Sesquilinear Forms: Parallelogram

Question

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

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

Suppose one has: $$q[\varphi+\psi]+q[\varphi-\psi]=2q[\varphi]+2q[\psi]$$

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 it is sesquilinear: $$s(\varphi+\varphi',\psi)=s(\varphi,\psi)+s(\varphi',\psi)\quad s(\kappa\varphi,\psi)=\kappa s(\varphi,\psi)$$ $$s(\varphi,\psi+\psi')=s(\varphi,\psi)+s(\varphi,\psi')\quad s(\varphi,\lambda \psi)=\lambda s(\varphi,\psi)$$

How can I check this?

Answer 1

Claim: There exists a function $f : \mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$, but which is not linear.

Proof of claim: To prove the existence of $f$, define an equivalence relation $\sim$ on $\mathbb{R}$ by $$ x \sim y \mbox{ iff } y-x \in\mathbb{Q}. $$ This defines equivalence classes $[x] = \{ y \in \mathbb{R} : y \sim x \}$. As with all equivalence relations, either $[x]\cap[y] = \emptyset$ or $[x]=[y]$. The collection of equivalence classes becomes a vector space over the field of rational numbers $\mathbb{Q}$ under the operations $$ [x]+[y]=[x+y],\;\;\; \alpha[x]=[\alpha x],\;\;\;\alpha\in\mathbb{Q},\;x,y\in\mathbb{R}. $$ These linear operations are well-defined because $x\sim x'$, $y\sim y'$ and $\alpha\in\mathbb{Q}$ give $$ x+y \sim x'+y',\;\;\; \alpha x \sim \alpha x'. $$ So the space $X$ consisting of all equivalence classes $[x]$ is a linear space over the field of rational numbers. Hence, $X$ has a Hamel basis $\{ [x_{\alpha}]\}_{\alpha\in\Lambda}$. Now, for any choice function $\{s_{\alpha}\}_{\alpha\in S}\subseteq \mathbb{R}$ from $\Lambda$ to $\mathbb{R}$, there is a unique function $F : X\rightarrow\mathbb{R}$ that is linear over $\mathbb{Q}$ and satisfies: $$ F([x_{\alpha}]) = s_{\alpha}. $$ The function $f(x) = F([x])$ is additive and is linear over $\mathbb{Q}$, but you can choose $\{ s_{\alpha} \}$ so that $f$ is not linear. $\;\;\Box$

Quadratic Form from an Additive Function: For any additive function $f : \mathbb{R}\rightarrow\mathbb{R}$, the following defines a quadratic form $$ q(x) = (f(x))^{2}. $$ This is easily checked: \begin{align} q(x+y)+q(x-y) & = (f(x)+f(y))^{2}+(f(x)-f(y))^{2}\\ & = 2f(x)^{2}+2f(y)^{2} = 2q(x)+2q(y). \end{align} An additive function on $\mathbb{R}$ is always linear over the rationals $\mathbb{Q}$. But without some kind of measurability or continuity, the function $f$ may not be linear over $\mathbb{R}$, which is equivalent to continuity over $\mathbb{R}$.

Theorem: An additive function $f : \mathbb{R}\rightarrow \mathbb{R}$ is linear over $\mathbb{R}$ iff it is Lebesgue measurable. Equivalently, $f$ is continuous.

What makes a norm different?: There is a classic theorem which states that a norm is generated by an inner product iff the norm satisfies the parallelogram law. The proof generally starts by showing that, if $\|\cdot\|$ satisfies the Parallelogram Law, the function $$ \rho(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2} $$ is additive in both coordinates $x$, $y$. Then it follows that $\rho(q x,y)=\rho(x,qy) = q\rho(x,y)$ for all $q \in \mathbb{Q}$. If you're dealing with a complex space, then $\rho(ix,y)=i\rho(x,y)$ and $\rho(x,iy)=-i\rho(x,y)$ follow from $\|ix\|=\|x\|$. What makes the norm different is that $\rho(x,x) \ge 0$; hence, the Cauchy-Schwarz inequality holds: $$ |\rho(x,y)|^{2} \le \rho(x,x)\rho(y,y) = \|x\|^{2}\|y\|^{2} $$ From this you get continuity of $\rho(x,y)$ with respect to the norm. So, for real $r$ and rational $\{ q_{n} \}$ converging to $r$, $$ \begin{align} |\rho(rx,y)-q_{n}\rho(x,y)| & \le \|rx-q_{n}x\|\|y\| \\ & =|r-q_{n}|\|x\|\|y\|\rightarrow 0. \end{align} $$ And that's enough to guarantee that $\rho(rx,y)=r\rho(x,ry)$ for all $r\in\mathbb{R}$.

The result below is useful for proving that a quadratic form arises from a sesquilinear form.

Theorem: [Bounded Quadratic Form Representation] Let $Q$ be a quadratic form on a complex linear space $X$. If there exists a seminorm $|\cdot|$ on $X$ such that $|Q(x)| \le |x|^{2}$, then $Q(x)= S(x,x)$ for a unique sesquilinear form $S(\cdot,\cdot)$ on $X\times X$.

Answer 2

As positive result...

Additivity

Introduce the forms: $$s_\Re(\varphi,\psi):=\frac{1}{4}\{q[\varphi+\psi]-q[\varphi-\psi]\}$$ $$s_\Im(\varphi,\psi):=\frac{1}{4}\{q[\varphi+i\psi]-q[\varphi-i\psi]\}$$

They reconstruct it: $$s(\varphi,\psi)=s_\Re(\varphi,\psi)+is_\Im(\varphi,\psi)$$

Note the relations: $$s_\Re(\psi,\varphi)=s_\Re(\varphi,\psi)\quad s_\Im(\varphi,\psi)=s_\Re(\varphi,i\psi)$$

By parallelogram law:*

$$s_\Re(\varphi+\varphi'+\psi)=s_\Re(\varphi,\psi)+s_\Re(\varphi',\psi)$$

That gives additivity.

*See the check: Sesquilinearity

Homogenity

Suppose Cauchy-Schwarz holds: $$|s(\varphi,\psi)|^2\leq q[\varphi]q[\psi]$$

By construction one has: $$s(i\varphi,\psi)=is(\varphi,\psi)\quad s(\varphi,i\psi)=-is(\varphi,\psi)$$

From additivity it follows: $$qs\left(\tfrac{p}{q}\varphi,\psi\right)=s\left(q\tfrac{p}{q}\varphi,\psi\right)=ps\left(\tfrac{q}{q}\varphi,\psi\right)$$ $$qs\left(\varphi,\tfrac{p}{q}\psi\right)=s\left(\varphi,q\tfrac{p}{q}\psi\right)=ps\left(\varphi,\tfrac{q}{q}\psi\right)$$

By continuity one obtains: $$0\leq|rs(\varphi,\psi)-s(r\varphi,\psi)|\leq2|r-r_n|\sqrt{q[\varphi]q[\psi]}\to0$$ $$0\leq|rs(\varphi,\psi)-s(\varphi,r\psi)|\leq2|r-r_n|\sqrt{q[\varphi]q[\psi]}\to0$$

That gives homogenity.

Caution

For real forms it is: $$q\subseteq\mathbb{R}:\quad\Re s(\varphi,\psi)=s_\Re(\varphi,\psi)\quad\Im s(\varphi,\psi)=s_\Im(\varphi,\psi)$$

Else, notation is symbolic!