This question is based on section 2.6.2 of the textbook Convex Optimization by Boyd. The specific mathematical statement I am referring to is the following:
$$ x \prec_{K} y \iff \lambda^{T}x < \lambda^{T}y \quad \forall \quad \lambda \succeq_{K^{*}} 0, \lambda \neq 0 $$
To get a feel for this inequality, I decided to try to demonstrate it geometrically in 2D using a cone $K$ such that the dual cone $K^{*} = R_{+}^{2}$. This should mean the cone K is a 45 deg cone centered about the y = x ray as shown in the image below.
Example of point X and Y chosen such that Y-X is not inside cone K
The image also shows an example pair of points X and Y I chose such that $(Y-X) = w \nsucc_{K} 0 $ (because $w \notin K$). As per the above mathematical expression then, I should be able to find at least one $\lambda \in K^{*} = R_{+}^{2} $ such that $\lambda^{T}x \gt \lambda^{T}y$. In other words some vector with only positive elements whose inner-product with $u$ would be greater than its inner product with $v$ but I don't see such a vector intuitively.
Where am I going wrong? I am new to this topic (Convex Sets) and I hope I got all my notation above correct.
I found my error. Just needed to take a break and return to the problem. The dual cone of the cone $K$ I used in my example is in fact not equal to $R_{+}^{2}$. What finally made me realize this is that the dual cone of $R_{+}^{2}$ is itself. But I know that calculating the dual cone twice is supposed to return you the original cone.
I was making a silly error. The dual cone of the 45 deg cone centered about the $x = y$ ray for $x > 0$ is actually the 135 deg cone centered about the same ray.
Here is the corrected drawing. Any vector along the green vector will clearly result in $\lambda^{T}x \gt \lambda^{T}y$.
Corrected Example