Gradient based bound on an L-smooth, strongly convex function

Question

Let $D \subset \mathbb{R}^n$ be a convex set. Let $||\cdot ||: \mathbb{R}^n \to \mathbb{R}$ denote the Euclidean norm.

Consider a smooth function $f: D \to \mathbb{R}$. Suppose that $f$ is $\mu$ strongly convex with $\mu > 0$, that is, $f(x) - \frac{\mu}{2}||x||^2$ is convex. Also suppose that $f$ is $L$-smooth with $L>0$, that is, for all $x,y \in \mathbb{R}^n$, $||\nabla f(x) - \nabla f(y)|| \le L||x-y||$. Both $L$ and $\mu$ are general parameters which may or may not be equal to each other.

Suppose that $f$ has a minimum (which is unique since $f$ is convex) at $m \in \mathbb{R}^n$. Define

$$D \ni x \mapsto v(x) := f(x) - f(m) \in \mathbb{R}.$$

Then is it true that $2\mu v(x) \le ||\nabla f(x)||^2 $ for all $x \in D$? And how does one prove it?

Answer 1

Yes, this is true, for a proof, refer to (4.12) and Appendix B of the paper (which is published at SIAM Review) "Optimization Methods for Large-Scale Machine Learning" written by Leon Bottou, Frank E. Curtis and Jorge Noceda. Bsically, this result says that one can bound the optimality gap at a given point in terms of the squared $\ell_2$-norm of the gradient of the (strongly convex) objective at that point.