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>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 whi

>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
    Functions with a derivative can definitely have the Lipschitz property. The “examples” section of https://en.m.wikipedia.org/wiki/Lipschitz_continuity https://en.m.wikipedia.org/wiki/Lipschitz_continuity begins with a subsection entitled, “Lipschitz continuous functions that are everywhere differentiable”. Most of the examples given there are differentiable in varying ways. (Unqualified “differentiable” normally means “everywhere differentiable”.) You're also contradicting your previous comment at https://news.ycombinator.com/item?id=44145078 https://news.ycombinator.com/item?id=44145078. I was trying very hard to ascribe some sort of sensible interpretation to your comments, but due to your aggression, now I regret having done so.
ajkjk 1y ago
You seem really confused about something. The article is talking about SDFs that do not have that property, which it is then trying to force to have that property via a hack of sorts (f/|del f| apparently), because once the property holds it is very useful for making the algorithm for more efficient. The arbitrary scaling is not arbitrary if the whole algorithm is defined around K=1. The Lipschitz property in analysis is "bounded by any K". The algorithm is specifically using "bounded by K=1". Also: for the purpose of numerical methods, it does not matter if a function "has a derivative" in a classical sense or not. Certainly distributions, step functions, etc are fine. Numerical methods cannot, after all, tell if a function is discontinuous or not. The article is using some known closed-form functions as examples, but it is trying to illustrate the general case, which is not going to be a closed form but just an arbitrary SDF that you constructed from somewhere---which it will be very useful to enforce the K=1 Lipschitz condition on.
- kragen 1y ago
  I think there are probably different populations commenting here with different backgrounds, and consequently different assumptions and different blind spots. For example, for Newton's Method, it does very much matter whether the function has a derivative in the classical sense everywhere in the relevant interval; the quadratic convergence proof depends on it. Discontinuous functions can break even binary chop.
- constantcrying 1y ago
  This is about differentiability. >which it is then trying to force to have that property via a hack of sorts (f/|del f| apparently) The functions he talks about do not have a derivative. So that hack is nonsensical.
  - ajkjk 1y ago
    Evidently they do? Maybe there is some domain knowledge you're missing here as you belligerent protest that it's nonsense.
  - kragen 1y ago
    Max, min, and the L1 norm? Those have derivatives (gradients) almost everywhere. The derivatives are undefined at a set of points of measure zero, and the functions are continuous across those gaps.

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
  Functions with a derivative can definitely have the Lipschitz property. The “examples” section of https://en.m.wikipedia.org/wiki/Lipschitz_continuity https://en.m.wikipedia.org/wiki/Lipschitz_continuity begins with a subsection entitled, “Lipschitz continuous functions that are everywhere differentiable”. Most of the examples given there are differentiable in varying ways. (Unqualified “differentiable” normally means “everywhere differentiable”.) You're also contradicting your previous comment at https://news.ycombinator.com/item?id=44145078 https://news.ycombinator.com/item?id=44145078. I was trying very hard to ascribe some sort of sensible interpretation to your comments, but due to your aggression, now I regret having done so.

ajkjk 1y ago

You seem really confused about something. The article is talking about SDFs that do not have that property, which it is then trying to force to have that property via a hack of sorts (f/|del f| apparently), because once the property holds it is very useful for making the algorithm for more efficient. The arbitrary scaling is not arbitrary if the whole algorithm is defined around K=1. The Lipschitz property in analysis is "bounded by any K". The algorithm is specifically using "bounded by K=1". Also: for the purpose of numerical methods, it does not matter if a function "has a derivative" in a classical sense or not. Certainly distributions, step functions, etc are fine. Numerical methods cannot, after all, tell if a function is discontinuous or not. The article is using some known closed-form functions as examples, but it is trying to illustrate the general case, which is not going to be a closed form but just an arbitrary SDF that you constructed from somewhere---which it will be very useful to enforce the K=1 Lipschitz condition on.

kragen 1y ago
I think there are probably different populations commenting here with different backgrounds, and consequently different assumptions and different blind spots. For example, for Newton's Method, it does very much matter whether the function has a derivative in the classical sense everywhere in the relevant interval; the quadratic convergence proof depends on it. Discontinuous functions can break even binary chop.
constantcrying 1y ago
This is about differentiability. >which it is then trying to force to have that property via a hack of sorts (f/|del f| apparently) The functions he talks about do not have a derivative. So that hack is nonsensical.
- ajkjk 1y ago
  Evidently they do? Maybe there is some domain knowledge you're missing here as you belligerent protest that it's nonsense.
- kragen 1y ago
  Max, min, and the L1 norm? Those have derivatives (gradients) almost everywhere. The derivatives are undefined at a set of points of measure zero, and the functions are continuous across those gaps.