Chapter 3 Geometry of convex functions - Meboo Publishing ...

234 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONS
3.6.0.0.1 Exercise. Epigraph sufficiency.
Prove that converse: Given any (X, u), (Y , v)∈ epi f , if for all µ∈[0, 1]
µ(X, u) + (1−µ)(Y , v)∈ epif holds, then f must be convex.
Sublevel sets of a real convex function are convex. Likewise, corresponding
to each and every ν ∈ R M
L ν f {X ∈ dom f | f(X) ≼
ν } ⊆ R p×k (537)
R M +
sublevel sets of a vector-valued convex function are convex. As for real
convex functions, the converse does not hold. (Figure 73)
To prove necessity of convex sublevel sets: For any X,Y ∈ L ν f we must
show for each and every µ∈[0, 1] that µX + (1−µ)Y ∈ L ν f . By definition,
f(µX + (1−µ)Y ) ≼
R M +
µf(X) + (1−µ)f(Y ) ≼
R M +
ν (538)
When an epigraph (534) is artificially bounded above, t ≼ ν , then the
corresponding sublevel set can be regarded as an orthogonal projection of
epigraph on the function domain.
Sense of the inequality is reversed in (534), for concave functions, and we
use instead the nomenclature hypograph. Sense of the inequality in (537) is
reversed, similarly, with each convex set then called superlevel set.
3.6.0.0.2 Example. Matrix pseudofractional function.
Consider a real function of two variables
f(A, x) : S n × R n → R = x T A † x (539)
on domf = S+× n R(A). This function is convex simultaneously in both
variables when variable matrix A belongs to the entire positive semidefinite
cone S n + and variable vector x is confined to range R(A) of matrix A .