For strongly convex functions, it is stated that for some $\mu>0$,
- $$f(y)\geq f(x)+\nabla f(x)^T(y−x)+\frac{\mu}{2}\|y−x\|^2, \quad \forall x,y.$$
- $$(\nabla f(x)−\nabla f(y))^T(x−y)≥\mu\|x−y\|^2, \quad \forall x,y.$$
How does one prove that 2) implies 1)?
Use Taylor expansion: Take $x,y$. Then $$\begin{split} f(y)& = f(x) + \int_0^1\nabla f(x + s(y-x))^T(y-x)\ ds \\ &= f(x) + \nabla f(x)^T(y-x)+ \int_0^1(\nabla f(x + s(y-x))-\nabla f(x))^T(y-x)\ ds\\ &\ge f(x) + \nabla f(x)^T(y-x)+ \int_0^1\mu s\|x-y\|^2\ ds\\ &= f(x) + \nabla f(x)^T(y-x)+ \frac\mu2 \|x-y\|^2. \end{split}$$