Let $f(x)=x^TQx+2b^Tx$, where $Q\succeq 0$. When does this function have a lower bound and when not?
I did some survey (especially from Prof. Lieven Vandenberghe’s slides on convex optimization) and the answer seems to be, the lower bound exists iff $b\in \mathcal R(Q)$, i.e. $b$ lies in the vector space spanned by $Q$'s row(column) vectors.
Intuitively I believe this is a constraint that guarentees that we can convert the function in a way like Completing the square, and I can complete this arithmetic when $Q\succ 0$ and in that case $f(x)=\|Q^{1/2}x+Q^{-1/2}b\|_2^2-b^TQ^{-1}b$, and it does have a lower bound. However, in general cases where $Q$ is singular, I don't know what to do. Hope somebody can help me, thanks!
Btw, this problem arises from a nonconvex problem with strong duality: $\min x^TAx+2b^Tx$ s.t. $x^Tx\le 1$, where the dual function is bounded when $b\in \mathcal R(A+\lambda I)$.
Proof:
When $b\not\in\mathcal R(Q)$, we can decompose $b$ into $b=b_1 + b_2$ where $b_1\in\mathcal R(Q)$ and $b_2\perp\mathcal R(Q)$. The direction given by $-b_2$ will lead the function to $-\infty$, as $f(-td_2)=t^2d_2^TQd_2-2tb^Td_2=-2tb^Td_2\to -\infty$ when $t\to\infty$.
when $b\in\mathcal R(Q)$, $\nabla f(x)=2Q^Tx+2b=0$ has a solution denoted by $x^*$. As $f(x)$ is convex, $x^*$ must be global minimum point.