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.

  • 2
    $\begingroup$ I changed $\alpha ln(x_1)$ to $\alpha\ln(x_1)$. That is standard. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 11 '14 at 13:52

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.

  • $\begingroup$ Thank you, in my example, $0<\alpha <1$ and both $x_1$ and $x_2$ are positive. In my question I was attempting to prove that $u(x_1,x_2)$ is convex and obviously the result I got was that it wasn't. $\endgroup$ – Jamzy Jun 12 '14 at 0:51

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