For a function $f$, it is said to be strongly convex if for all $x,y$
\begin{equation} (\nabla f(x) - \nabla f(y) )^T (x-y) \ge m \|x-y\|_2^2 \end{equation}
for a constant $m \ge 0$
Is it called strongly affine if
\begin{equation} (\nabla f(x) - \nabla f(y) )^T (x-y) = m \|x-y\|_2^2 \text{ ?} \end{equation}
In addition as we know if a function is strongly convex it satisfy
\begin{equation} f(y) \ge f(x) + \nabla f(x)^T (y-x) + \frac{m}{2} \|y-x\|_2^2 \end{equation}
For strongly affine function can I say
\begin{equation} f(y) = f(x) + \nabla f(x)^T (y-x) + \frac{m}{2} \|y-x\|_2^2 \text{ ?} \end{equation}
I never heard about "strongly affine" functions. However (as suggested by @LittleO to show that a "strongly affine" function is not necessarily strongly convex and strongly concave) consider the function $$f(x)= \frac{1}{2}\|x\|_2^2 \quad \Longrightarrow \quad \nabla f(x) = x.$$ Then $$(\nabla f(x)-\nabla f(y))^T(x-y) = \|x-y\|_2^2,$$ but $$\begin{array}{rcl}f(x) + \nabla f(x)^T(y-x) + \frac{1}{2}\|y-x\|_2^2 &=& \frac{1}{2}(\|x\|_2^2-\|x\|_2^2+x^Ty+\|x\|_2^2+\|y\|_2^2-2x^Ty)\\ &=& \frac{1}{2}(\|x\|^2+\|y\|^2-x^Ty) \\ &\neq & \frac{1}{2}\|y\|_2^2.\end{array} $$