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I have a function of the following type:

$f(x_1,x_2,...,x_n)$ Each $x_i$ has domain $[0,\infty)$. The function is continuous and differentiable in each variable (It is an expectation of several continuous probability distributions). My numerical experiments show that the function is convex, and has an optimal solution for which the gradient is 0 in every direction, i.o.w. $x_i=0$, and $x_i=\infty$ are not optimal.

Now, my goal is to prove that there exists an optimal solution, which is unique.

I tried to proving convexity by showing that the hessian is positive definite.

I already proved that the function is strictly convex w.r.t. each variable seperately, by showing for each i that $\frac{\partial ^2 f}{(x_i)^2}>0$

These terms form the diagonal of the Hessian. For the nondiagonal elements I know the following properties:

Each element not on the diagonal is negative. $$ a_{ij}<0, j\neq i $$

the absolute value is is smaller than the value of the element on the diagonal. $$a_{ii}>|a_{ij}|, \forall j \neq i$$

My questions: 1) Are there other ways to prove convexity other than computing the determinant of the hessian?

2) Do I really need convexity to show that my function has one global minimum or are there other ways to show this?

Exponential
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    why don't you show us the function :) ? With the properties you have given it is difficult to judge if Hessian is positive definite. See if $a_{ii} >= \sum_{j}|a_{ij}|$ holds for you, then you are good to go. Convexity doesn't guarantee unique global solution, it just says any down-hill algorithm will hit a global optimum. I am not sure if strict convexity in each variable guarantees strict convexity of the function. – dineshdileep Jul 13 '16 at 08:12
  • Checking strict convexity is a good idea, since it will give yield uniqueness of a solution. Checking dineshdileep's condition on the off-diagonals would be a good way to show the matrix is diagonally dominant. Strong convexity is also sufficient to get uniqueness, and although it's a stronger condition it may be easier to check... To show strong convexity, you just need to show there exists $\beta>0$ such that for every $(x,y)$ that $\langle x-y | \nabla f(x) - \nabla f(y)\rangle \geq \beta |x-y|$ (for a precise statement, check Exercise 17.5 from Bauschke/Combettes' book, vol. 2 – Zim May 30 '20 at 23:24

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