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:

Show (weak) concavity of a function-example_of_f.jpg

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