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

In article <1141715475.115592.190660@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
tsy@xxxxxxxxxxxx writes:
I need to prove that there is no increasing function f such that the
function f[sqrt(x+y^2)+y] (x,y>=0) is convex.

sounds much like homework.
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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