Orthogonality in non-inner product spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:47:12Z http://mathoverflow.net/feeds/question/90129 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90129/orthogonality-in-non-inner-product-spaces Orthogonality in non-inner product spaces Uday 2012-03-03T17:47:13Z 2012-03-04T11:10:12Z <p>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. </p> <p>1) What is the rational behind the definition above? I guess, it has got something to do with minimum overlap between $x$ and $y$. </p> <p>2) Is this unique generalization of the concept of orthogonality from inner product spaces?</p> <p>Thank you. </p> http://mathoverflow.net/questions/90129/orthogonality-in-non-inner-product-spaces/90136#90136 Answer by András Bátkai for Orthogonality in non-inner product spaces András Bátkai 2012-03-03T19:24:58Z 2012-03-03T19:24:58Z <p>Well, it depends what do you need it for. You may also have a look at <a href="http://en.wikipedia.org/wiki/Semi-inner-product" rel="nofollow">semi-inner-product spaces</a>, which are natural generalizations of inner product spaces.</p> http://mathoverflow.net/questions/90129/orthogonality-in-non-inner-product-spaces/90140#90140 Answer by Ralph for Orthogonality in non-inner product spaces Ralph 2012-03-03T19:55:09Z 2012-03-03T19:55:09Z <p>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 $\perp_1$). </p> <p>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).</p> <p>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 </p> <p>$\qquad (0,1) \perp_2 (2,1)$ but <strong>not</strong> $(0,1) \perp_1 (2,1)\quad$ (take $t=-1/4$)</p> <p>$\qquad (1,1) \perp_1 (2,0)$ but <strong>not</strong> $(1,1) \perp_2 (2,0).$ </p> http://mathoverflow.net/questions/90129/orthogonality-in-non-inner-product-spaces/90177#90177 Answer by Valerio Capraro for Orthogonality in non-inner product spaces Valerio Capraro 2012-03-04T03:08:39Z 2012-03-04T10:37:57Z <p>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.</p> <p>Birkhoff-James' orthogonality is not the unique notion of orthogonality for normed space. </p> <p>Some references:</p> <p>In the following paper</p> <p><a href="http://arxiv.org/pdf/0907.1813.pdf" rel="nofollow">http://arxiv.org/pdf/0907.1813.pdf</a></p> <p>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 </p> <p>Diminnie, C.R. A new orthogonality relation for normed linear spaces, Math. Nachr. 114 (1983), 192-203</p> <p>and the survey by Alonso and Benitez <a href="http://dmle.cindoc.csic.es/pdf/EXTRACTAMATHEMATICAE_1989_04_03_03.pdf" rel="nofollow">http://dmle.cindoc.csic.es/pdf/EXTRACTAMATHEMATICAE_1989_04_03_03.pdf</a></p> <p>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.</p> http://mathoverflow.net/questions/90129/orthogonality-in-non-inner-product-spaces/90187#90187 Answer by J.J. Green for Orthogonality in non-inner product spaces J.J. Green 2012-03-04T09:58:44Z 2012-03-04T11:10:12Z <p>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. </p>