Chapter 3 Geometry of convex functions - Meboo Publishing ...
Chapter 3 Geometry of convex functions - Meboo Publishing ... 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 .
3.6. EPIGRAPH, SUBLEVEL SET 235 To explain this, we need only demonstrate that the function epigraph is convex. Consider Schur-form (1484) fromA.4: for t ∈ R [ ] A z G(A, z , t) = z T ≽ 0 t ⇔ z T (I − AA † ) = 0 t − z T A † z ≥ 0 A ≽ 0 (540) Inverse image of the positive semidefinite cone S n+1 + under affine mapping G(A, z , t) is convex by Theorem 2.1.9.0.1. Of the equivalent conditions for positive semidefiniteness of G , the first is an equality demanding vector z belong to R(A). Function f(A, z)=z T A † z is convex on convex domain S+× n R(A) because the Cartesian product constituting its epigraph epif(A, z) = { (A, z , t) | A ≽ 0, z ∈ R(A), z T A † z ≤ t } = G −1( S n+1 + ) (541) is convex. 3.6.0.0.3 Exercise. Matrix product function. Continue Example 3.6.0.0.2 by introducing vector variable x and making the substitution z ←Ax . Because of matrix symmetry (E), for all x∈ R n f(A, z(x)) = z T A † z = x T A T A † Ax = x T Ax = f(A, x) (542) whose epigraph is epif(A, x) = { (A, x, t) | A ≽ 0, x T Ax ≤ t } (543) Provide two simple explanations why f(A, x) = x T Ax is not a function convex simultaneously in positive semidefinite matrix A and vector x on domf = S+× n R n . 3.6.0.0.4 Example. Matrix fractional function. (confer3.8.2.0.1) Continuing Example 3.6.0.0.2, now consider a real function of two variables on domf = S n +×R n for small positive constant ǫ (confer (1830)) f(A, x) = ǫx T (A + ǫI) −1 x (544)
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3.6. EPIGRAPH, SUBLEVEL SET 235<br />
To explain this, we need only demonstrate that the function epigraph is<br />
<strong>convex</strong>. Consider Schur-form (1484) fromA.4: for t ∈ R<br />
[ ] A z<br />
G(A, z , t) =<br />
z T ≽ 0<br />
t<br />
⇔<br />
z T (I − AA † ) = 0<br />
t − z T A † z ≥ 0<br />
A ≽ 0<br />
(540)<br />
Inverse image <strong>of</strong> the positive semidefinite cone S n+1<br />
+ under affine mapping<br />
G(A, z , t) is <strong>convex</strong> by Theorem 2.1.9.0.1. Of the equivalent conditions for<br />
positive semidefiniteness <strong>of</strong> G , the first is an equality demanding vector z<br />
belong to R(A). Function f(A, z)=z T A † z is <strong>convex</strong> on <strong>convex</strong> domain<br />
S+× n R(A) because the Cartesian product constituting its epigraph<br />
epif(A, z) = { (A, z , t) | A ≽ 0, z ∈ R(A), z T A † z ≤ t } = G −1( S n+1<br />
+<br />
)<br />
(541)<br />
is <strong>convex</strong>.<br />
<br />
3.6.0.0.3 Exercise. Matrix product function.<br />
Continue Example 3.6.0.0.2 by introducing vector variable x and making<br />
the substitution z ←Ax . Because <strong>of</strong> matrix symmetry (E), for all x∈ R n<br />
f(A, z(x)) = z T A † z = x T A T A † Ax = x T Ax = f(A, x) (542)<br />
whose epigraph is<br />
epif(A, x) = { (A, x, t) | A ≽ 0, x T Ax ≤ t } (543)<br />
Provide two simple explanations why f(A, x) = x T Ax is not a function<br />
<strong>convex</strong> simultaneously in positive semidefinite matrix A and vector x on<br />
domf = S+× n R n .<br />
<br />
3.6.0.0.4 Example. Matrix fractional function. (confer3.8.2.0.1)<br />
Continuing Example 3.6.0.0.2, now consider a real function <strong>of</strong> two variables<br />
on domf = S n +×R n for small positive constant ǫ (confer (1830))<br />
f(A, x) = ǫx T (A + ǫI) −1 x (544)
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