Nonlinear Optimization I
Dr. Thomas M. Surowiec
Humboldt University of Berlin Department of Mathematics
Summer 2013
Convexity
Definition
1 A setX⊂Rnis called convex, if for allx,y∈X and allλ∈(0,1)
λx+ (1−λ)y∈X, i.e. the segment[x,y]lies completely inX.
2 LetX⊂Rnbe convex. A functionf :X →Ris called
1 (strictly) convex (inX), if for allx,y ∈X and for allλ∈(0,1) the following holds true:
f(λx+ (1−λ)y)≤λf(x) + (1−λ)f(y) (f(λx+ (1−λ)y)< λf(x) + (1−λ)f(y))
3 f is uniformly convex, if there existsµ >0with
f(λx+ (1−λ)y) +µλ(1−λ)||x−y||2≤λf(x) + (1−λ)f(y),
∀x,y∈X,∀λ∈(0,1).
A Few Remarks
Geometrically speaking convexity of a function implies that the line segment between function values lies above the function’s graph.
Clearly: Uniform convexity→strict convexity→convexity.
Example
Example
Letf:Rn→Rbe a quadratic function, i.e., f(x) + 1
2xTAx+bTx+c withA∈ Sn,b∈Rn,c∈R. Then
1 f is convex iffAis positive semi-definite.
2 f is strictly convex ifffunif. conv. iffApos. def.
These concepts can be extended in some cases to differentiabe convex functions...
Supporting Hyperplanes/Quadratic Functions
Lemma
LetX⊂Rnbe open and convex, andf :X →Rcontinuously differentiable.
Then the following assertions hold
1 f is convex (onX) iff for allx,y∈Xit holds that f(x)≥f(y) +∇f(y)T(x−y).
2 f is strictly convex (onX) iff for allx,y∈X, withx6=yit holds that f(x)>f(y) +∇f(y)T(x−y).
3 f is uniformly convex (onX) iff there existsµ >0such that f(x)≥f(y) +∇f(y)T(x−y) +µ||x−y||2 for allx,y∈X.
Proof.
On the board.
Second-Order Properties
Theorem
LetX⊂Rnbe open and convex, andf :X →Rtwice continuously differentiable. Then the following assertions hold
1 f is convex (onX) iff∇2f(x)is positive semi-definite for allx∈X.
2 If∇2f(x)is positive definite for allx∈X, thenfis strictly convex (onX).
3 f is uniformly convex (onX) iff∇2f(x)is uniformly positive definite on X, i.e., if there existsµ >0such that
dT∇2f(x)d≥µ||d||2 forx∈X and for alld ∈Rn.
Proof.
On the board.
Existence of Solutions
Lemma
Letf:Rn→Rbe continuously differentiable andx0∈Rnarbitrary.
Furthermore, assume that the level set L(x0) :=n
x∈Rn
f(x)≤f(x0)o
is convex and thatfis uniformly convex inL(x0). Then the setL(x0)is compact.
Proof.
On the board.
Existence of Solutions
Theorem
Letf:Rn→Rbe continuously differentiable andX⊂Rnbe convex.
Consider the optimization problem min
x∈Xf(x), (1)
then the following statements hold
1 Iff is convex onX, then the solution set is convex (possibly empty).
2 Iff is strictly convex onX, then(1)has at most one solution.
3 Iff is uniformly convex (onX) andXis non-empty and closed, then(1) has exactly one solution.
Proof.
On the board.
A Few Cautionary Remarks
1 f(x) =exp(x)is strictly convex onX =R, yet it has no minimimum!
2 X must be closed for a solution to exist. Any ideas?
Consequences I
Lemma
Letf:Rn→Rbe continuously differentiable,x0∈R, and let the level set L(x0)be convex andf be uniformly convex onL(x0). In addition, suppose thatx∗∈Rnis the unique global minimizer off. Then there existsµ >0with
µ||x−x∗||2≤f(x)−f(x∗),∀x∈L(x0).
Consequences II
Theorem
Letf:Rn→Rbe a continuously differentiable and convex function. and let x∗∈Rnbe a stationary point off. Thenx∗is a global minimizer off inRn.