Show that $g(x)=f(x)+\varepsilon \| x \|^2$ is coercive

Question

Let $f : \Bbb R^n \to \Bbb R$ be a continuously differentiable convex function. Show that for any $\varepsilon \gt 0$ the function $$g(x)=f(x)+\varepsilon \| x \|^2$$ is coercive, i.e., $$\lim_{\|x\| \to \infty }f(x) = \infty$$

Does anyone have any hints?

Answer 1

By convexity, it holds $$ f(x) \geq f(0) + \nabla f(0) \cdot x, \qquad \forall x\in\mathbb{R}^n. $$ (You can choose any point other than $0$.)

Hence $$ g(x) \geq f(0) + \nabla f(0) \cdot x + \epsilon \|x\|^2 \geq f(0) - \|\nabla f(0)\| \, \|x\| + \epsilon \|x\|^2, $$ so that the claim follows.