1. Determine whether the following functions are convex or strictly
convex over the stated
domains. Justify all answers.
(a) f(x, y, z) = ex2+y+z ? ln(x + y) + 3z2,  over the set S = R3.
(b) f(x, y) = x2 ? 4xy + 5y2 ? ln(xy) over S = {x belongsto R2 : x >
0, y > 0} where x =matrix
x
y
.

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Clarification of Question by shaherome-ga on 04 Mar 2005 18:21 PST

question 1. is f(x,y,z) = e^(x^2+y+z)-ln(x+y) + 3^(z^2)


question 2. f(x,y) =x^2 - 4xy +5y^2 - ln(xy)

I believe that if all the second partials exist then you need the
Determinant of the matrix of the partials (remember maximization?).

So for Q2:
f=x^2-4xy+5y^2-ln(xy)

fxx=2+1/x^2
fxy=-4
fyy=5-1/y^2

D = fxxfyy-(fxy)^2 = (2+1/x^2)(5+1/y^2)-(-4)^2 = 10+5/x^2+2/y^2+1/(xy)^2-16
  = 5/x^2 + 2/y^2 + 1/(xy)^2 - 6

You can obviously see for large x or y this is negative  So it is NOT convex

You can proceed in a similar fashion with the first problem... just
calculate the determinant of the Jacobian and see if its always
positive

In Q1: f(x,y,z) = e^(x^2+y+z)-ln(x+y) + 3^(z^2).

Firstly, if two functions are convex, then the sum of these two
functions is also convex. So we will examine the three terms
separately.

e^(x^2+y+z): Using the 2nd derivative test (i.e. g'' > 0 then g is
convex), we quickly realize that if a function g is convex then so is
the function e^g. Therefore, we first try to prove convexity of
x^2+y+z. Once again, we look at each term separately and verify that
each of them (x^2, y and z) are convex functions. To conclude, the
first term is a convex function.

-ln(x+y): For this we use the second order partial derivatives.

g = -ln(x+y)
gxx = 1 / (x+y)^2
gyy = 1 / (x+y)^2
gxy = 1 / (x+y)^2

Therefore the determinant D = gxx gyy - (gxy)^2 = 0. 
Hence, -ln(x+y) is also convex.

3^(z^2): This is quite easy. The second derivative is 3^(z^2) ln3 (2 +
4 ln3 z^2). This is always positive as 3^(z^2) >0 and z^2 >= 0.

To sum up, the function f is convex as all the three terms are convex. 
Moreover, f is strictly convex, since the first and third terms are strictly conex.

Hope this helps.
--k

Dont do it that way... 

its possible that two non convex functions add to a convex function

for example

take h = f + g

f = 2x^2+lnx

f is not convex

g = -x^2-lnx

g is not convex

but f+g = x^2 is convex...

now if f and g are convex then f+g is convex... but splitting it into
a sum and if one is not convex doesnt mean they arent convex.

how did you get 

fyy =5-1/y^2?
i think it should be 
fyy= 10+1/y^2?
also fxy=second derivative of xy?

Ok... the - instead of plus was just a typo on the one line... and yes
it should be +10... which actually changes the answer... just plugin
and recalculate...

fxy = d^2f / dxdy

i should probably do these things on paper first :P