v2007.09.13 - Convex Optimization

196 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONSepif ∆ = {(X , t)∈ R p×k × R M | X ∈ domf , f(X) ≼t } (459)R M +id est,f convex ⇔ epi f convex (460)Necessity is proven: [46, exer.3.60] Given any (X, u), (Y , v)∈ epif , wemust show for all µ∈[0, 1] that µ(X, u) + (1−µ)(Y , v)∈ epif ; id est, wemust showf(µX + (1−µ)Y ) ≼R M +µu + (1−µ)v (461)Yet this holds by definition because f(µX+(1−µ)Y ) ≼ µf(X)+(1−µ)f(Y ).The converse also holds.3.1.7.0.1 Exercise. Epigraph sufficiency.Prove that converse: Given any (X, u), (Y , v)∈ epif , 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, correspondingto each and every ν ∈ R ML ν f ∆ = {X ∈ domf | f(X) ≼ν } ⊆ R p×k (462)R M +sublevel sets of a vector-valued convex function are convex. As for realfunctions, the converse does not hold. (Figure 57)To prove necessity of convex sublevel sets: For any X,Y ∈ L ν f we mustshow 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 (463)