What I find weird is that the derivative, if it exists, cannot have a jump discontinuity. That means that the only other kind of discontinuity it can have is infinite oscillation like sin(1/x) near zero.
This one is a bit of an obscure property of derivatives, corollary to theorem 5.12 in Baby Rudin.
You mean the derivative of a specific function you have in mind, like perhaps the field equations? Or do you mean something other than what I understand by a jump discontinuity in a derivative, such as one gets for f(x) = {-x for x<0, x for x>=0}?
Tone: Clarification request for my own understanding, not a "gotcha" post; I strongly believe you are saying something true but there's just too many details elided because they are trivial to you for me to quite follow, and I'm intrigued enough to want to be able to follow up, if you'd be so kind as to indulge me.
f'(x) being discontinuous is not the same thing as f(x) not having a derivative at some point; for example, the absolute value function |x| is simply not differentiable at x=0. The following function IS differentiable at x=0 but its derivative at x=0 is discontinuous:
f(x) = x^2 sin(1/x) if x != 0 else 0
Darboux's theorem implies in particular that if f'(x) exists at some point x and is discontinuous at that x then the discontinuity is not a jump discontinuity.
Link to Darboux's theorem: https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)
A jump discontinuity is a discontinuity where intermediate values are not attained, such as in the heaviside or signum functions. If a derivative exists at a point x, then it cannot have a jump discontinuity at x. However, it can have a discontinuity like the one exemplified by f(x) = 2x sin(1/x) - cos(1/x), with f(0) = 0, as that's the derivative of g(x) = x^2 sin (1/x)
Darboux's theeorem says that there is no way to create a jump in the derivative, in part because a derivative at a point is defined in terms of limits from both sides, so the limits must be the same.
(I'm pretty decent in mathematics in the general sense, reasoning from axioms, proofs, etc. But as I came up on the computer science side, I'm very lopsided into discrete mathematics, which is a bit unusual. Almost every other way to become a good mathematician makes you lopsided into real analysis and the fields that build on that.)
So when we get pedantic about what things and use early 20th century formalism, we do so kind of as a reaction to historical misunderstandings. When we ask, "what is the derivative of f at 0?", we're trying to show holes in understanding that have been patched by more modern frameworks.
This is definitely wrong. The derivative of |x| is -1 where x < 0, and 1 where x > 0, and doesn't exist where x = 0. That is a perfect match to the definition of a jump discontinuity -- the limit from the left is not equal to the limit from the right.
It's not at all necessary for the function to exist at x = 0 in order for it to have a discontinuity at x = 0.
But hey, don't take my word for it; why not check the definition on Wolfram?
http://mathworld.wolfram.com/JumpDiscontinuity.html
The original claim was "the derivative, if it exists, cannot have a jump discontinuity." This is badly stated. You're defending the idea that if the derivative exists at a particular point, then there is no jump discontinuity in the derivative at that point. But there can be a function f which satisfies both of these properties:
- f is the derivative of some other function F. ("The derivative of F exists.")
- f has a jump discontinuity, somewhere. ("The derivative of F has a jump discontinuity.")