If the Jacobian matrix is positive definite, does that imply that the optimization problem has a unique solution? My PhD adviser told me that
if the Jacobian matrix of the optimality conditions is positive definite,
then it implies that the optimization problem has a unique solution.
I was wondering what is the basis for this statement,
in other words, what theorem or proposition gives this statement.
Here is a simple example which illustrates the statement.
Let $a, b > 0$ be positive constants.
Our goal is to compute the $(p, q)$ which maximizes the objective
$$
ap - bp^2 + bpq - bq^2.
$$
The first-order optimality conditions are:
$$
a - 2bp + bq = 0, \\
bp - 2bq = 0.
$$
If we take the Jacobian matrix of the vector-valued function
which is the optimality conditions,
it gives us the matrix
$$
\begin{bmatrix}
-2b & b \\
b & -2b
\end{bmatrix}.
$$
If I were to change the signs of the optimality conditions,
I could get the Jacobian to be the positive definite matrix
$$
\begin{bmatrix}
2b & -b \\
-b & 2b
\end{bmatrix}.
$$
What mathematical theorem or proposition tells us that
because the Jacobian matrix is positive definite,
therefore the optimization problem has a unique solution?
 A: Positive definiteness of the Hessian is equivalent to convexity for twice differentiable functions, and convexity is what's really at work here. So, the theorem you want is 

Theorem Let $f:U\to\mathbb{R}$ be strictly convex, where $U\subset \mathbb{R}^n$ is convex. Then if $f$ has a local minimum in $U$, it has a global minimum.

You'll observe that looks weaker than your advisor's claim might have sounded: there are convex functions with no minimum, the simplest example being $e^x$.
But it's easy to find local minima of differentiable functions, certainly in your case when the first derivatives form a linear system, so then you can be sure of a global minimum.
A: So, let me explain you. 
If a function is defined on a convex set and derivative(on its domain) and also its second derivation is positive, then this function is convex.
An extremely easy example is $f(x)=x^2, x \in \mathbb{R}$
So for multivariate functions this theorem results in the fact that Jacobian is positive definite.
Roughly speaking, positive definite in matrices (in $\mathbb{R}^m$) is equivalent to being positive in $\mathbb{R}$.
Convex optimization by Steve Boyd, Click here is the best reference for your question.
