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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?

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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.

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  • $\begingroup$ Thanks! I think my mistake was thinking that from increasing it follows unbounded. But as I see, it is not like this. arctan(x) would be another example $\endgroup$
    – Trb2
    Apr 12, 2021 at 22:29

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