suppose that $f(x)=x^TAx+2b^Tx+c$ and $A$ is positive semidefinite. $A$ is a $n\times n$ matrix in $\Bbb R$, $b \in \Bbb R^n$ and $c\in \Bbb R$.
if $f$ is bounded below then does it have a global minimum? I think it should be correct but I could not prove it. Can any one help?
The point of minimum should be the center of symmetry of this function. That is, find $m$ a point so that $$f(m+x) = f(m-x)$$ for all $x$. We calculate a bit $$(m+x)^TA(m+x) + 2b^T(m+x) - (m-x)^TA(m-x)-2b^T(m-x)=4 (m^TA+b^T)x$$
Assume that the equation $m^TA+b^T$ has a solution $m$. Then we have $$f(x) = (x-m)^TA(x-m) -m^T A m +c$$ (this is "completing the square"). Now this function is beounded below if and only if $A$ is positive semidefinite -- in that case $m$ will be a point of minimum.
Assume that the equation (in $m$) $m^TA+b^T$ has no solution. Then $b$ is not in the image of $A$. Recall that for symmetric matrices $A$ we have $$Im A=(\ker A)^{\perp}$$ Therefore, $b$ is not perpendicular to $\ker A$. Take $x_0$ in $\ker A$, $b^Tx_0\ne 0$. Then we have $$f(\alpha x_0) = b^T x_0\cdot \alpha + c$$ so $f$ restricted to $\mathbb{R}x_0$ is not bounded.