Assume we have a Hilbert space $X$ with some inner product $\langle \cdot, \cdot \rangle$ defined on it. Let $X_{c}$ be the subset of $X$ which is a convex cone. Next, we define a projection $P$ of $x \in X$ onto the cone $X_{c}$ as $$ y = Px = \underset{z \in X_{c}}{\text{argmin}} \langle x-z, x-z \rangle. $$

The question: is $y$ unique?

  • $\begingroup$ If there is one, it is unique. Drop the "cone", add "closed" and it is verbatim a classical theorem. $\endgroup$
    – user228113
    Aug 29, 2017 at 9:41

1 Answer 1


There is a theorem that says that if $C$ is a closed and convex set in a Hilbert space, then there exists a metric projection $P$ onto $C$, defined by the property that for each $x\in H$ there is a unique $Px=y\in C$ such that $|\!|Px-x|\!|$ minimizes the function $|\!|z-x|\!|$ over $z\in C$. Therefore, if your convex cone is also closed, then the answer to your question is affirmative. However, if the cone is not closed, then there is no guarantee that $Px$ even belongs to the set. For example, take a cone with vertex at the origin, and then remove the origin. The resulting cone is convex, but not closed, and there is no point in it which is closest to the origin. This shows that it does not make much sense to speak about a "projection onto $X_c$" unless you assume that $X_c$ is closed, and then - as indicated above - it is well known that there is a (unique) metric projection onto it. The additional cone structure does not contribute much here.


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