 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: $(\nabla f(x)-\nabla f(y){)}^T(x-y)\ge \Vert \nabla f(x)-\nabla f(y){\Vert }^2/\beta$. Apply this to $\varphi =f-\alpha \Vert x{\Vert }^2/2$, whose Hessian lies between $0$ and $(\beta -\alpha )I$, and rearrange to obtain the displayed strengthened inequality. With $y=x^{\ast }$, $\nabla f(x^{\ast })=0$, and $\eta =2/(\alpha +\beta )$, expansion of the squared update gives contraction factor $((\beta -\alpha )/(\beta +\alpha ){)}^2$ per step and the stated bound.