$$\min \frac{1}{2}x^tQx-c^tx\\ Ax=b$$ 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?
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.