Hessian matrix to establish convexity I have a function, $u(x_1,x_2)=\alpha \ln(x_1)+(1-\alpha)\ln(x_2)$.
where $0<\alpha <1$
I want to prove that it is convex.
The Hessian matrix I have constructed is:
$$ \left( \begin{array}{ccc} -\alpha/x_1^2 & 0 \\ 0 & -(1-\alpha)/x_2^2)\end{array} \right)$$
From here I found that it determinant is positive but the leading sub minor is negative. Am I making a silly mistake?
From what I can tell for the function to be convex, the Hessian needs to be positive definite. Which, by my calculation it is not.
Since I have, with help, worked out the correct answer, I will write it down here:
The Hessian is calculated correctly. The leading first order principal submatrix is $-\alpha/x_1^2$ which in my case is negative. The 2nd order principal submatrix is the determinant of the 2x2 Hessian. This is a positive value. The Hessian is negative semidefinite.
Using this link, http://www.economics.utoronto.ca/osborne/MathTutorial/CVNF.HTM
and this quote from that page:


*

*f  is concave if and only if H(x) is negative semidefinite for all x ∈ S

*if H(x) is negative definite for all x ∈ S then  f  is strictly concave

*f  is convex if and only if H(x) is positive semidefinite for all x ∈ S

*if H(x) is positive definite for all x ∈ S then  f  is strictly convex.


I concluded the function is concave. 

If $\alpha$ is real than either $\alpha>0$ or $1-\alpha>0$ and if $0<\alpha<1$ then both are positive.  If $\alpha>0$, consider the two points $(x_1,x_2)$ and $(w_1,x_2)$.  If $x_1$, $w_1$ are distinct positive numbers then we would have
$$
u(p(x_1,x_2) + (1-p)(w_1,x_2)) > pu(x_1,x_2)+(1-p)u(w_1,x_2).
$$
If by "convex" you mean that the opposite inequality holds, then this function is not convex.
