- Convex Optimization Tutorial
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- Convex Set
- Affine Set
- Convex Hull
- Caratheodory Theorem
- Weierstrass Theorem
- Closest Point Theorem
- Fundamental Separation Theorem
- Convex Cones
- Polar Cone
- Conic Combination
- Polyhedral Set
- Extreme point of a convex set
- Direction
- Convex & Concave Function
- Jensen's Inequality
- Differentiable Convex Function
- Sufficient & Necessary Conditions for Global Optima
- Quasiconvex & Quasiconcave functions
- Differentiable Quasiconvex Function
- Strictly Quasiconvex Function
- Strongly Quasiconvex Function
- Pseudoconvex Function
- Convex Programming Problem
- Fritz-John Conditions
- Karush-Kuhn-Tucker Optimality Necessary Conditions
- Algorithms for Convex Problems
- Convex Optimization Resources
- Convex Optimization - Quick Guide
- Convex Optimization - Resources
- Convex Optimization - Discussion
Convex Optimization - affine Set
A set $A$ is said to be an affine set if for any two distinct points, the line passing through these points lie in the set $A$.
Note −
$S$ is an affine set if and only if it contains every affine combination of its points.
Empty and singleton sets are both affine and convex set.
For example, solution of a linear equation is an affine set.
Proof
Let S be the solution of a linear equation.
By definition, $S=\left \{ x \in \mathbb{R}^n:Ax=b \right \}$
Let $x_1,x_2 \in S\Rightarrow Ax_1=b$ and $Ax_2=b$
To prove : $A\left [ \theta x_1+\left ( 1-\theta \right )x_2 \right ]=b, \forall \theta \in\left ( 0,1 \right )$
$A\left [ \theta x_1+\left ( 1-\theta \right )x_2 \right ]=\theta Ax_1+\left ( 1-\theta \right )Ax_2=\theta b+\left ( 1-\theta \right )b=b$
Thus S is an affine set.
Theorem
If $C$ is an affine set and $x_0 \in C$, then the set $V= C-x_0=\left \{ x-x_0:x \in C \right \}$ is a subspace of C.
Proof
Let $x_1,x_2 \in V$
To show: $\alpha x_1+\beta x_2 \in V$ for some $\alpha,\beta$
Now, $x_1+x_0 \in C$ and $x_2+x_0 \in C$ by definition of V
Now, $\alpha x_1+\beta x_2+x_0=\alpha \left ( x_1+x_0 \right )+\beta \left ( x_2+x_0 \right )+\left ( 1-\alpha -\beta \right )x_0$
But $\alpha \left ( x_1+x_0 \right )+\beta \left ( x_2+x_0 \right )+\left ( 1-\alpha -\beta \right )x_0 \in C$ because C is an affine set.
Therefore, $\alpha x_1+\beta x_2 \in V$
Hence proved.
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