This in place to keep some study notes, and other items to remember while taking this hard course.
Let \(G\left ( u\right ) =g\left ( u\right ) +f\left ( u\right ) \) where we know \(g,f\) are two convex functions. We need to show \(G\left ( u^{\lambda }\right ) \) is also convex. Then, pick point \(u^{\lambda }\in U\,\ \)therefore
But the set \(U\) is convex (it must be, these are convex functions, so their domain is convex by definition). Pick a point \(u^{\lambda }=\left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\) where \(\lambda \in \left [ 0,1\right ] \) and \(u^{1},u^{2}\in U\). Hence we can write
\begin{align*} G\left ( u^{\lambda }\right ) & =g\left ( u^{\lambda }\right ) +f\left ( u^{\lambda }\right ) \\ G\left ( \left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\right ) & =g\left ( \left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\right ) +f\left ( \left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\right ) \end{align*}
But \(g\left ( \left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\right ) \leq \left ( 1-\lambda \right ) g\left ( u^{1}\right ) +\lambda g\left ( u^{2}\right ) \) and the same for \(f\). Then the above reduces to
\begin{align*} G\left ( \left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\right ) & \leq \left ( 1-\lambda \right ) g\left ( u^{1}\right ) +\lambda g\left ( u^{2}\right ) +\left ( 1-\lambda \right ) f\left ( u^{1}\right ) +fg\left ( u^{2}\right ) \\ & =\left ( 1-\lambda \right ) \left ( g\left ( u^{1}\right ) +f\left ( u^{1}\right ) \right ) +\lambda \left ( g\left ( u^{2}\right ) +f\left ( u^{2}\right ) \right ) \end{align*}
But \(G\left ( u\right ) =g\left ( u\right ) +f\left ( u\right ) \), then the RHS above becomes
\[ G\left ( \left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\right ) \leq \left ( 1-\lambda \right ) G\left ( u^{1}\right ) +\lambda G\left ( u^{2}\right ) \]
Therefore \(G\) is a convex function.
Smallest set that contains all the sets inside it, such that it is also convex. (put a closed convex ”container” around everything)
No. Polytope is region which has straight edges (flat sides) and also be convex. But http://mathworld.wolfram.com/Polytope.html says ”The word polytope is used to mean a number of related, but slightly different mathematical objects. A convex polytope may be defined as the convex hull of a finite set of points” And https://en.wikipedia.org/wiki/Polytope says ”In elementary geometry, a polytope is a geometric object with flat sides, and may exist in any general number of dimensions n as an n-dimensional polytope or n-polytope”
https://en.wikipedia.org/wiki/Polytope says ”In elementary geometry, a polyhedron (plural polyhedra or polyhedrons) is a solid in three dimensions with flat polygonal faces, straight edges and sharp corners or vertices.”
Polyhedron can be convex or not. But polyhedron can be open? While polytope not. Need to check.