# Nonexistence of projection

Following is a doubt I got when I was reading Csiszar's Annals of Statistics paper "Why least squares and maximum entropy: An axiomatic approach to inference for linear inverse problems". He comes up with a set of axioms to justify $\ell_2$-norm minimisation in $\mathbb{R}^n$.

He also argues that $\ell_2$-norm minimization would not be appropriate if the space we consider consists of vectors with positive components,say, $\mathbb{R}_+^n$. Further he says that there exists sets of the form

$$L=\{v \in \mathbb{R}_+^n :Av=b\}$$

where $A$ is some $k\times n$ matrix and $u\notin L$ such that $\min_{v\in L}\|v-u\|_2$ is not attained in $L$.

I can't think of such an example.

Can someone think of such an example?

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This has nothing to do with algebraic geometry. –  Angelo Mar 13 '11 at 5:43
sorry, I have removed that tag now. –  Ashok Mar 13 '11 at 12:14

HI Ashok, I think $\mathbb{R}^n_+$ requires strictly positive components. Then it's easy to come up with the situation above, namely consider $L$ to be the part of the line $x+y = 1$ in $\mathbb{R}_+^2$. And $u = (2,1)$. Then there is no closest point in $L$ to $u$, because it wants to be $(1,0)$, which has nonpositive component. So I guess $n=2$ and $A = (1,1)$ is a $1\times 2$ matrix, and $b=2$.

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Thanks John Jiang. I didn't think that I would get such an easy example. –  Ashok Mar 13 '11 at 12:32
Let $A=(1 -1)$ and $b=(0)$. Then $L=\left\{{a\choose a}\vert a\in \mathbb R_+\right\}$, so for instance for $u={\ \ 1\choose -1}$ $\min_{v\in L}\|v-u\|_2=\sqrt{2}$ is not attained in $L$.
The point of the example is that the projection of $u$ to the linear space genrated by $L$ is is on the closure of $L$ but not in $L$. In other words, the problem is that $L$ is not closed.