First draft of quadratic forms.
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Bokuan Li
@@ -369,4 +369,160 @@ A significant property of Hilbert spaces is that every closed subspace is comple
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\end{proof}
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\begin{definition}[Quadratic Function]
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\label{definition:hilbert-quadratic-function}
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Let $H$ be a complex Hilbert space and $q: H \to \complex$, then $q$ is \textbf{quadratic} if:
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\begin{enumerate}
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\item For each $\lambda \in \complex$ and $\xi \in H$, $q(\lambda x) = |\lambda|^2 q(x)$.
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\item (\textbf{Parallelogram Law}) For each $\xi, \eta \in H$,
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\[
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q(\xi + \eta) + q(\xi - \eta) = 2q(\xi) + 2q(\eta)
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\]
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\end{enumerate}
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and \textbf{bounded} if
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\begin{enumerate}[start=2]
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\item There exists $M \ge 0$ such that $|q(\xi)| \le M\norm{\xi}_H^2$ for all $\xi \in H$.
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\end{enumerate}
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\end{definition}
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\begin{proposition}
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\label{proposition:quadratic-function-sesquilinear}
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Let $H$ be a complex Hilbert space, $q: H \to \complex$ be a bounded \hyperref[quadratic function]{definition:hilbert-quadratic-function}, and
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\[
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H \times H \to \complex \quad (\xi, \eta) \mapsto \dpn{\xi, \eta}{q} := \frac{1}{4}\sum_{k = 0}^3i^kq(\xi + i^k\eta)
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\]
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then
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\begin{enumerate}
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\item $\dpn{\cdot, \cdot}{q}$ is the unique bounded sesquilinear form on $H$ such that $\dpn{\xi, \xi}{q} = q(\xi)$ for all $\xi \in H$.
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\item There exists a unique $T \in B(H)$ such that $q(\xi) = \dpn{T\xi, \xi}{H}$ for all $\xi \in H$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(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 $\xi \in H$.
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\begin{enumerate}[label=(1.\roman*)]
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\item For each $\xi, \eta, \theta \in H$,
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\[
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\dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q} = 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q}
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\]
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and
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\[
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\dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2 \angles{\xi, \frac{\eta + \theta}{2}}_{q}
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\]
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\item For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in [0, 1]$,
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\[
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\dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
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\]
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\item For each $\xi, \eta, \theta \in H$,
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\[
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\dpn{\xi + \theta, \eta}{H} = \dpn{\xi, \eta}{H} + \dpn{\theta, \eta}{H}
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\]
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and
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\[
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\dpn{\xi, \eta + \theta}{H} = \dpn{\xi, \eta}{H} + \dpn{\xi, \theta}{H}
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\]
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\item For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in \mathbb{D}$,
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\[
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\dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
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\]
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\item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le M(\norm{\xi}_H^2 +\norm{\eta}_H^2)$.
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\item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le 3M \norm{\xi}_H\norm{\eta}_H$.
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\item $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form on $H$.
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\item For all $\xi \in H$, $\dpn{\xi, \xi}{q} = q(\xi)$.
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\end{enumerate}
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(1.i): Let $\xi, \eta, \theta \in H$, then
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\[
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4\dpn{\xi, \eta}{q} + 4 \dpn{\theta, \eta}{q} = \sum_{k = 0}^3 i^k\braks{q(\xi + i^k\eta) + q(\theta + i^k\eta)}
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\]
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For each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $((\xi + \theta)/2 + i^k\eta, (\xi - \eta)/2)$,
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\[
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q(\xi + i^k\eta) + q(\theta + i^k\eta) = 2\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}}
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\]
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so
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\begin{align*}
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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)} \\
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&= 2\sum_{k = 0}^3 i^k\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}} \\
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&= 2\sum_{k = 0}^3 i^k{q\paren{\frac{\xi + \theta}{2} + i^k\eta}} = 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q}
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\end{align*}
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Similarly, for each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $(\xi + i^k(\eta + \theta)/2, i^k(\eta - \theta)/2)$,
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\[
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q(\xi + i^k\eta) + q(\xi + i^k\theta) = 2\braks{q\paren{\xi + i^k\frac{\eta + \theta}{2}} + q\paren{i^k\frac{\eta - \theta}{2}}}
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\]
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and
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\[
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\dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2\angles{\xi, \frac{\eta + \theta}{2}}_q
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\]
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so $\dpn{\cdot, \cdot}{q}: H \times H \to \complex$ is a $\real$-bilinear form.
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(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),
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\begin{align*}
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\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} \\
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&= \frac{\mu + \nu}{2}\dpn{\xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q}
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\end{align*}
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By induction, $\dpn{\lambda \xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$.
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Similarly, $\dpn{\xi, \lambda \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$.
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(1.iii): Let $\xi, \eta, \theta \in H$, then by (1.i) and (1.ii),
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\[
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\frac{1}{2}(\dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q}) = \angles{\frac{\xi + \theta}{2}, \eta}_q = \frac{1}{2}\dpn{\xi + \theta, \eta}{q}
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\]
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(1.iv): By (1.ii) and repeated application of (1.iii),
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\[
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\dpn{\lambda \xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
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\]
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for all $\lambda \in \mathbb{D}$.
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(1.v): Let $\xi, \eta \in H$, then
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\begin{align*}
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|\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 \\
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&\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)
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\end{align*}
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(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),
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\begin{align*}
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|\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 \\
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&\le M\norm{\xi}_H\norm{\eta}_H + 2M\norm{\xi}_H\norm{\eta}_H = 3M\norm{\xi}_H\norm{\eta}_H
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\end{align*}
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(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 $q$,
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\begin{align*}
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\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) \\
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&= \frac{i}{4}\sum_{k = 0}^3 i^{k-1} q(\xi + i^{k-1}\eta) = \dpn{\xi, \eta}{q} \\
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\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) \\
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&= \dpn{-i\xi, \eta}{q} = -i \dpn{\xi, \eta}{q}
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\end{align*}
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Therefore $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form.
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(1.viii): Let $\xi \in H$, then by absolute homogeneity of $q$,
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\begin{align*}
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\dpn{\xi, \xi}{q} &= \frac{1}{4}\braks{q(\xi + \xi) - q(\xi - \xi) + iq(\xi + i\xi) - iq(\xi - i\xi)} \\
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&= \frac{1}{4}\braks{4q(\xi) + iq(\xi + i\xi) - iq(i\xi + \xi)} = q(\xi)
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\end{align*}
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(2): By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}.
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\end{proof}
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