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Examples of Diminnie orthogonality

  1. Dec 29, 2013 #1
    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. jcsd
  3. Dec 29, 2013 #2
    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:

    [tex]\textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\} = \|x\|\|y\|[/tex]

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

    Very related is the following quantity:

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

    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

    [tex]\psi(a,b) = <a,x><b,y> - <b,x><a,y>[/tex]

    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

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

    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##

    [tex]\|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>[/tex]

    Now, what does this have to do with our quantity

    [tex]\textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\} = \|x\|\|y\|[/tex]

    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

    [tex]
    \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*}
    [/tex]

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

    [tex]<a,x><b,y> - <b,x><a,y> = 1 - <y,x><x,y>[/tex]

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

    [tex]\textrm{sup} \{<a,x><b,y> - <b,x><a,y>~\vert~\|a\|,\|b\|\leq 1\} = 1 - <x,y><y,x>[/tex]

    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.
     
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