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Given that cones $K_1 = \{ (x_1, x_2) \in \mathbb{R}^2 \mid |x_1| \leq x_2\}$ and $K_2 = \{ (x_1, x_2) \in \mathbb{R}^2 \mid x_1 + x_2 = 0 \}$..

I think that $K$ will be like the below picture:

enter image description here

What will be the $K^*$? Is it supposed to be an empty set, since the union of the hyperplanes drawn from the two boundary rays of $K$ do not have any common element? Any explanation will be helpful. Thanks.

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    Cones are normally specified by a set of generating vectors, which form the edges of the cone. In $\mathbb{R}^2$ we always need exactly two such vectors. Your cone $K_1$ can be specified as $\{a\vec{v}_1+b\vec{v}_2\,|a,b\in \mathbb{R}_+\}$ with $\vec{v}_1=(-1,1)$ and $\vec{v}_2=(1,1)$, i.e. all possible linear combinations of $(-1,1)$ and $(1,1)$ with positive real coefficients. Your description of $K_2$ does not describe a cone but rather a line $y=-x$. I am not sure what to make of the diagrams, especially as they have also been edited by others since you posted your question. – 2017-02-14
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    The dual cone is made up of the set of all vectors $\vec{w}$ that have scalar product $\vec{w}.\vec{v}$ greater than or equal to zero with every vector $\vec{v}$ in the original cone. In the case of the cone $K_1$, the dual cone is actually the same as the original cone. As we make $K_1$ narrower, the dual cone gets wider, and vice versa. – 2017-02-14
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    @diracula, i have updated the picture? my understanding is that $K^*$ for $K_2$ will be a straight line $x_1 - x_2 =0$. Is that right? – 2017-02-16
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    I believe that is correct, given the definition of the dual cone in terms of scalar products. – 2017-02-16

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