Proposition 14.7.16 | Jerry's Digital Garden

Proposition 14.7.16.label Let $H$ be a complex Hilbert space, $q: H \to \complex$ be a bounded quadratic function, and

\[H \times H \to \complex \quad (\xi, \eta) \mapsto \dpn{\xi, \eta}{q}:= \frac{1}{4}\sum_{k = 0}^{3}i^{k}q(\xi + i^{k}\eta)\]

then

(1)
$\dpn{\cdot, \cdot}{q}$ is the unique bounded sesquilinear form on $H$ such that $\dpn{\xi, \xi}{q}= q(\xi)$ for all $\xi \in H$.
(2)
There exists a unique $T \in B(H)$ such that $q(\xi) = \dpn{T\xi, \xi}{H}$ for all $\xi \in H$.

Proof. (1): Uniqueness is given by polarisation. 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 $\xi \in H$.

(1.i)
For each $\xi, \eta, \theta \in H$,
\[\dpn{\xi, \eta}{q}+ \dpn{\theta, \eta}{q}= 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q}\]

and
\[\dpn{\xi, \eta}{q}+ \dpn{\xi, \theta}{q}= 2 \angles{\xi, \frac{\eta + \theta}{2}}_{q}\]
(1.ii)
For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in [0, 1]$,
\[\dpn{\lambda\xi, \eta}{q}= \lambda\dpn{\xi, \eta}{q}= \dpn{\xi, \lambda \eta}{q}\]
(1.iii)
For each $\xi, \eta, \theta \in H$,
\[\dpn{\xi + \theta, \eta}{q}= \dpn{\xi, \eta}{q}+ \dpn{\theta, \eta}{q}\]

and
\[\dpn{\xi, \eta + \theta}{q}= \dpn{\xi, \eta}{q}+ \dpn{\xi, \theta}{q}\]
(1.iv)
For each dyadic rational number $\lambda \in \mathbb{D}$,
\[\dpn{\lambda\xi, \eta}{q}= \lambda\dpn{\xi, \eta}{q}= \dpn{\xi, \lambda \eta}{q}\]
(1.v)
For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le M(\norm{\xi}_{H}^{2} +\norm{\eta}_{H}^{2})$.
(1.vi)
For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le 3M \norm{\xi}_{H}\norm{\eta}_{H}$.
(1.vii)
$\dpn{\cdot, \cdot}{q}$ is a sesquilinear form on $H$.
(1.viii)
For all $\xi \in H$, $\dpn{\xi, \xi}{q}= q(\xi)$.

(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 $((\xi + \theta)/2 + i^{k}\eta, (\xi - \theta)/2)$,

\[q(\xi + i^{k}\eta) + q(\theta + i^{k}\eta) = 2\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}}\]

\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}}= 8 \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

\[\dpn{\xi, \eta}{q}+ \dpn{\xi, \theta}{q}= 2\angles{\xi, \frac{\eta + \theta}{2}}_{q}\]

(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)| \le \frac{M}{4}\sum_{k = 0}^{3}\normn{\xi + i^k\eta}_{H}^{2} \\&\le \frac{M}{4}\sum_{k = 0}^{3} [\norm{\xi}_{H}^{2} + \norm{\eta}_{H}^{2} + 2\text{Re}\dpn{i^{-k}\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^{n}M\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) = i\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 Riesz Representation Theorem.$\square$

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