# Show (weak) concavity of a function

• Oct 21st 2011, 08:32 AM
jens
Show (weak) concavity of a function
I have a real-valued function $f(\boldsymbol{x}) = f(x_1,\dotsc,x_n)$ of $n$ variables, defined on the non-negative orthant $\forall i \colon x_i \geq 0$, which fulfills the following properties:
• $f(\boldsymbol{x}) \geq \boldsymbol{0}$
• The function vanishes on the axes only, i.e., $f(\dotsc,0,\dotsc) = 0$ and $\boldsymbol{x} \neq \boldsymbol{0} \Rightarrow f(\boldsymbol{x}) > 0$
• $f$ is continuous and twice differentiable
• $f$ is marginally concave in all $x_i$, that is, $\frac{\partial^2 f}{\partial x_i^2} < 0$, and marginally bounded in x_i, that is, $f(\dotsc,x_i,\dotsc)$ is bounded in $x_i$ if all $x_j$ with $j \neq i$ are held fixed
• For any $\alpha \geq 0$, one has $f(\alpha \boldsymbol{x}) = \alpha f(\boldsymbol{x})$. Note that one consequence of this is that the concavity can only be weak!

A direct computation of the Hessian does not seem possible. So my question is: can you prove (weak) concavity of this function $f$ by knowing only the above properties? (Note that I might have listed more properties than actually needed)

Here is an examplary plot of such a function:

Attachment 22641

You see very clearly that it is concave, and considering the above properties, it seems like it can't be otherwise.

Thanks for any hints!

Jens