Invertibility of bordered Hessian I have an optimization problem: $max_{x \in C} f(x)$ s.t. $Ax=b$, where $x \in  R^n$ and $b \in  R^m$, $m \le n$, adn $C$ compact. I know that $f$ is strictly quasi-concave, and that $A$ has rank $m$ (linearly independent, linear constraints). My question is: can I guarantee that the Hessian matrix bordered with the Jacobian of the constraints is invertible --when all constraints are binding, i.e, the multipliers positive--? That is, can I guarantee that the solution (in $x$ and Lagrange multipliers, which exists and is unique) is differentiable with respect to $b$?
Thanks!!!
 A: Let $f:C\rightarrow \mathbb{R}$ where $C$ is a convex subset of $\mathbb{R}^n$ and $f$ is SQC. We consider the extrema of $f$ over $D=C\cap \{x|Ax=b\}$. I assume that you look for a maximum of $f$ over the convex $D$.
What do you mean by "can I guarantee that the solution (in x and Lagrange multipliers) is differentiable?" ? 
On the other hand, why do you write that " the solution (in x and Lagrange multipliers, which exists and is unique)" ? The function $f(x)=\sqrt{x}$ is strictly concave and not upper bounded over $\mathbb{R}^+$.


*

*The Lagrange method gives the candidates $(x_i)_{i\in I}$ that are solutions of: 


$Ax=b$ and there is $\Lambda \in M_{1,m}$ s.t. $Df_x+\Lambda A=0$.


*You seek THE $x_i$ satisfying, for every $j\not= i$, $f(x_i)>f(x_j)$. If this unique $x_i$ does not exist, then $f$ is not upper bounded or does not reach its upper bound. 

*Otherwise, there are two posibilities: either $f(x_i)$ is the required maximum or $f$ is not upper bounded or does not reach its upper bound.
About your question, consider the case $m=0,n=1$ and the strictly concave function $f(x)=-x^4$. The Hessian in the critical point $x=0$ is $0$.
EDIT. Answer to Rob. You are very kind (rare quality) but you do not understand precisely these issues.
In my post, I assumed that $D$ has a void edge. Since you assume that $D$ is compact, it may have a non-void edge and you are faced with a different kind of problem. Indeed, the max can be reached on the edge in a point that does not satisfy the Lagrange conditions (take $\sqrt{x}$ over $[1,2]$); on the other hand, your claim about the unicity (when a solution exists) is clearly false; consider the function $y=x^3$ over $[-1,1]$ and extend to the right by a bell curve (like a parabola $y=-x^2$); you obtain a SQC function and Lagrange method gives its max and a critical point, in $x=0$, which is not associated to an extremum of it.
Of course, when your compact is defined by a system of inequalities, then you must write Kuhn-Kutter conditions and it is no more the same question...
The last but not the least, you are very unlucky about your last sentence "So, your example $f(x)=−x^4$ is not an issue"; simply see it as a restriction:
$f(x,y)=-x^4+x(y^2-1),[0,1][x,y]^T=[1]$. Then the Hessian in the unique critical (Lagrange) point $(0,1)$ is $\begin{pmatrix}0&2\\2&0\end{pmatrix}$ and is invertible. The bordered Hessian is $\begin{pmatrix}0&0&1\\0&0&2\\1&2&0\end{pmatrix}$ and is singular.
PS. I have not checked if my function $f$ above is SQC; otherwise we have to change a little the counter-example.
