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?
Best Answer
This is not true. A function can be increasing and bounded. For example, the sigmoid function $\sigma(x) = 1/(1 + \exp(-x))$. Integrating this yields a function $f(x) = \log(1 + \exp(x))$, which is strictly convex and Lipschitz.