Prove that $f(x) - f(x^{*}) \geq \frac{m}{2} \| x - x^{*} \|^{2}$.

Question

Suppose $MI \succeq \nabla^{2}f(x) \succeq mI $ for $ M > m > 0$ and that $ \nabla f(x^{*}) = 0 $.

Prove that $f(x) - f(x^{*}) \geq \frac{m}{2} \| x - x^{*} \|^{2}$.

Since $ \nabla f(x^{*}) = 0 $, we know that $f(x) \geq f(x^{*})$. Also the Hessian is positive semi-definite and $f$ is convex.

Also, this is a property of strongly convex functions?

I am not sure how to proceed from here.

Answer 1

Use the fundamental theorem of calculus and partial integration to get $$ f(x+v)=f(x)+\int_0^1f'(x+tv)[v]\,dt=f(x)+f'(x)[v]+\int_0^1(1-t)f''(x+tv)[v,v]\,dt $$ and apply the the restrictions on the spectrum of the Hessian.