TITLE: How to determine whether a function of many variables is convex or non-convex?
QUESTION [26 upvotes]: A function $f$ is convex if $$f(\theta x + (1 − \theta)y) \leq \theta f(x) + (1 − \theta)f(y)$$ for all $x, y \in \mathcal{D}(f)$, the domain of $f$, and $\theta \in [0, 1]$.

How do I determine whether a function of many variables is convex or non-convex?

If I find the Hessian of my function and they are not all positive, is it a non-convex function?

REPLY [10 votes]: This is a proof that if $f:U \rightarrow \mathbb{R}$, $U$ is an open and convex set of $\mathbb{R}^n$ and f is twice differentiable, then $f''(x)$ positive definite imply $f$ convex.
Let $f''(x)(h²) \geq 0 $ $\forall x \in U, \forall h \in \mathbb{R}^n. $
If $f$ isn't convex, $\exists x_{1},x_{2} \in U , \exists \lambda $ st. $ f((1- \lambda)x_{1} + \lambda x_{2}) > (1-\lambda)f(x_{1}) + \lambda f(x_{2})$
Let $\phi:[0,1] \rightarrow \mathbb{R}, \phi(t) = f((1-t)x_{1} + tx_{2})$
Then $ \phi(\lambda) > (1- \lambda) \phi(0) + \lambda \phi(1)$
Using the Mean Value Theorem, $\exists t_{1}, t_{2}$ with $0<t_{1}<\lambda<t_{2}<1$ st.
$\phi(\lambda) - \phi(0) = \lambda \frac{d}{dt} \phi(t_{1}) $ and $\phi(1)-\phi(\lambda) = (1-\lambda) \frac{d}{dt} \phi(t_{2}) $
Observe that 
$\lambda ( \phi(1) - \phi(0)) < \phi(\lambda) - \phi(0) = \lambda \frac{d}{dt} \phi(t_{1})$  
and that
$ (1-\lambda) \frac{d}{dt} \phi(t_{2}) = \phi(1) - \phi(\lambda) < (1-\lambda) [ \phi(1) - \phi(0) ]$
Then   $ \frac{d}{dt} \phi (t_{2}) < \phi(1) - \phi(0) < \frac{d}{dt} \phi(t_{1}).$
Again, using MVT,
$\exists t_{3} \in (t_{1}, t_{2})$ st. $\frac{d²}{dt²} \phi(t_3) < 0$
This is an absurd, since by hypothesis $\frac{d²}{dt²} \phi(t) \geq 0, \frac{d²}{dt²} \phi(t) = f''((1-t)x_{1} + tx_{2})(x_{2}-x_{1})²$
$\square$