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I am reading the paper Coordinate Descent Algorithms by Wright (see here) and in Assumption 1 on page 12 it is assumed that the objective function $f$ is uniformly Lipschitz continuously differentiable. I find this assumption confusing. It seems that Lipschitz continuity implies uniform continuity (see here) so the word uniformly is superfluous here, right? Are there any examples of Lipschitz continuous functions that are not uniformly continuous? Is this just a mistake or is there some meaning of this assumption?

Any help is much appreciated!

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    $\begingroup$ You are right: the article does not explain what "uniformly Lipschitz" means. Maybe it means simply that it is more than locally Lipschitz, in the sense that it is globally Lipschitz. $\endgroup$
    – Crostul
    Commented Nov 24, 2019 at 18:55

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Every continuously differentiable function is locally lipschitz. However, the function $f(x)=e^x$ is continuously differentiable, but not uniformly lipschitz.

So we are essentially assuming that the derivative exists and is globally bounded.

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  • $\begingroup$ Thank you for your response. I thought that Lipschitz continuity means that there exists $K\ge0$ such that $d_Y(f(x_1), f(x_2)) \le K d_X(x_1, x_2)$ (see here). It still seems that just saying Lipschitz continuously differentiable would be enough unless the author means something else. $\endgroup$
    – Cm7F7Bb
    Commented Nov 24, 2019 at 19:02
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    $\begingroup$ Sure, that depends on your definition. Some authors like to stress (globally/uniformly) lipschitz vs locally lipschitz. This is just a matter of giving names. $\endgroup$ Commented Nov 24, 2019 at 19:05
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    $\begingroup$ Uniformly lipschitz is the same what you call lipschitz (uniformly means here that we can pick the lipschitz constant uniformly). $\endgroup$ Commented Nov 24, 2019 at 19:07

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