We consider a real number $\alpha \in \mathbb{R}_+^{\star}$ and a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ differentiable on $\mathbb{R}^n$. Recall that $f$ is $\alpha$-convex if for all $x, y \in \mathbb{R}^n$, $$f(y) \geqslant f(x) + \langle \nabla f(x), y-x \rangle + \frac{\alpha}{2}\|y-x\|^2.$$ a. We consider the function $g_{\alpha} : \mathbb{R}^n \rightarrow \mathbb{R}$ defined by $g_{\alpha}(x) = f(x) - \frac{\alpha}{2}\|x\|^2$ for all $x \in \mathbb{R}^n$. Calculate $\nabla g_{\alpha}(x)$ for all $x \in \mathbb{R}^n$, and show that $f$ is $\alpha$-convex if and only if $g_{\alpha}$ is convex. b. Deduce that $f$ is $\alpha$-convex if and only if for all $x, y \in \mathbb{R}^n$, $$\langle \nabla f(y) - \nabla f(x), y-x \rangle \geqslant \alpha \|y-x\|^2.$$
We consider a real number $\alpha \in \mathbb{R}_+^{\star}$ and a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ differentiable on $\mathbb{R}^n$. Recall that $f$ is $\alpha$-convex if for all $x, y \in \mathbb{R}^n$,
$$f(y) \geqslant f(x) + \langle \nabla f(x), y-x \rangle + \frac{\alpha}{2}\|y-x\|^2.$$
a. We consider the function $g_{\alpha} : \mathbb{R}^n \rightarrow \mathbb{R}$ defined by $g_{\alpha}(x) = f(x) - \frac{\alpha}{2}\|x\|^2$ for all $x \in \mathbb{R}^n$. Calculate $\nabla g_{\alpha}(x)$ for all $x \in \mathbb{R}^n$, and show that $f$ is $\alpha$-convex if and only if $g_{\alpha}$ is convex.\\
b. Deduce that $f$ is $\alpha$-convex if and only if for all $x, y \in \mathbb{R}^n$,
$$\langle \nabla f(y) - \nabla f(x), y-x \rangle \geqslant \alpha \|y-x\|^2.$$