Let $f(\boldsymbol{x})$ be a continuous function defined on $\mathbb{R}^n$. I suppose the following is not a sufficient condition for $f(\boldsymbol{x})$ to be concave.
$$ \frac{\partial^2 f}{\partial x_i^2} \leq 0, \quad \forall \boldsymbol{x} $$
Can someone give me a counterexample?
The necessary and sufficient condition for a bivariate and twice continuously differentiable function to be concave is that the Hessian matrix be negative semi definite, i.e.: $$ \begin{bmatrix} {\partial^2 f\over \partial x^2}&{\partial^2 f\over \partial x\partial y}\\ {\partial^2 f\over \partial x\partial y}&{\partial^2 f\over \partial y^2}\\ \end{bmatrix}\preceq 0 $$which is originated from the Taylor expansion of $f(x,y)$.