Relationship between convex functions and coercivity

Question

I was struggling with this problem:

Let $f : \mathbb{R}^n \to \mathbb{R}$ be continuously differentiable convex function. Show that for any $\epsilon > 0$ the function $g_\epsilon (x) = f(x) + \epsilon \|x\|^2$ is coercive.

I'm a little confused as to the relationship between a continuously differentiable convex function and coercivity. I know the definitions of a convex function and a coercive function, but I'm unsure if they're related in some way that would help me prove this. Any hints?

Thanks

Answer 1

We have $f(x) \geq f(z)+\langle \nabla f (z),x-z\rangle$ for a fixed $z$. Now estimate below using Cauchy-Schwarz as follows: \begin{align*} \epsilon\|x\|^2 + f(x) &\geq \epsilon\|x\|^2+ f(z)+\langle \nabla f (z),x-z\rangle\\ &\geq \epsilon \|x\|^2 +f(z) - \|\nabla f(z)\|\cdot\|x-z\|\\ &\geq \epsilon \|x\|^2 +f(z) - \|\nabla f(z)\|\cdot \left(\|x\|-\|z\|\right)\\ &\to \infty \quad \text{as $\|x\|\to\infty$.} \end{align*}

Answer 2

If $x_1, x_2 \in \mathbb R^n$ then \begin{align} \langle x_1 - x_2, \nabla g_\epsilon(x_1) - \nabla g_\epsilon(x_2) \rangle &= \langle x_1 - x_2, \nabla f(x_1) + 2 \epsilon x_1 - (\nabla f(x_2) + 2 \epsilon x_2) \rangle \\ &= \underbrace{\langle x_1 - x_2, \nabla f(x_1) - \nabla f(x_2) \rangle}_{\geq 0} + 2 \epsilon \underbrace{\langle x_1 - x_2, x_1 - x_2 \rangle}_{\|x_1 - x_2 \|_2^2} \\ &\geq 2 \epsilon \|x_1 - x_2 \|_2^2. \end{align} This shows that $g_\epsilon$ is coercive.

(The inequality $$ \langle x_1 - x_2, \nabla f(x_1) - \nabla f(x_2) \rangle \geq 0 $$ expresses the fact that $\nabla f$ is monotone.)