0

By definition,

  • symmetry: $\phi(x, y) = \overline{\phi(y, x)}$
  • positive: $\phi(x, x) \ge 0$

So far I have,

$|\phi(x, y)|^2$ $= \phi(x, y)\overline{\phi(x, y)}$ $= \phi(x, y)\phi(y, x)$

But I'm unsure how to show this is $\le \phi(x, x)\phi(y, y)$.

Letting $x = y + z \iff y = x - z$, I have $\phi(x, y)\phi(y, x)$

$= \phi(y + z, y)\phi(x - z, x)$

$= (\phi(y, y) + \phi(z, y))(\phi(x, x) - \phi(z, x))$

$= \phi(y, y)\phi(x, x) + \phi(z, y)\phi(x, x) - \phi(y, y)\phi(z, x) - \phi(z, y)\phi(z, x)$

$= \phi(y, y)\phi(x, x) + \phi(z, y)\phi(x, x) - \phi(y, y)\phi(z, x) - \phi(z, y)\phi(z, x)$

From here, I am not sure how to show that $\phi(z, y)\phi(x, x) \le (\phi(y, y) + \phi(z, y))\phi(z, x)$

I think I need to use the positivity of $\phi$ somehow, but I don't know how.

william_grisaitis
  • 419

2 Answers2

3

Suppose $V$ is a complex vector space, $\phi:V\times V\to\Bbb{C}$ is a conjugate bilinear, positive functional. The following proof then amounts to the proof of the standard Cauchy-Schwarz inequality. For each $t\in\Bbb{R}$, we have \begin{align} 0&\leq \phi(x+ty,x+ty)=\phi(x,x)+2t\text{Re}[\phi(x,y)]+t^2\phi(y,y). \end{align} So, the discriminant of this quadratic polynomial in $t$ must be non-positive, hence \begin{align} [2\text{Re}[\phi(x,y)]]^2-4\phi(y,y)\phi(x,x)\leq 0, \end{align} or equivalently, \begin{align} \left(\text{Re }\phi(x,y)\right)^2\leq \phi(x,x)\phi(y,y). \end{align} Since this inequality holds for all $x,y$, it must also hold for the pair $e^{i\theta}x,y$. Hence, \begin{align} \left(\text{Re}\left(e^{i\theta}\cdot \phi(x,y)\right)\right)^2\leq \phi(x,x)\phi(y,y) \end{align} Note that if $\phi(x,y)=0$, then the inequality in question $|\phi(x,y)|^2\leq \phi(x,x)\phi(y,y)$ is trivially satisfied. Otherwise, choose $\theta\in\Bbb{R}$ so as to cancel the argument of $\phi(x,y)$. Then, we find that $|\phi(x,y)|^2\leq \phi(x,x)\phi(y,y)$ in this case as well.

So as you can see, in the case of real vector spaces and actually symmetric bilinear functionals, the proof is very easy. In the complex case, with conjugate bilinear positive functionals, one just needs to amplify the argument slightly (another proof, as described in the link is to start with $\phi(x-y,x-y)\geq 0$, then exploit the phase imbalance, and also a scaling imbalance… these are very cute but powerful tricks of analysis).

peek-a-boo
  • 55,725
  • 2
  • 45
  • 89
0

We have $$ 0\leq \varphi(tx-y,tx-y)=|t|^2\varphi(x,x)-2\text{Re}(t\varphi(x,y))+\varphi(y,y). $$ If $\varphi(x,x)=0$ and $\varphi(x,y)\neq 0$ we can take $t=s\overline{\varphi(x,y)}$ and the function becomes linear with respect to the real variable $s,$ thus is unbounded from below. Hence we may assume $\varphi(x,x)>0.$ Let $$t={\overline{\varphi(x,y)}\over \varphi(x,x)}$$ Then $$0\le {|\varphi(x,y)|^2\over \varphi(x,x)}-2{|\varphi(x,y)|^2\over \varphi(x,x)}+\varphi(y,y)\\ = -{|\varphi(x,y)|^2\over \varphi(x,x)}+\varphi(y,y)$$ which implies the statement.

Ryszard Szwarc
  • 30,319