Re: Help with a problem: nonconvexifiability 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
.
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