Problem of Sequential quadratic programming Help me with this problem of Sequential quadratic programming
Given the problem:
\begin{align}
   \text{min}&\quad f(x)&& \\
   \text{s.t}& \quad c(x)=0 && 
\end{align}
Consider the application of a sequential quadratic programming algorithm, with trust region,to this problem. Suppose that, at each iteration k, given the point $x_{k}$ is approximated by
\begin{align}
   \text{min}&\quad \frac{1}2 p^T\nabla^2f(x_k)p+\nabla f(x_k)^Tp && \\
   \text{s.t}& \quad A(x_k)p+c(x_k)=0 &&\\
             & \quad \Vert p \Vert_{\infty}\le \Delta_k &&\\
\end{align}
Where A is the Jacobian matrix of constraints. Suppose that $\nabla^2f(x_k)\gt 0$ and that  $A(x_k)$ rank full. Show or give a counterexample for the following statements:
a) There is p that solves the above problem.
b) if $(x_k)$ is feasible, then $p$ is a direction of descent for $f$.
c) if $p=0$, then  $(x_k)$ is a stationary point of the original problem.
d) if $(x_k)$ is feasible and $p=0$, then $(x_k)$ is a local minimizer of the original problem.
_
 A: I will assume the following fact, which can be deduced from Farkas's lemma.
FACT: The point $x_k$ is stationary for the problem if and only if the system 
$$\nabla f(x_k)^Tp<0,\; A(x_k)p=0$$ have no solution. 
a) Yes. The set 
$$X=\{p\in \mathbb{R}^n: A(x_k)p+ c(x_k)=0,\; \|p\|_\infty\leq \Delta_k\}$$ is compact and the objective function is continuous. Just apply Weirstrass's Theorem.
b) No. Choose, for a given problem with local minimizer $x^*$ (also that LICQ holds at $x^*,$ so that KKT holds at $x^*,$ and that $\nabla^2 f(x^*) >0$ ), the SQP associated problem to $x_k=x^*.$(For example, $f(x)=x^2, c(x)=x$) Then, any feasible $p$ for the auxiliary problem must satisfy $A(x_k)p=0.$ In particular, $p=0$ is feasible, so that 
$$\min_{p\in X} \frac{1}{2}p^T\nabla^2f(x_k)p +\nabla f(x_k)^Tp\leq 0. $$
Since $x_k$ is stationary by construction, there cannot be (by FACT) $p\in \mathbb{R}^n$ such that 
$$\nabla f(x_k)^Tp<0,\; A(x_k)p=0.$$ So any $p\in X$ must satisfy $$\nabla f(x_k)^Tp\geq0.$$ But then, since $\nabla^2f(x_k)>0,$ we would have
$$\frac{1}{2}p^T\nabla^2f(x_k)p +\nabla f(x_k)^Tp\geq 0 \;\forall p\in X.$$  This means that $p=0$ is the solution of the auxiliary problem and it is obviously not a descent direction.
c) Yes. If $p=0$ is a solution of this problem, we have $c(x_k)=0$ (so $x_k$ is feasible) and 
$$\frac{1}{2}p^T\nabla^2f(x_k)p +\nabla f(x_k)^Tp\geq 0 $$ for all $p\in X.$ Choose any $p\in X.$ Then, for $\alpha >0$ and sufficiently small, $\alpha p\in X.$ From this we find 
$$\frac{\alpha^2}{2}p^T\nabla^2f(x_k)p + \alpha\nabla f(x_k)^Tp\geq 0 \;\forall \alpha >0\textrm{ (and small)}.$$ Divide by $\alpha$ and let $\alpha \to 0,$ to find then $\nabla f(x_k)^Tp\geq 0.$  We have proved that $\nabla f(x_k)^Tp\geq 0$ whenever $A(x_k)p=0.$ This means that the system 
$$\nabla f(x_k)^Tp<0,\; A(x_k)p=0$$ have no solution. Now apply FACT.
d) Not sure about this one, but I think that this is false. It will suffice to find a problem with a KKT point $x^*$ which is not a minimizer and at which $\nabla^2 f(x^*)>0.$ 
