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

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248 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONS<br />

3.7.2 first-order <strong>convex</strong>ity condition, real function<br />

Discretization <strong>of</strong> w ≽0 in (476) invites refocus to the real-valued function:<br />

3.7.2.0.1 Theorem. Necessary and sufficient <strong>convex</strong>ity condition.<br />

[59,3.1.3] [129,I.5.2] [384,1.2.3] [41,1.2] [324,4.2] [300,3] For real<br />

differentiable function f(X) : R p×k →R with matrix argument on open<br />

<strong>convex</strong> domain, the condition (conferD.1.7)<br />

f(Y ) ≥ f(X) + 〈∇f(X) , Y − X〉 for each and every X,Y ∈ domf (585)<br />

is necessary and sufficient for <strong>convex</strong>ity <strong>of</strong> f .<br />

⋄<br />

When f(X) : R p →R is a real differentiable <strong>convex</strong> function with vector<br />

argument on open <strong>convex</strong> domain, there is simplification <strong>of</strong> the first-order<br />

condition (585); for each and every X,Y ∈ domf<br />

f(Y ) ≥ f(X) + ∇f(X) T (Y − X) (586)<br />

From this we can find ∂H − a unique [364, p.220-229] nonvertical [195,B.1.2]<br />

hyperplane [ ](2.4), expressed in terms <strong>of</strong> function gradient, supporting epif<br />

X<br />

at : videlicet, defining f(Y /∈ domf) ∞ [59,3.1.7]<br />

f(X)<br />

[ ] Y<br />

∈ epif ⇔ t ≥ f(Y ) ⇒ [ ∇f(X) T −1 ]([ ] [ ])<br />

Y X<br />

− ≤ 0<br />

t<br />

t f(X)<br />

(587)<br />

This means, for each and every point X in the domain <strong>of</strong> a real <strong>convex</strong><br />

[ function ] f(X) , there exists a hyperplane ∂H − [ in R p ] × R having normal<br />

∇f(X)<br />

X<br />

supporting the function epigraph at ∈ ∂H<br />

−1<br />

f(X)<br />

−<br />

{[ ] [ ] Y R<br />

p [<br />

∂H − = ∈ ∇f(X) T −1 ]([ ] [ ]) }<br />

Y X<br />

− = 0 (588)<br />

t R<br />

t f(X)<br />

One such supporting hyperplane (confer Figure 29a) is illustrated in<br />

Figure 75 for a <strong>convex</strong> quadratic.<br />

From (586) we deduce, for each and every X,Y ∈ domf<br />

∇f(X) T (Y − X) ≥ 0 ⇒ f(Y ) ≥ f(X) (589)

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