strict convexity and Lipschitz continuity [closed]

Question

Consider a continuously differentiable function $f: \mathbb{R}^n \mapsto \mathbb{R}$. If $f$ is strictly convex, does it imply that it is not Lipschitz on $\mathbb{R}^n$?

Because if $f$ is strictly convex, the derivative is monotonically increasing and hence not bounded, which makes impossible to find a constant $L$ for which $f$ is Lipschitz. Is this true and can it be proven in a rigorous manner using the definition of Lipschitz and strict convexity?

Answer 1

No it does not. $f(x)=x\arctan x -\frac{1}{2}\ln(1+x^2)$ is strictly convex and Lipschitz.