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Given a Hilbert space $\mathcal{H}$.

Consider a quadratic form: $$q:\mathcal{H}\to\mathbb{C}:\quad q[\lambda\varphi]=|\lambda|^2q[\varphi]$$

Suppose it satisfies: $$q[\varphi+\psi]+q[\varphi-\psi]=2q[\varphi]+2q[\psi]$$

Define a sesquilinear form: $$s:\mathcal{H}\times\mathcal{H}\to\mathbb{C}:\quad s(\varphi,\psi):=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha q[\varphi+i^\alpha\psi]$$

Then for positive forms: $$q\geq0:\quad|s(\varphi,\psi)|\leq q[\varphi]q[\psi]$$

How can I check this?

Rodrigo de Azevedo
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C-star-W-star
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1 Answers1

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For this, you need only assume $Q$ is real and non-negative on a complex space with $$ Q(x+y)+Q(x-y)=2Q(x)+2Q(y),\\ Q(ix)=Q(x). $$ Then it is true that $B(x,y)=\frac{1}{4}\sum_{n=0}^{3}i^{n}Q(x+i^{n}y)$ satisfies $$ |B(x,y)| \le Q(x)^{1/2}Q(y)^{1/2}. $$ Therefore $|x|_{Q}=Q(x)^{1/2}$ satisfies the triangle inequality. Then, if $Q(\lambda x)=|\lambda|^{2}Q(x)$ holds for all scalars $\lambda$, it is also true that $B$ is sesquilinear.

Outline

  1. $B(x+x',y)=B(x,y)+B(x',y)$. So $B(\alpha x,y)=\alpha B(x,y)$ for all complex scalars with rational real and imaginary parts.

  2. $4|B(x,y)|=\lim_{n}4B(\alpha_n x,y)= \lim_{n}Q(\alpha_n x + y)-Q(\alpha_n x-y) \le 2Q(x)+2Q(y)$ if $Q(x) \ge 0$ for all $x$.

  3. Use rational positive real scalars to obtain $|B(rx,sy)| \le 1$ and $|B(x,y)| \le Q(x)^{1/2}Q(y)^{1/2}$.

  4. Continuity: $|\lambda B(x,y)-B(\lambda x,y)|=\lim_{n}|B((\lambda_n-\lambda)x,y)|\le\lim_{n}|\lambda_{n}-\lambda|Q(x)^{1/2}Q(y)^{1/2}=0$ if $Q(\alpha x)=|\alpha|^{2}Q(x)$. So $\lambda B(x,y)=B(\lambda x,y)$ for general $\lambda\in\mathbb{C}$. That makes $B$ fully sesquilinear.

Disintegrating By Parts
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  • Yes and how can I check Cqauchy-Schwarz: $|B(x,y)|\leq Q(x)^{1/2}Q(y)^{1/2}$ – C-star-W-star Jul 07 '15 at 13:04
  • @Freeze_S I've given you the outline, and will let you fill in the pieces. – Disintegrating By Parts Jul 07 '15 at 15:13
  • Number '3' is the normalization trick?? – C-star-W-star Jul 09 '15 at 02:54
  • Number '2' restricts to hermitian forms?? – C-star-W-star Jul 09 '15 at 02:56
  • Decomposing arbitrary form into: $s=s_\Re^++s_\Re^-+is_\Im^++is_\Im^-$ – C-star-W-star Jul 09 '15 at 03:00
  • Ok I'm way to tired.. Let me just believe it for now. :) – C-star-W-star Jul 09 '15 at 03:02
  • @Freeze_S : #1 is general for $B$ coming from a general $Q$. The additivity gives linearity over rational scalars because $B(nx,y)=nB(x,y)$ and $mB(\frac{1}{m}x,y)=B(x,y)$ gives $B(\frac{1}{m}x,y)=\frac{1}{m}B(x,y)$ for positive integers $m$, $n$.

    #2 is specific to $Q$ real and non-negative.

    #3 uses (1) and limits of rationals along with (2) to get the tight Cauchy-Schwarz bound.

    #4 uses (3) and $Q(\lambda x)=|\lambda|^{2}Q(x)$ to derive continuity. Or instead assume $Q(x) \le C|x|$ for some seminorm $|\cdot|$, which you can see directly. So $Q \ge 0$ bounded is enough.     – Disintegrating By Parts Jul 09 '15 at 03:13
  • Ya I guessed already that taking first positive quadratic forms is necessary to obtain the 'tight' Cauchy-Schwarz. Otherwise one probably obtains a factor 2 or 4 to much that cannot be killed afterwards. Right? – C-star-W-star Jul 09 '15 at 03:15
  • @Freeze_S : Yes, that's right. But you get Cauchy-Schwarz only having linearity over the rational complex scalars IF $Q \ge 0$. That, however, is not enough to give full linearity over $\mathbb{Q}$. But, $Q(\lambda x)=|\lambda|^{2}Q(x)$ then turns $|x|=Q(x)$ into a full-blow seminorm. And you really only need $Q(x) \le |x|$ for some seminorm $|\cdot|$ with $Q \ge 0$ to also force $Q$ to be a seminorm. Either condition forces continuity over the scalar field and, sesquiinearity for $B$. This stuff is tricky and wrapped up with the axiom of choice. – Disintegrating By Parts Jul 09 '15 at 03:42
  • Mhh yes. Like measurability and continuity as you mentioned earlier.. – C-star-W-star Jul 09 '15 at 03:55
  • By the way isn't there the hypothesis in ZF: $AC\iff WE\iff Zorn\iff Hamel\iff Baire\stackrel{???}{\iff} UBP\iff OMT\iff CGT\iff Zabreiko$ – C-star-W-star Jul 09 '15 at 03:58
  • @Freeze_S : I'm not sure of all of those. What is Zabreiko? – Disintegrating By Parts Jul 09 '15 at 16:19
  • See on wiki: Baire vs. AC See on MSE: Zabreiko – C-star-W-star Jul 09 '15 at 16:29
  • @Freeze_S : In this case, you end up with additive functions on $\mathbb{R}$, which is to say that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. And you want to know if $f$ is linear. $f$ is linear iff it is continuous. $f$ is linear over every set $x\mathbb{Q}={ x q : q \in \mathbb{Q}}$ because $f(xq)=qf(x)$. You can partition $\mathbb{R}\setminus{0}$ into sets by the equivalent relation $x \sim y$ iff $x/y \in \mathbb{Q}$. This is very closely related to the classical construction of a non-measurable set. Consequently, if $f$ is meausrable, then it is continuous. – Disintegrating By Parts Jul 09 '15 at 16:59