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Hmm, I guess you're right. But those functions (highly relevant here!) are differentiable almost everywhere; is that enough for Lipschitz continuity to be equi
by kragen 1y ago
Hmm, I guess you're right. But those functions (highly relevant here!) are differentiable almost everywhere; is that enough for Lipschitz continuity to be equivalent to the derivative being bounded?
I still think there's a point in talking about Lipschitz continuity for differentiable functions because the algorithms they're talking about depend on the Lipschitz property, not the differentiability.
- constantcrying 1y ago>because the algorithms they're talking about depend on the Lipschitz property, not the differentiability. That is why the definition should not contain the differentiability condition. It is violated by the examples and irrelevant. My point is that the Lipschitz property, as defined e.g. in the Wikipedia article, is the important part. Also assuming differentiability is unhelpful, because it misses the point and is even violated by the rest of the article.
- kragen 1y agoYou seem to have sedulously avoided answering my question.