# Gradient descent step size for strongly convex functions

Suppose we are optimizing a strongly convex function $$f(x)$$ via gradient descent $$x_{t+1} = x_t - \eta_t \nabla f(x_t)$$. By strongly convex I mean that $$f(x+h) \ge f(x) + \langle \nabla f(x), h \rangle + \frac{\alpha}{2}||h||^2$$.

This nice survey paper by Bansal and Gupta (Section 2.2) suggests using a step size $$\eta_{t} = \frac{1}{\alpha (t+1)}$$ and mentions it is optimal. I am curious how does the $$1/(t+1)$$ term pop up? Is there a simple-to-state reason for it?

For comparison, when optimizing a smooth function (i.e., upper bounded by some quadratic) the step size $$\eta_t = \frac{1}{\beta}$$ is very natural because it minimizes the upper bound.