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I have unit balls defined by the $1$, $2$ and $\infty$ norm in $\mathbb{R}^2$. I want to find the orthogonal projection of a vector $(x,y)$ onto the balls.

How could it be done? I only know how to project vectors orthogonally but have no idea how to do it over the balls.

Thanks a lot.

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    $\begingroup$ I'm not sure what an "orthogonal projection over a ball" means. You usually project over a vector space (geometrically, a line, a plane, etc). Perhaps we mean a non-orthogonal projection, just the scaled vector that lies on the ball. $\endgroup$
    – leonbloy
    May 5, 2012 at 23:10
  • $\begingroup$ @leonbloy, I mean the orthogonal projection, if you look at the 2-norm in $\mathbb{R}^2$, it is a circle. and for $\infty$-norm it is a square... I hope it clarifies. $\endgroup$
    – Shan
    May 6, 2012 at 10:07
  • $\begingroup$ not really for me $\endgroup$
    – leonbloy
    May 6, 2012 at 12:34
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    $\begingroup$ I think Shan calls "orthogonal projection" the point on the sphere which is closest to the given vector. $\endgroup$
    – Siminore
    May 18, 2012 at 9:08
  • $\begingroup$ @Siminore, Yup this is correct! $\endgroup$
    – Shan
    May 21, 2012 at 9:53

1 Answer 1

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Recall that the orthogonal projection of a vector $u$ on a vector subspace $V$ is defined as the unique vector $v$ such that (1) $v$ is in $V$, and (2) $u-v$ is orthogonal to $V$, that is, $\langle u,w\rangle=\langle v,w\rangle$ for every $w$ in $V$.

What could be the orthogonal projection of $u$ on a sphere $S$? (In the question you write ball but in the comments it seems clear you mean the unit sphere for a given norm.) Following the classical definition, one should look for $s$ such that (1) $s$ is in $S$ and (2) $u-s$ is orthogonal to... what exactly? Orthogonal to $S$? Alas, no vector $r$ is orthogonal to $S$ in the sense that $\langle r,t\rangle=0$ for every $t$ in $S$, except the null vector. (For example, $t=\langle r,r\rangle^{-1/2}r$ is in $S$ and $\langle r,t\rangle=\langle r,r\rangle^{1/2}\ne0$ for every $r\ne0$.)

This remark shows that you really need to clarify the question.

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