From d698b9b7f1942442a7c58ef6c1dc2c9714970c5c Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 18 Oct 2026 21:44:10 -0400 Subject: [PATCH] First draft of quadratic forms. --- src/fa/norm/hilbert.tex | 156 ++++++++++++++++++++++++++++++++++++++++ 1 file changed, 156 insertions(+) diff --git a/src/fa/norm/hilbert.tex b/src/fa/norm/hilbert.tex index 9e949f0..a971d5f 100644 --- a/src/fa/norm/hilbert.tex +++ b/src/fa/norm/hilbert.tex @@ -369,4 +369,160 @@ A significant property of Hilbert spaces is that every closed subspace is comple \end{proof} +\begin{definition}[Quadratic Function] +\label{definition:hilbert-quadratic-function} + Let $H$ be a complex Hilbert space and $q: H \to \complex$, then $q$ is \textbf{quadratic} if: + \begin{enumerate} + \item For each $\lambda \in \complex$ and $\xi \in H$, $q(\lambda x) = |\lambda|^2 q(x)$. + \item (\textbf{Parallelogram Law}) For each $\xi, \eta \in H$, + \[ + q(\xi + \eta) + q(\xi - \eta) = 2q(\xi) + 2q(\eta) + \] + \end{enumerate} + + and \textbf{bounded} if + \begin{enumerate}[start=2] + \item There exists $M \ge 0$ such that $|q(\xi)| \le M\norm{\xi}_H^2$ for all $\xi \in H$. + \end{enumerate} +\end{definition} + +\begin{proposition} +\label{proposition:quadratic-function-sesquilinear} + Let $H$ be a complex Hilbert space, $q: H \to \complex$ be a bounded \hyperref[quadratic function]{definition:hilbert-quadratic-function}, and + \[ + H \times H \to \complex \quad (\xi, \eta) \mapsto \dpn{\xi, \eta}{q} := \frac{1}{4}\sum_{k = 0}^3i^kq(\xi + i^k\eta) + \] + + then + + + \begin{enumerate} + \item $\dpn{\cdot, \cdot}{q}$ is the unique bounded sesquilinear form on $H$ such that $\dpn{\xi, \xi}{q} = q(\xi)$ for all $\xi \in H$. + \item There exists a unique $T \in B(H)$ such that $q(\xi) = \dpn{T\xi, \xi}{H}$ for all $\xi \in H$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): Uniqueness is given by \hyperref[polarisation]{proposition:polarisation-complex}. Existence requires a large number of checkpoints, which are detailed below. Let $M \ge 0$ such that $|q(\xi)| \le M\norm{\xi}_H^2$ for all $\xi \in H$. + \begin{enumerate}[label=(1.\roman*)] + \item For each $\xi, \eta, \theta \in H$, + \[ + \dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q} = 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q} + \] + + and + \[ + \dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2 \angles{\xi, \frac{\eta + \theta}{2}}_{q} + \] + \item For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in [0, 1]$, + \[ + \dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q} + \] + + \item For each $\xi, \eta, \theta \in H$, + \[ + \dpn{\xi + \theta, \eta}{H} = \dpn{\xi, \eta}{H} + \dpn{\theta, \eta}{H} + \] + + and + \[ + \dpn{\xi, \eta + \theta}{H} = \dpn{\xi, \eta}{H} + \dpn{\xi, \theta}{H} + \] + + \item For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in \mathbb{D}$, + \[ + \dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q} + \] + + \item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le M(\norm{\xi}_H^2 +\norm{\eta}_H^2)$. + \item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le 3M \norm{\xi}_H\norm{\eta}_H$. + + \item $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form on $H$. + \item For all $\xi \in H$, $\dpn{\xi, \xi}{q} = q(\xi)$. + \end{enumerate} + + + (1.i): Let $\xi, \eta, \theta \in H$, then + \[ + 4\dpn{\xi, \eta}{q} + 4 \dpn{\theta, \eta}{q} = \sum_{k = 0}^3 i^k\braks{q(\xi + i^k\eta) + q(\theta + i^k\eta)} + \] + + For each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $((\xi + \theta)/2 + i^k\eta, (\xi - \eta)/2)$, + \[ + q(\xi + i^k\eta) + q(\theta + i^k\eta) = 2\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}} + \] + + so + \begin{align*} + 4\dpn{\xi, \eta}{q} + 4 \dpn{\theta, \eta}{q} &= \sum_{k = 0}^3 i^k\braks{q(\xi + i^k\eta) + q(\theta + i^k\eta)} \\ + &= 2\sum_{k = 0}^3 i^k\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}} \\ + &= 2\sum_{k = 0}^3 i^k{q\paren{\frac{\xi + \theta}{2} + i^k\eta}} = 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q} + \end{align*} + + Similarly, for each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $(\xi + i^k(\eta + \theta)/2, i^k(\eta - \theta)/2)$, + \[ + q(\xi + i^k\eta) + q(\xi + i^k\theta) = 2\braks{q\paren{\xi + i^k\frac{\eta + \theta}{2}} + q\paren{i^k\frac{\eta - \theta}{2}}} + \] + + and + \[ + \dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2\angles{\xi, \frac{\eta + \theta}{2}}_q + \] + + + so $\dpn{\cdot, \cdot}{q}: H \times H \to \complex$ is a $\real$-bilinear form. + + (1.ii): By definition of $\dpn{\cdot, \cdot}{q}$, the claim holds for $0$ and $1$. Let $n \in \natz$ and suppose inductively that the claim holds for all $\lambda \in \mathbb{D}_n \cap [0, 1]$. Take $\lambda \in \mathbb{D}_{n+1} \cap [0, 1]$, then there exists $\mu, \nu \in \mathbb{D}_n \cap [0, 1]$ such that $\lambda = (\mu + \nu)/2$. In which case, by (1.i), + \begin{align*} + \dpn{\lambda \xi, \eta}{q} &= \angles{\frac{\mu + \nu}{2}\xi, \eta}_{q} = \frac{1}{2}\dpn{\mu\xi, \eta}{q} + \frac{1}{2}\dpn{\nu\xi, \eta}{q} \\ + &= \frac{\mu + \nu}{2}\dpn{\xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q} + \end{align*} + + By induction, $\dpn{\lambda \xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$. + + Similarly, $\dpn{\xi, \lambda \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$. + + (1.iii): Let $\xi, \eta, \theta \in H$, then by (1.i) and (1.ii), + \[ + \frac{1}{2}(\dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q}) = \angles{\frac{\xi + \theta}{2}, \eta}_q = \frac{1}{2}\dpn{\xi + \theta, \eta}{q} + \] + + (1.iv): By (1.ii) and repeated application of (1.iii), + \[ + \dpn{\lambda \xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q} + \] + + for all $\lambda \in \mathbb{D}$. + + (1.v): Let $\xi, \eta \in H$, then + \begin{align*} + |\dpn{\xi, \eta}{q}| &\le \frac{1}{4}\sum_{k = 0}^3 |q(\xi + i^k\eta)| = \frac{M}{4}\sum_{k = 0}^3\normn{\xi + i^k\eta}_H \\ + &\le \frac{M}{4}\sum_{k = 0}^3 [\norm{\xi}_H^2 + \norm{\eta}_H^2 + i^{-k}\dpn{\xi, \eta}{H}] = M(\norm{\xi}_H^2 + \norm{\eta}_H^2) + \end{align*} + + + + (1.vi): Let $\xi, \eta \in H$ and assume without loss of generality that $\xi, \eta \ne 0$. Since $\norm{\eta}_H/\norm{\xi}_H > 0$, there exists $n \in \integer$ such that $\norm{\eta}_H/\norm{\xi}_H \in [2^n, 2^{n+1}]$. In which case, by (1.iv), + \begin{align*} + |\dpn{\xi, \eta}{q}| &= 2^{-n}|\dpn{2^n\xi, \eta}{q}| \le 2^{-n}M(4^n \norm{\xi}_H^2 + \norm{\eta}_H^2) = 2^nM\norm{\xi}_H^2 + 2^{-n}M\norm{\eta}_H^2 \\ + &\le M\norm{\xi}_H\norm{\eta}_H + 2M\norm{\xi}_H\norm{\eta}_H = 3M\norm{\xi}_H\norm{\eta}_H + \end{align*} + + (1.vii): By (1.iv) and (1.vi), $\dpn{\cdot, \cdot}{q}$ is a $\real$-bilinear form on $H$. For any $\xi, \eta \in H$, by absolute homogeneity of $q$, + \begin{align*} + \dpn{i\xi, \eta}{q} &= \frac{1}{4}\sum_{k = 0}^3 i^k q(i\xi + i^k\eta) = \frac{1}{4}\sum_{k = 0}^3 i^k q(\xi + i^{k-1}\eta) \\ + &= \frac{i}{4}\sum_{k = 0}^3 i^{k-1} q(\xi + i^{k-1}\eta) = \dpn{\xi, \eta}{q} \\ + \dpn{\xi, i\eta}{q} &= \frac{1}{4}\sum_{k = 0}^3 i^k q(\xi + i^{k+1}\eta) = \frac{1}{4}\sum_{k = 0}^3 i^k q(i^{-1}\xi + i^{k}\eta) \\ + &= \dpn{-i\xi, \eta}{q} = -i \dpn{\xi, \eta}{q} + \end{align*} + + Therefore $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form. + + (1.viii): Let $\xi \in H$, then by absolute homogeneity of $q$, + \begin{align*} + \dpn{\xi, \xi}{q} &= \frac{1}{4}\braks{q(\xi + \xi) - q(\xi - \xi) + iq(\xi + i\xi) - iq(\xi - i\xi)} \\ + &= \frac{1}{4}\braks{4q(\xi) + iq(\xi + i\xi) - iq(i\xi + \xi)} = q(\xi) + \end{align*} + + (2): By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}. +\end{proof}