# Examples of Diminnie orthogonality

1. Dec 29, 2013

### PonyBarometer

Definition of this orthogonality goes like this:
$x, y \in X$, where $X$ - normed space and $X^*$ - its dual space. Then $x$ is orthogonal $y$, if

$$\sup\{f(x)g(y)-f(y)g(x)|, \, f,g\in X^*, \|f\|,\|g\|≤1\}=\|x\|\|y\|$$

From what I understand $f$ and $g$ are linear functionals from the dual space.
I was wondering if someone could provide some example of Diminnie orthogonality and its usage, because I have difficulty understanding how it works.

2. Dec 29, 2013

### R136a1

Orthogonality is perfectly well-defined in Hilbert spaces. Indeed, we say that $x\bot y$ iff $<x,y> = 0$. The idea of Diminnie orthogonality is to extend the notion of orthogonality to more general Hilbert spaces.

The Riesz representation theorem says that any continuous functional on Hilbert space $f:X\rightarrow \mathbb{C}$ has the form $f(x) = <a,x>$.

So on Hilbert space, we get the following:

$$\textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\} = \|x\|\|y\|$$

if $x$ and $y$ are Diminnie orthogonal.

Very related is the following quantity:

$$\textrm{sup}\{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|^2 + \|b\|^2 \leq 1\}$$

This quantity is in Hilbert spaces somewhat better behaved. Indeed, we can take the Hilbert space $X\times X$ with inner product $<(a,b),(c,d)> = <a,c> + <b,d>$. Then we can look at the following operator

$$\psi(a,b) = <a,x><b,y> - <b,x><a,y>$$

We often use the notation $x\otimes y$ for the operator $(x\otimes y)(a,b) = <x,a><y,b>$, so we have $\psi = x\otimes y - y\otimes x$. The quantity

$$\|x\otimes y - y\otimes x\|= \textrm{sup}\{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|^2 + \|b\|^2 \leq 1\}$$

is the norm of this functional.

Now, the space generated by all the $x\otimes y$ is called the tensor product $X\otimes Y$ and is a Hilbert space under the inner product $<x\otimes y, z\otimes w> = <x,z><y,w>$. The associated norm is denoted as $\|~\|_2$ and we have $\|~\|\leq \|~\|_2$.

In particular, if we have $<x,x> = <y,y>=1$

$$\|x\otimes y - y\otimes x\|^2 \leq \|x\otimes y - y\otimes x\|_2^2 = <x\otimes y - y\otimes x, x\otimes y - y\otimes x> = 2 - 2<x,y><y,x>$$

Now, what does this have to do with our quantity

$$\textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\} = \|x\|\|y\|$$

Well, let's take $\|x\| = \|y\|= 1$ (this is the general case since we can just divide by $\|x\|\|y\|$).

Then if we have $\|a\|,\|b\|\leq 1$, then $\|a\|^2 + \|b\|^2 \leq 2$. Thus we see that

$$\begin{eqnarray*} & & \textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\}\\ & \leq & \textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|^2+\|b\|^2\leq 2\}\\ & = & \frac{1}{2}\textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|^2 + \|b\|^2 \leq 1\}\\ & = & 1 - <x,y><y,x> \end{eqnarray*}$$

In fact, equality holds since we can take $a=x$ and $b=y$ and then

$$<a,x><b,y> - <b,x><a,y> = 1 - <y,x><x,y>$$

Thus we get that for $\|x\|= \|y\| = 1$ that

$$\textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\} = 1 - <x,y><y,x>$$

Thus if $x$ and $y$ are Diminnie orthogonal, then

$1 - <x,y><y,x> = \textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\} =1$

and thus easily follows that $<x,y> = 0$.

Conversely if $<x,y>= 0$, then we see easily that $x$ and $y$ are Diminnie orthogonal.