= Solution
<Gradient descent> is $x_{t+1}=x_t-\eta\nabla f(x_t)$. Integrating the Hessian along a segment gives the descent lemma. Applying it to the hinted $z$ and convexity yields cocoercivity: $(\nabla f(x)-\nabla f(y))^T(x-y)\ge\|\nabla f(x)-\nabla f(y)\|^2/\beta$. Apply this to $\phi=f-\alpha\|x\|^2/2$, whose Hessian lies between $0$ and $(\beta-\alpha)I$, and rearrange to obtain the displayed strengthened inequality. With $y=x^*$, $\nabla f(x^*)=0$, and $\eta=2/(\alpha+\beta)$, expansion of the squared update gives contraction factor $((\beta-\alpha)/(\beta+\alpha))^2$ per step and the stated bound.
Back to article page
