Lecture15

Lecture15 - C&O 355 Lecture 15 N. Harvey Topics...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Lecture 15 N. Harvey
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Topics Subgradient Inequality Characterizations of Convex Functions Convex Minimization over a Polyhedron (Mini)-KKT Theorem Smallest Enclosing Ball Problem
Background image of page 2
Subgradient Inequality Prop: Suppose f : R ! R is differentiable. Then f is convex iff f(y) ¸ f(x) + f’(x)(y -x) 8 x,y 2 R Proof: ( : See Notes Section 3.2. ) : Exercise for Assignment 4. ¤
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Convexity and Second Derivative Prop: Suppose f : R ! R is twice-differentiable. Then f is convex iff f’’(x) ¸ 0 8 x 2 R . Proof: See Notes Section 3.2.
Background image of page 4
Subgradient Inequality in R n Prop: Suppose f : R n ! R is differentiable. Then f is convex iff f(y) ¸ f(x) + r f (x) T (y-x) 8 x,y 2 R Proof: ( : Exercise for Assignment 4. ) : See Notes Section 3.2. ¤
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Minimizing over a Convex Set Prop: Let C μ R n be a convex set. Let f : R n ! R be convex and differentiable. Then x minimizes f over C iff r f(x) T (z-x) ¸ 0 8 z 2 C. Proof: ( direction Direct from subgradient inequality. f(z) ¸ f(x) + r f(x) T (z-x) ¸ f(x) Subgradient inequality Our hypothesis
Background image of page 6
Minimizing over a Convex Set Prop: Let C μ R n be a convex set. Let f : R n ! R be convex and differentiable. Then x minimizes f over C iff r f(x) T (z-x) ¸ 0 8 z 2 C. Proof: ) direction Let x be a minimizer, let z 2 C and let y = z-x. Recall that r f(x) T y = f’( x;y) = lim t ! 0 f(x+ty)-f(x). If limit is negative then we have f(x+ty)<f(x) for some t 2 [0,1], contradicting that x is a minimizer. So the limit is non-negative, and r f(x) T y ¸ 0. ¥ t
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Positive Semidefinite Matrices (again) Assume M is symmetric Old definition: M is PSD if 9 V s.t. M = V T V. New definition: M is PSD if y T My ¸ 0 8 y 2 R n . Claim: Old ) New. Proof: y T My = y T V T Vy = k Vy k 2 ¸ 0. Claim: New ) Old. Proof: Based on spectral decomposition of M.
Background image of page 8
Convexity and Hessian Prop: Let f: R n ! R be a C 2 -function. Let H(x) denote the Hessian of f at point x.
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 24

Lecture15 - C&amp;O 355 Lecture 15 N. Harvey Topics...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online