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I want to know "if the metric projection onto one dimensional (closed) subspace in $L_p (p\neq 2)$ is linear? I think it is not linear, but I can not give a strict proof.

Thanks for any answer!

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Welcome to MSE! How did you come to the conclusion? It helps to add those details as that might be as important to answering the question. Regards – Amzoti Apr 3 '13 at 15:15
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No. Consider space $\mathbb{R}_1^2$ and its one dimensional subspace $X=\{(t,t): t\in \mathbb{R}\}$. Then for all $a\in \mathbb{R}$ the distance from $(a,0)$ to $X$ is given by $$ d((0,a), X) =\min_{x\in X} d((a,0), x) =\min_{t\in\mathbb{R}}d((a,0),(t,t)) =\min_{t\in\mathbb{R}}(|a-t|+|t|) $$ You can see that this function is minimized when $t\in[0,a]$. Hence solution of this minimization problem is not unique. So you can not even define metric projection, not to mention its linearity.

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Thanks for your answer, I do hope there are some examples in $L_p (p\neq 2)$. – user70795 Apr 4 '13 at 8:17
By strict convexity there is a unique point that minimizes the distance for $1 \lt p \lt \infty$, so you can define a nearest point projection. The issue is its linearity if $p \neq 2$. – Martin Apr 6 '13 at 21:31
@Martin fixed $\phantom{}$ – Norbert Apr 6 '13 at 22:09
Well, that was a cheap fix :-) I don't think it actually answers the question since it leaves the more interesting cases $p \neq 1,2, \infty$ open. However, since OP seems satisfied, I won't insist... – Martin Apr 7 '13 at 7:14
@Martin I thought a little, it is linear. If you want I can write the proof – Norbert Apr 7 '13 at 8:15

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