# How to get regularity condition from smooth and strong convex conditions?

Given a function $$f:\mathbb{R^n}\rightarrow \mathbb{R}$$ and f is twice differentiable. We say $$f$$ is $$l$$-smooth if $$||\nabla f(x)-\nabla f(y)|| \leq l||x-y||$$ for all $$x,y \in \mathbb{R}^n$$. And we say f is $$\alpha$$-strongly convex($$\alpha >0$$) if for all $$x\in \mathbb{R}^n$$, $$\lambda_{min}(\nabla^2 f(x)) \geq \alpha$$

Now we suppose the function $$f$$ satisfies both $$l$$-smooth and $$\alpha$$-strongly convex conditions. How can we get the $$(\alpha,l)$$- regularity condition, which says if for any $$x\in \mathbb{R}^n$$, we have $$\langle \nabla f(x), x - x^*\rangle \geq \frac{\alpha}{2}||x - x^*||^2+\frac{1}{2\beta}||\nabla f(x)||^2$$ Here $$x^*$$ is the global minimum of the function $$f$$.

This is regularity condition is defined in Assumption 3b in this paper.

Since your estimate only contains gradients and differences between points in $$\mathbb{R}^n$$, we can assume that $$f(x^*)=0$$ and hence $$f(x)\ge 0$$ everywhere (otherwise you add $$-\min f$$).
Given that $$f$$ is twice differentiable, you have the Taylor expansion around $$x$$: $$f(x^*)=f(x)+\langle\nabla f(x),x^*-x\rangle+\frac{1}{2}(x^*-x)^T\nabla^2(\xi)(x^*-x)$$ Given your lower estimate on the Hessian matrix, you have $$f(x^*)\ge f(x)+\langle\nabla f(x),x^*-x\rangle+\frac{\alpha}{2}\|x^*-x\|^2$$ or $$\frac{\alpha}{2}\|x^*-x\|^2-\langle\nabla f(x),x-x^*\rangle\le f(x^*)-f(x)=-f(x)\quad (*)$$ by our initial assumption.
Given that $$f$$ is $$l$$-smooth, you have the Taylor-type estimate $$f(v)-f(w)\le \langle\nabla f(w),v-w\rangle+\frac{l}{2}\|v-w\|^2$$ for any $$v,w$$. If you set $$w=x$$ and $$v=x-\frac{1}{l}\nabla f(x)$$ you get $$f(v)-f(x)\le -\frac{1}{l}\|\nabla f(x)\|^2+\frac{l}{2l^2}\|\nabla f(x)\|^2=-\frac{1}{2l}\|\nabla f(x)\|^2.$$ Our function satisfies $$f(v)\ge 0$$ for every $$v$$ and the above inequality implies $$-f(x)\le -\frac{1}{2l}\|\nabla f(x)\|^2,$$ which combined with the inequality $$(*)$$ gives the desired result.