# Math Help - Concave / Convex function

1. ## Concave / Convex function

hi

is the following function concave or convex:
f{x,y,z} = x*y + z

I tried to find the hessian of this but it is coming out to be 0 which means that the Hessian test doesnt say anything about convexity or concavity. So, is there any other method that can be used which can give conclusive results?

Thanks

(p.s. MF rocks as always thanks guys for ur replies to all my posts! )

2. The Hessian is $\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}$ in any point $(x,y,z)\in\mathbb{R}^3$, and it is neither positive semidefinite, nor negative semidefinite,
which implies that the function is neither convex nor concave.

3. Hessian is a sufficient condition. So when |H| = 0, we cannot say anything if the function is concave or convex. Hessian can be 0 even for strictly convex functions. For example : Hessian of f=x^2 + y^2 is 0 at (0,0) but then also the function is STrictly convex

4. If you have $X\subset\mathbb{R}^n$ convex with $X^\circ\neq\emptyset$, $f:X\to\mathbb{R}$, $f\in\mathrm{C}^2$ and denote by $Hf(x)$ the Hessian of $f$ at $x\in X$ i.e. $Hf(x)=\bigg(\frac{\partial^2 f}{\partial x_i\partial x_j}(x)\bigg)_{i,j=1,\dots,n}$, then:
$f$ is convex if and only if $Hf(x)$ is a positive semidefinite matrix, $\forall x\in X$.

5. in ur definition it says "If and only if the hessian is positive semidefinite" .

so that means that when Hessian is 0 there still exists a possibility of the function being strictly convex. We cannot conclude anything by using Hessian in this case.

6. I didn't say anything about strict convexity. And my last post is a proposition (a characterisation of convex functions that belong to $\mathrm{C}^2$), not a definition.

Check and see that the Hessian is not positive semidefinite (in every/any point). That implies that the function is not convex. Also, the Hessian is not negative semidefinite - and that implies that the function is not concave.