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Let $K$ be a proper cone in $\mathbb{R}^{n}$, which is a cone that is convex, closed, pointed, and has a nonempty interior. If $x \in \operatorname{int}(K)$, and $y \in K$, must $x+y \in \operatorname{int}(K)$?Or similarly, if $x \in \operatorname{relint}(K)$, $y \in K$, does $x+y \in \operatorname{relint}(K)$?

Note here it's clear that convex is necessary for both cases, and pointed is necessary for the interior case. However, I'm not sure if it's necessary with the relative interior case. Also, I think the closure of the convex set has the same boundary points as the original set, which might be useful.

The reason I wonder about this is because of generalized inequalities, where I wonder if $x \leq_{K} y <_{K} z$ implies $x <_{K} z$, which reduces this problem.

Very intuitively speaking, I think this is true as once x leads the point to the interior, we can shift the entire cone so that it starts at x. And afterward, if we wanted to go the boundary of the original cone, I felt we need to choose a direction that is not in the cone. However, this is just my vague intuition, and it hasn't led me anywhere.

patchouli
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1 Answers1

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Note that you can write $$ x + y = \frac{1}{2} \left(2x + 2y\right). $$ This is a convex combination of $2x \in \mathrm{int}(K)$ and $2y \in K$. Therefore by the accessibility Lemma of convex analysis, the point $x + y \in \mathrm{int}(K)$.

The same is true if you replace interior with relative interior.

VHarisop
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