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I have come across a notion of orthogonality of two vectors in a normed space not necessarily inner product space. Two vectors $x$ and $y$ in a normed space are said to be orthogonal (represented $x\perp y$) if $||x||\leq ||x+\alpha y||,$ for every $\alpha,$ a scalar.

1) What is the rational behind the definition above? I guess, it has got something to do with minimum overlap between $x$ and $y$.

2) Is this unique generalization of the concept of orthogonality from inner product spaces?

Thank you.

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  • $\begingroup$ Here's another paper (seems not to be referenced in the ones mentioned in Valerio's answer): projecteuclid.org/… $\endgroup$
    – Ralph
    Commented Mar 4, 2012 at 22:31

4 Answers 4

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Concerning question 1: The rational is that in an inner product space $$x\perp y \Leftrightarrow \forall \alpha \in K: ||x||\leq ||x+\alpha y|| \qquad(K = \mathbb{R} \text{ or } K = \mathbb{C})$$ Now, if no inner product is available (but a norm), the idea is, to just take the right hand side as definition of orthogonality (call it $\perp_1$).

Concerning question 2: No, there are other -non-equivalent - generalizations as well. As an example, note that in an inner product space over the reals $$\langle x,y \rangle = \frac{1}{4}( ||x+y||^2 - ||x-y||^2).$$ Hence $x\perp y \Leftrightarrow ||x+y|| = ||x-y||$. So the definition $$ x\perp_{\scriptstyle 2}\; y : \Leftrightarrow ||x+y|| = ||x-y||$$ generalizes the orthogonality from an inner product space to any normed space (over the reals).

Now let's show that $\perp_1, \perp_2$ aren't equivalent. Let $E = \mathbb{R}^2$ with norm $||(a,b)|| = \max(|a|, |b|)$. Then

$\qquad (0,1) \perp_2 (2,1)$ but not $(0,1) \perp_1 (2,1)\quad$ (take $t=-1/4$)

$\qquad (1,1) \perp_1 (2,0)$ but not $(1,1) \perp_2 (2,0).$

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  • $\begingroup$ Follow up questions: (i) Are these two notions of orthogonality symmetric? $\perp_2$ certainly is, but what about $\perp_1$? (ii) Are these notions invariant under scaling? For example, if $x \perp_1 y$, then $x \perp_1 \beta y$ for any $\beta \in K$. Most importantly, (iii): In which contexts are these two notions of orthogonality useful? $\endgroup$
    – shuhalo
    Commented Mar 4, 2012 at 1:43
  • $\begingroup$ Another follow up question I have is: Since $\perp_{1}$ and $\perp_{2}$ are equivalent in inner-product spaces, in what way normed spaces are deficient so that these orthogonality concepts don't agree? $\endgroup$
    – Uday
    Commented Mar 4, 2012 at 2:07
  • $\begingroup$ Another question: Are there any more non-equivalent orthogonality definitions? $\endgroup$
    – Uday
    Commented Mar 4, 2012 at 2:16
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    $\begingroup$ @Martin: (i) $\perp_1$ is not symmetric: From above $(1,1) \perp_1 (2,0)$, but not $(2,0) \perp_1 (1,1)$ (take $t = -1$). (ii) $\perp_1$ is invariant under scalar multiplication, as is aparant from the definition. $\perp_2$ ist not: Again from above, $(0,1) \perp_2 (2,1)$ but not $3 \cdot (0,1) \perp_2 (2,1)$. $\endgroup$
    – Ralph
    Commented Mar 4, 2012 at 10:00
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    $\begingroup$ @Uday: Yes, there are more notions of orthogonality in normed spaces - see the survey article quoted by Valerio. In principle you can take any norm-expression that is equivalent to orthogonality in an inner product space and use it as definition of orthogonality in a normed space. $\endgroup$
    – Ralph
    Commented Mar 4, 2012 at 10:04
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The definition you gave is called Birkhoff-James orthogonality and the intuition is the following: suppose you have $x,y\in\mathbb R^2$ and construct a triangle with sides $x$ and $y$. Now let $x$ be fixed and consider the same triangle with $-\alpha y$ instead of $y$. Observe that $||x+\alpha y||$ is the length of the third side of this triangle. If you try to write down a picture, you figure out in a moment that the condition $||x||\leq||x+\alpha y||$ can be true for all $\alpha$ iff $x$ and $y$ are orthogonal (looking at the picture, if they are not orthogonal and the inequality is true for some $\alpha$, then it is false for $-\alpha$). Birkhoff-James' orthogonality is a tentative to capture orthogonality through this geometric property.

Birkhoff-James' orthogonality is not the unique notion of orthogonality for normed space.

Some references:

In the following paper

http://arxiv.org/pdf/0907.1813.pdf

you can find some recent very easy application of BJ's orthogonality, as well, if you go through the bibliography, some references about other notions of orthogonality are given. In particular I suggest the paper of Diminnie

Diminnie, C.R. A new orthogonality relation for normed linear spaces, Math. Nachr. 114 (1983), 192-203

and the survey by Alonso and Benitez http://dmle.cindoc.csic.es/pdf/EXTRACTAMATHEMATICAE_1989_04_03_03.pdf

P.s. Bikhoff-James orthogonality is not symmetric in general. Some interesting remarks about symmetric orthogonalities can be found in the paper(s) by Partington in the bibliography of the arxiv paper cited above.

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Concerning your follow-up question (iii) there is the following very nice result: For Birkhoff-James orthogonality it is easy to find examples where $y\perp x$ but $\left\|x\right\|/\left\|x+\alpha y\right\| > 1$ for some real $\alpha$, and so natural to investigate the largest such value $\left\|x\right\|/\left\|x+\alpha y\right\|$ over $X$. In "R. L. Thele, Some results on the radial projection in Banach spaces. Proc. Amer. Math. Soc., 42(2):484--486", it is it is shown that this quantity is exactly the Lipshitz constant for the radial projection onto the unit ball in this norm.

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Well, it depends what do you need it for. You may also have a look at semi-inner-product spaces, which are natural generalizations of inner product spaces.

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