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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$\begingroup$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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