I have a real-valued function $\displaystyle f(\boldsymbol{x}) = f(x_1,\dotsc,x_n)$ of $\displaystyle n$ variables, defined on the non-negative orthant $\displaystyle \forall i \colon x_i \geq 0$, which fulfills the following properties:

- $\displaystyle f(\boldsymbol{x}) \geq \boldsymbol{0}$
- The function vanishes on the axes only, i.e., $\displaystyle f(\dotsc,0,\dotsc) = 0$ and $\displaystyle \boldsymbol{x} \neq \boldsymbol{0} \Rightarrow f(\boldsymbol{x}) > 0$
- $\displaystyle f$ is continuous and twice differentiable
- $\displaystyle f$ is marginally concave in all $\displaystyle x_i$, that is, $\displaystyle \frac{\partial^2 f}{\partial x_i^2} < 0$, and marginally bounded in x_i, that is, $\displaystyle f(\dotsc,x_i,\dotsc)$ is bounded in $\displaystyle x_i$ if all $\displaystyle x_j$ with $\displaystyle j \neq i$ are held fixed
- For any $\displaystyle \alpha \geq 0$, one has $\displaystyle f(\alpha \boldsymbol{x}) = \alpha f(\boldsymbol{x})$. 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 $\displaystyle f$ 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:

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