} \Omega(y_1, y_2)^2 $$ More specifically $$ \boxed{ \mu(y_1, y_1) \mu(y_2, y_2)\geq \frac{1}{4} \Omega(y_1, y_2)^2 } $$

Now, here is the statement that confuses me

Indeed, since $K$ is one-to-one and onto and since the Schwarz inequality on ${\cal H}$ always can be ``saturated", we obtain the following stronger version of the last inequality: For each $y_1 \in {\cal S}$ we have $$ \mu(y_1, y_1) = \frac{1}{4} \max_{y_2 \neq 0} \frac{ \Omega(y_1, y_2)^2}{\mu(y_2, y_2)} $$ Here's my question Q. Where did he get the above expression from?

He seems to be claiming that the boxed inequality is always saturated for some vector $y_2 \in {\cal S}$. Is that true? Why?

PS - I will understand if some people think that this question is more of a math question than a physics one. But, I thought that it might be possible that the answer relies on some of the assumptions we make in physics, so I asked it here. Any comments will be helpful

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Wow - much more detailed than the early version :) I think you could get a good answer either here or on math - and I'm personally not against leaving the question here. – Chris White Jul 16 '13 at 3:48
As for addressing the question, this is just a passing thought: I don't have the text, and I'm not entirely familiar with what's going on, but should $\max_{y_2\neq0}$ be replaced with $\sup_{y_2\neq0}$? As in, is it possible that no $y_2$ actually saturates, but that there always exists a sequence that converges as needed? Of course this question is moot if the underlying space is nice and compact or something, but again I also don't really know what I'm talking about... – Chris White Jul 16 '13 at 4:06
The book specifically says $\max$. Also, I think I have figured it out. Given a $y_1 \in {\cal S}$, any choice $y_2 \in {\cal S}$ such that $Ky_2 = \lambda K y_1$ with $\lambda^* = - \lambda$ saturates the boxed inequality. Now, we must only confirm that given the relation for $y_2$ above, one can always find a $y_2$ given any $y_1$. But this is obviously possible since $K$ is an invertible map. – Prahar Jul 16 '13 at 4:10
up vote 1 down vote accepted

Let $y_1, y_2$ $2$ complex vectors and let $<,>$ be a complex inner product defined by $<y_1,y_2> = \vec y_1^*.\vec y_2$.

Let $\vec a$ and $\vec b$ the real and imaginary part of $\vec y$ :
$\vec y = \vec a + i \vec b$

Then :

$$<y_1,y_2> = (\vec a_1 .\vec a_2 + \vec b_1 .\vec b_2) + i (\vec a_1 .\vec b_2 - \vec b_1 .\vec a_2) = u(y_1,y_2) + iv(y_1,y_2)$$

The Cauchy-Schwartz inequality gives :

$$<y_1,y_1><y_2,y_2> ~~\ge ~~|<y_1,y_2>|^2$$

We note that : $<y_1,y_1> = u(y_1, y_1)$, so we have :

$$u(y_1,y_1)~~\ge ~~ \frac{u^2(y_1,y_2) + v^2(y_1,y_2)}{u(y_2,y_2)}$$

Now, fixing a particular $y_1$, we limit the set of $y_2$ to those which respect $u(y_1,y_2) =0$. So, we have now :

$$u(y_1,y_1)~~\ge ~~ \frac{ v^2(y_1,y_2)}{u(y_2,y_2)}~~ ~~ ~~ ~~ ~~ (1)$$

Now, take explicitely $y_2$ defined by $ \vec a_2 = - \vec b_1, \vec b_2 =\vec a_1,$, we see that $\vec a_1 .\vec a_2 + \vec b_1 .\vec b_2 = 0$, that is $u(y_1,y_2) = 0$, so this choice is coherent with our previous hyphothesis.

Morevoer, we have $v(y_1,y_2) = \vec a_1^2 + \vec b_1^2 $, and $u(y_2,y_2) = \vec a_1^2 + \vec b_1^2 $, so we have, for this particular $y_2$.

$$u(y_1,y_1)~~= ~~ \frac{ v^2(y_1,y_2)}{u(y_2,y_2)}~~ ~~ ~~ ~~ ~~ (1)$$

So, we see, that the inequality $(1)$ is effectively saturated by our choice of this particular $y_2$

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