Consider the compact convex set $X = \{ x \in \mathbb{R}^n \mid x \geq 0, \ \underline{1}^\top x = 1 \}$.

I am wondering if the projection onto $X$ is the composition of the projection on $[0,1]^n$ and the projection onto $\{x \mid \underline{1}^\top x = 1\}$, namely if $$ P_X(\cdot) = P_{ [0,1]^n }\left( P_{ \{x \in \mathbb{R}^n \mid \underline{1}^\top x = 1\} }(\cdot) \right). $$

Comment: the projection onto $X$ is defined as $P_X(x) := \arg \min_{y \in X} \left\| x-y\right\| $.


No, it is not.

$H=\{x\;|\;\underline 1^\top x=1\}$ is a hyperplane, $[0,1]^n$ a hypercube, and $X$ is a simplex which results from intersecting these two. But the problem is that the second projection may well leave the hyperplane and therefore result in a point outside the simplex.

Here is an example for $n=3$: Consider the point $(8,8,-15)$ which already lies in $H$. Its projection onto $X$ would be $(\frac12,\frac12,0)$ which has distance $\frac{15}2\sqrt6\approx18.4$. But the corner of the cube at $(1,1,0)$ has distance $\sqrt{323}\approx18.0$ so it is closer.

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    $\begingroup$ @Adam: Adding such a requirement changes nothing, since you can still find points which project onto the example point. E.g. $P_H(-2, -2, -25)=(8,8,-15)$ so there you are back in the example case I stated, even from an all-negative starting point. $\endgroup$ – MvG Apr 27 '14 at 8:42
  • $\begingroup$ Right, thanks. Is the claim still false if the argument is non-positive? Namely, $P_X(x)=P_{[0,1]^n}\left(P_H(x) \right)$ for all x s.t. $x \leq 0$? $\endgroup$ – user693 Apr 27 '14 at 8:42
  • $\begingroup$ Thanks also for this last comment. Now I am thinking about the projection of non-positive vectors onto $H = \{x \mid 1^\top x = 1\}$ and then onto the simplex $S = \{ x \mid x \geq 0, 1^\top x \leq 1 \}$. Namely I am thinking whether $P_X(x) = P_{ S }\left( P_X(x) \right)$ for all $x$ such that $x \leq 0$. $\endgroup$ – user693 Apr 27 '14 at 19:39

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