From Convex Optimization by Boyd and Vandenberghe:
Let $f: \Bbb R^n \rightarrow \Bbb R$ be continuous and twice differentiable.
Assume $$\|\nabla f(x)\|_2 \le \eta \le 3(1-2 \alpha)m^2/L$$
where $m$ is the number that satisfies $mI \preceq \nabla ^2 f(x)$ (meaning $f$ is strongly convex).
Let $\lambda(x)^2 = \Delta x_{\text{nt}}^T \nabla^2 f(x) \Delta x_{\text{nt}} = – \nabla f(x)^T \Delta x_{\text{nt}}$ and $\Delta x_{\text{nt}} = – \nabla ^2 f(x)^{-1} \nabla f(x)$
Then, by strong convexity, $\lambda(x) \le 3(1-2 \alpha)m^{3/2}/L$
How does strong convexity imply the final inequality?
Best Answer
I think you are confusing yourself with symbols.
Let $g, H$ be the gradient & Hessian respectively, then $\lambda(x)^2 = (-H^{-1}g)^T H (-H^{-1} g) = g^T H^{-1} g$.
Note that if $A \ge mI$, then $A^{-1} \le {1 \over m} I$.
Then $\lambda(x)^2 \le {1 \over m} \|g\|^2$, or $\lambda(x) \le {1 \over \sqrt{m}} \|g\|$.
Now substitute the bound on $\|g\|$ to get the result.