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In the definition of a convex cone, given that $x,y$ belong to the convex cone $C$,then $\theta_1x+\theta_2y$ must also belong to $C$, where $\theta_1,\theta_2 > 0$. What I don't understand is why there isn't the additional constraint that $\theta_1+\theta_2=1$ to make sure the line that crosses both $x$ and $y$ is restricted to the segment in between them.

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    $\begingroup$ The cone, by definition, contains rays, i.e. half-lines that extend out to the appropriate infinite extent. Adding the constraint that $\theta_1 + \theta_2 = 1$ would only give you a convex set, it wouldn't allow the extent of the cone. $\endgroup$
    – postmortes
    Sep 15, 2015 at 17:07
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    $\begingroup$ If $\theta_1,\theta_2 \geq 0 $ this means in particular that all $\theta_1,\theta_2$ with $\theta_1+\theta_2 =1$ are also included $\endgroup$
    – asterisk
    Sep 15, 2015 at 17:09
  • $\begingroup$ @postmortes But how will you check for convexity if you don't? Wouldn't the set then have to be affine to satisfy that condition? $\endgroup$ Sep 15, 2015 at 17:11
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    $\begingroup$ @postmortes I mean that if you don't restrict the line to the segment in between them, then the whole line would have to be in the set, so it would not just be convex, it would have to be an affine set to satisfy those conditions, right? I don't know. $\endgroup$ Sep 15, 2015 at 17:15
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    $\begingroup$ @Undertherainbow A good explanation is given here youtube.com/watch?v=QV5qtTq1Tro in minute 16:28 .. hope this helps :) $\endgroup$ Feb 12, 2019 at 11:48

2 Answers 2

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It is sufficient that $\theta_1 x+ \theta_2y \in C, \theta_1,\theta_2\ge 0$ implies $C$ is convex, which includes the case $\theta x+(1-\theta)y,\theta \in [0,1]$.

Conversely, is it necessary? Say that a cone is convex implies $\theta_1 x+ \theta_2y \in C, \theta_1,\theta_2\ge 0$. Convexity means $\theta x+(1-\theta)y \in C$. For a cone, $x\in C$ requires $\lambda x \in C, \lambda \ge 0$. We can then replace $x,y$ with $\lambda x,\lambda \ge 0$ and $\mu y,\mu \ge 0$ that both belong to $C$, like $\theta \lambda x+(1-\theta)\mu y \in C$. Since $\lambda,\mu$ can be any non-negative real number, we can conclude that $\theta_1 x +\theta_2 y \in C, \theta_1, \theta_2 \ge 0$ is necessary, under the convexity condition.

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Given that $x, y$ belong to the convex cone $C$, then $θ_1x+θ_2y$ must also belong to C, where $θ_1,θ_2 \geq 0 $.

Some visualization helped me to understand the equation. As you can see below, a point z within the convex cone can be represented as a linear combination of x and y with non-negative coefficients $θ_1$ and $θ_2$.

To be able to reach all points (i.e. write them as linear combinations of x and y with with non-negative coefficients $θ_1$ and $θ_2$) within the boundaries of the convex cone, we need to vary $θ_1$ and $θ_2$ without the constraint that they sum to 1.

convex cone interior point as a linear combination of x and y

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