I want to prove that the solution space for the $\vec{x}$ of the following equations is convex:
$$1. \quad\vec{x} \cdot \vec{a} \geq b \quad \text{and} \quad 2. \quad \sum_ix_i=N$$
$\vec{x},\vec{a} \in \mathbb{R}^N$, $b \in \mathbb{R}$. The value of $b$ should not be important here, it shifts the hyperplane parallel.
The solutions $\vec{x}$ have to lie on the hyperplane definded by (2). Equation (1) defines a halfspace.
So the solution space is the part of one hyperplane (2) on one side of the other hyperplane (1).
How can i prove, that this unbounded solution space (if the hyperplanes intersect) is convex?
Thanks for helping :)
As Michael points out in the comments, the easiest way would be to use intersections. Half-spaces and hyperplanes are both convex, so the intersection is too.
If you want to first-principles it, that's also not too bad and all you need is linearity of both the dot-product and sum.
If $x = (x_1,\dots, x_n)$ and $y = (y_1, \dots, y_n)$ are two solutions to both (1) and (2), you need to prove that any point of the segment $L_{xy} = \{(1-t)x + (t)y: t\in [0,1]\}$ is itself a solution to both.
(1) $((1-t)x + (t)y)\cdot a = (1-t)x\cdot a + (t)y\cdot a \geq (1-t)b + (t)b \geq b$.
(2) $\sum ((1-t)x + (t)y)_i = \sum (1-t)x_i + \sum (t)y_i = (1-t)\sum x_i + t\sum y_i = (1-t)N + tN = N$