v2010.10.26 - Convex Optimization
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v2010.10.26 - Convex Optimization

v2010.10.26 - Convex Optimization v2010.10.26 - Convex Optimization

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242 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONSq(x)f(x)quasiconvexconvexxxFigure 75: Quasiconvex function q epigraph is not necessarily convex, butconvex function f epigraph is convex in any dimension. Sublevel sets arenecessarily convex for either function, but sufficient only for quasiconvexity.all invariant properties of convex sets carry over directly to convex functions.Generalization of epigraph to a vector-valued function f(X) : R p×k →R M isstraightforward: [294]epi f {(X , t)∈ R p×k × R M | X ∈ domf , f(X) ≼t } (556)R M +id est,f convex ⇔ epif convex (557)Necessity is proven: [61, 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 ) ≼µu + (1−µ)v (558)Yet this holds by definition because f(µX+(1−µ)Y ) ≼ µf(X)+(1−µ)f(Y ).The converse also holds.R M +3.5.0.0.1 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.

3.5. EPIGRAPH, SUBLEVEL SET 243Sublevel sets of a convex real function are convex. Likewise, correspondingto each and every ν ∈ R ML ν f {X ∈ dom f | f(X) ≼ν } ⊆ R p×k (559)R M +sublevel sets of a convex vector-valued function are convex. As for convexreal functions, the converse does not hold. (Figure 75)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 (560)When an epigraph (556) is artificially bounded above, t ≼ ν , then thecorresponding sublevel set can be regarded as an orthogonal projection ofepigraph on the function domain.Sense of the inequality is reversed in (556), for concave functions, and weuse instead the nomenclature hypograph. Sense of the inequality in (559) isreversed, similarly, with each convex set then called superlevel set.3.5.0.0.2 Example. Matrix pseudofractional function.Consider a real function of two variablesf(A, x) : S n × R n → R = x T A † x (561)on domf = S+× n R(A). This function is convex simultaneously in bothvariables when variable matrix A belongs to the entire positive semidefinitecone S n + and variable vector x is confined to range R(A) of matrix A .To explain this, we need only demonstrate that the function epigraph is

242 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONSq(x)f(x)quasiconvexconvexxxFigure 75: Quasiconvex function q epigraph is not necessarily convex, butconvex function f epigraph is convex in any dimension. Sublevel sets arenecessarily convex for either function, but sufficient only for quasiconvexity.all invariant properties of convex sets carry over directly to convex functions.Generalization of epigraph to a vector-valued function f(X) : R p×k →R M isstraightforward: [294]epi f {(X , t)∈ R p×k × R M | X ∈ domf , f(X) ≼t } (556)R M +id est,f convex ⇔ epif convex (557)Necessity is proven: [61, 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 ) ≼µu + (1−µ)v (558)Yet this holds by definition because f(µX+(1−µ)Y ) ≼ µf(X)+(1−µ)f(Y ).The converse also holds.R M +3.5.0.0.1 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.

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