$$\min \frac{1}{2}x^tQx-c^tx\\ Ax=b$$ where $Q\in \mathbb{R}^{n\times
n}$ is symmetric, $c\in \mathbb{R}^n$, $A\in \mathbb{R}^{m\times n}$
and $b\in\mathbb{R}^m$. Prove that if ${x}^*$ is a local
minimizer $\iff$ it is a global minimizer
We can transform the problem with constraints to a problem without constraints by doing
$$\phi(\gamma) = f(\overline{x} + Z\gamma)$$
where $f=\frac{1}{2}x^tQx-c^T x$, $\overline{x}$ is a solution of $Ax=b$ and $Z$ is a matrix elements from the basis of $\ker A$ as columns. Then the problem is transformed into minimizing the unconstrained problem $\min \phi(\gamma)$.
Global minimizer implies local, now,
If $\overline{x}$ is a local minimizer, it means $\nabla \phi(\gamma^*) = Z^T\nabla f(x^*)=0$ (I think), and $\nabla^2 \phi(\gamma)=Z^t\nabla^2 f(x^*)Z\ge 0$ (I think).
I'm thinking about what can be used now. Does somebody have an idea or can find some error in my thinking?
Best Answer
As you pointed out, the problem is equivalent to: $$ \begin{align} \phi(y) = f(\overline{x} + Zy) &= (\overline{x} + Zy)^TQ(\overline{x} + Zy) - c^T(\overline{x} + Zy) \\ & = y^T(Z^TQZ)y + (2Z^TQ\overline{x} + c)^Ty + \alpha \end{align} $$ for some constant $\alpha$ (whose value is irrelevant for the optimal $y$). If an optimum exists, the matrix $Z^TQZ$ must be positive semidefinite, and $\phi$ must therefore be convex. For a convex function, any local optimum is also a global optimum.