Exercises
[188] Topics:subdifferential.Prerequisites:[184].Difficulty:*.
Let \(Cβ β^ n\) be an open convex set, and \(f:Cββ\) a convex function; Given \(zβ C\), we define the subdifferential \(β f(z)\) as the set of \(v\) for which the relation [(14.24)] is valid (i.e., \(β f(z)\) contains all vectors \(v\) defining the support planes to \(f\) in \(z\)).
\(β f(z)\) enjoys interesting properties.
\(β f(z)\) is locally bounded: if \(zβ C\) and \(r{\gt}0\) is such that \(B(z,2r)β C\), then \(L{\gt}0\) exists such that \(β yβ B(z,r)\), \(β vβ β f(x)\) you have \(|v|β€ L\). In particular, for every \(zβ C\), we have that \(β f(z)\) is a bounded set.
Show that \(β f\) is upper continuous in this sense: if \(zβ C\) and \((z_ n)_ nβ C\) and \(v_ nββ f(z_ n)\) and if \(z_ nβ_ n z\) and \(v_ nβ_ n v\) then \(vββ f(z)\). In particular, for every \(zβ C\), \(β f(z)\) is a closed set.
1