v2009.01.01 - Convex Optimization

232 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONS
3.1.11.0.1 Exercise. Real fractional function. (confer3.1.4,3.1.7.0.4)
Prove that real function f(x,y) = x/y is not convex on the first quadrant.
Also exhibit this in a plot of the function. (f is quasilinear (p.242) on
{y > 0} , in fact, and nonmonotonic even on the first quadrant.)
3.1.11.0.2 Exercise. Stress function. (confer (1213))
Define |x −y| ∆ = √ (x −y) 2 and
X = [x 1 · · · x N ] ∈ R 1×N (68)
Given symmetric nonnegative data [h ij ] ∈ S N ∩ R N×N
+ , consider function
f(vec X) =
N−1
∑
i=1 j=i+1
N∑
(|x i − x j | − h ij ) 2 ∈ R (555)
Take the gradient and Hessian of f . Then explain why f is not a convex
function; id est, why doesn’t second-order condition (554) apply to the
constant positive semidefinite Hessian matrix you found. For N = 6 and
h ij data from (1284), apply the line theorem from3.2.3.0.1 to plot f along
some arbitrary lines through its domain.
3.1.11.1 second-order ⇒ first-order condition
For a twice-differentiable real function f i (X) : R p →R having open domain,
a consequence of the mean value theorem from calculus allows compression
of its complete Taylor series expansion about X ∈ dom f i (D.1.7) to three
terms: On some open interval of ‖Y ‖ 2 , so that each and every line segment
[X,Y ] belongs to domf i , there exists an α∈[0, 1] such that [342,1.2.3]
[37,1.1.4]
f i (Y ) = f i (X) + ∇f i (X) T (Y −X) + 1 2 (Y −X)T ∇ 2 f i (αX + (1 − α)Y )(Y −X)
(556)
The first-order condition for convexity (545) follows directly from this and
the second-order condition (554).