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> It's true that there are also functions that are Lipschitz-continuous but not differentiable, but in this context they aren't interesting. They absolutely ar

> It's true that there are also functions that are Lipschitz-continuous but not differentiable, but in this context they aren't interesting. They absolutely are, even in the article. E.g. the max function or the 1-norm are non-differentiable, but continuous. There is literally no point in talking about Lipschitz continuity for differentiable functions, because it is equivalent to the derivative being bounded.

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 ago
    You seem to have sedulously avoided answering my question.
ajkjk 1y ago
The article is talking about functions whose changes over an displacement dx are not just bounded by K |dx| for some K, but specifically by the value K=1. So it's not the same thing as the Lipschitz property in analysis, although I guess it's inspired by it--but as a result it's concrete and useful for algorithms.
- constantcrying 1y ago
  >The article is talking about functions whose changes over an displacement dx are not just bounded by K |dx| for some K, It is also talking about functions which do not have a derivative at all and therefore could not possibly have that property. The point of Lipschitz functions is that they need not have a derivative. >So it's not the same thing as the Lipschitz property in analysis, although I guess it's inspired by it--but as a result it's concrete and useful for algorithms. That is just some arbitrary scaling. Only relevant to have the math look nicer. E.g. if f is Lipschitz for L=10, then f/10 is Lipschitz for L=1.
  - kragen 1y ago
    > It is also talking about functions which do not have a derivative at all and therefore could not possibly have that property [that their changes over an displacement dx are bounded by K |dx| for some K] Functions that do not have a derivative at all can certainly have that property. Perhaps you were confused by the notation "dx", suggesting a differential, for the displacement, rather than the conventional Δx, but the sentence makes no sense if we try to read it as a statement about differentials. Possibly your correspondent is not a Greek Keyboard Achiever. Moreover, the functions they are discussing, while indeed not differentiable, are continuous and differentiable almost everywhere, so you can also prove Lipschitz continuity if you can prove a bound on their gradients at the points where they do exist. > That is just some arbitrary scaling. Only relevant to have the math look nicer. E.g. if f is Lipschitz for L=10, then f/10 is Lipschitz for L=1. It's a relevant difference if you don't know the Lipschitz constant L or a bound on it, because then you don't know what to divide by.
    - constantcrying 1y ago
      >Functions that do not have a derivative at all can certainly have that property. Perhaps you were confused by the notation "dx", suggesting a differential, for the displacement, rather than the conventional Δx, but the sentence makes no sense if we try to read it as a statement about differentials. Possibly your correspondent is not a Greek Keyboard Achiever. Moreover, the functions they are discussing, while indeed not differentiable, are continuous and differentiable almost everywhere, so you can also prove Lipschitz continuity if you can prove a bound on their gradients at the points where they do exist. The Lipschitz property is not a property of functions with a derivative. I really do dislike this kind of drive by commenting, where you ignore an entire conversation and pretend someone does not understand what a gradient is.

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 ago
  You seem to have sedulously avoided answering my question.

ajkjk 1y ago

The article is talking about functions whose changes over an displacement dx are not just bounded by K |dx| for some K, but specifically by the value K=1. So it's not the same thing as the Lipschitz property in analysis, although I guess it's inspired by it--but as a result it's concrete and useful for algorithms.

constantcrying 1y ago
>The article is talking about functions whose changes over an displacement dx are not just bounded by K |dx| for some K, It is also talking about functions which do not have a derivative at all and therefore could not possibly have that property. The point of Lipschitz functions is that they need not have a derivative. >So it's not the same thing as the Lipschitz property in analysis, although I guess it's inspired by it--but as a result it's concrete and useful for algorithms. That is just some arbitrary scaling. Only relevant to have the math look nicer. E.g. if f is Lipschitz for L=10, then f/10 is Lipschitz for L=1.
- kragen 1y ago
  > It is also talking about functions which do not have a derivative at all and therefore could not possibly have that property [that their changes over an displacement dx are bounded by K |dx| for some K] Functions that do not have a derivative at all can certainly have that property. Perhaps you were confused by the notation "dx", suggesting a differential, for the displacement, rather than the conventional Δx, but the sentence makes no sense if we try to read it as a statement about differentials. Possibly your correspondent is not a Greek Keyboard Achiever. Moreover, the functions they are discussing, while indeed not differentiable, are continuous and differentiable almost everywhere, so you can also prove Lipschitz continuity if you can prove a bound on their gradients at the points where they do exist. > That is just some arbitrary scaling. Only relevant to have the math look nicer. E.g. if f is Lipschitz for L=10, then f/10 is Lipschitz for L=1. It's a relevant difference if you don't know the Lipschitz constant L or a bound on it, because then you don't know what to divide by.
  - constantcrying 1y ago
    >Functions that do not have a derivative at all can certainly have that property. Perhaps you were confused by the notation "dx", suggesting a differential, for the displacement, rather than the conventional Δx, but the sentence makes no sense if we try to read it as a statement about differentials. Possibly your correspondent is not a Greek Keyboard Achiever. Moreover, the functions they are discussing, while indeed not differentiable, are continuous and differentiable almost everywhere, so you can also prove Lipschitz continuity if you can prove a bound on their gradients at the points where they do exist. The Lipschitz property is not a property of functions with a derivative. I really do dislike this kind of drive by commenting, where you ignore an entire conversation and pretend someone does not understand what a gradient is.