Consider a quadratic function $f(x)=\frac{1}{2}x^{T}Ax+b^{T}x+c$, where $A$ is an $n \times n$ symmetric positive definite matrix.
My question is:
(1) Does $f(x)$ must be strongly convex?
(2a) If the answer to (1) is yes, assume that $f(x)$ is $\mu$-strongly convex. What is the condition of $\mu$?
(2b) If the answer to (1) is no, what is the condition of $f(x)$ to be strongly convex?
First, are you sure you mean strongly convex and not strictly convex?
Assuming that you mean strongly convex then all you need is the Hessian of the $f$-$mI$ i.e.
$\bigtriangledown^{2}f(x)-mI$ is positive semidefinite,for some $m>0$. Taking into account that
$\bigtriangledown\,f(x)$=$Ax+b$ and $\bigtriangledown^{2}f(x)=A$ which is positive definite, we proceed by
taking $m$ to be the smallest eigenvalue of $A$ (which has only positive eigenvalues),
then $A-mI$ is positive semidefinite and the function $f$ is strongly convex with parameter
$m=min\lambda_{i}$.!!