# Re: Help with a problem: non-convexifiability of a function

*From*: spellucci@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx (Peter Spellucci)*Date*: Tue, 7 Mar 2006 18:44:26 +0000 (UTC)

In article <1141715475.115592.190660@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

tsy@xxxxxxxxxxxx writes:

I need to prove that there is no increasing function f such that thesounds much like homework.

function f[sqrt(x+y^2)+y] (x,y>=0) is convex.

hint: consider x>0, y>0 .

compute the Hessian matrix of g(x,y)=f(sqrt(x+y^2)+y)

clearly this depends also on the derivatives of f and

g is convex there if and only if the Hessian is positive semidefinite

f is increasing means f'(z)>=0 .

a two by two matrix H is positive semidefinite if and only if

1) it is the zero matrix

or

2) H(1,1)>0 or H(2,2)>0 and H(1,1)*H(2,2)-H(1,2)^2>=0

hth

peter

.

**Follow-Ups**:

**References**:

- Prev by Date:
**Re: adjusting an image** - Next by Date:
**Re: Help with a problem: non-convexifiability of a function** - Previous by thread:
**Help with a problem: non-convexifiability of a function** - Next by thread:
**Re: Help with a problem: non-convexifiability of a function** - Index(es):