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by ogogmad 601 days ago
I think that the problem is that theoretical real analysis is often presented like it's nothing but a validation of things people already knew to be true -- but maybe it's not?

The example you gave concerns differentiation. Differentiation is messy in real analysis because it's messy in numerical computing. How real analysis fixes this mess parallels how numerical computing must fix the mess. How do we make differentiation - or just derivatives, perhaps - computable?

The rock-bottom condition for computability is continuity. All discontinuous functions are uncomputable. It turns out that it is sufficient, to make your theorem hold, to have the 2nd partial derivatives f_{xy} and f_{yx} be continuous. They wouldn't even be computable otherwise!

One of the proofs provided uses integration. In numerical contexts, it is integration which is considered "easy", and "differentiation" which is considered hard. This is totally backwards to symbolic calculus.

The article also mentions Distribution Theory. This is important in the theory of linear PDEs. I suspect it is implicit in the algorithmic theory as well, whether practitioners have spelled this out or not. This is a theory that makes the differentiation operator itself computable, but at the cost of making the derivatives weaker than ordinary functions. How so? On the one hand, it allows to obtain things like the Dirac delta as derivatives, but those aren't even functions. On the other hand, these objects behave like functions - let's say f(x,y) - but we can't evaluate them at points; instead, we can take their inner product with test functions, which we can use to approximate evaluation. This is important because PDE solvers may only be able to provide solutions in the weak, distribution-theoretic sense.

1 comments

Do I understand properly that in a different universe distributions could have been called prefunctions?
A distribution is a function, on the space of test functions.
OK, so if we have a distribution D (less nice than the average function) and a test function T (nicer than the average function), we have ⟨D,T⟩ = c: ℂ, so ⟨D,—⟩: test fn→ℂ and ⟨—,T⟩: distribution→ℂ ?
Wait i thought functions are predistributions..

[My bad, it was Matvei, not Manuel, no idea how i mixed that up..

Checkout his childrens books, as well as

https://archive.is/eaYRs

Note how the independent diagonals are what i consider interesting]

if there are no interiors (maybe edges but no faces nor volumes) then the vertices on the diagonals are truly independent: eg QM on small scales, GR on large ones.

[I'm currently pondering how the "main diagonal" of a transition matrix provides objects, while all the off-diagonal elements are the arrows. This implies that by rotating into an eigenframe (diagonalising), we're reducing the diversion to -∞ (generalised eigenvectors have nothing to lose but their Jordan chains) and hence back in the world of classical boolean logic?]

EDIT: https://mmozgovoy.dev/posts/solar-matter/

[Righhht, maybe you can excite me even more by relating this to quantales?? Or maybe expand on fns vs distributions a bit more?]

L: quantal (quasiparticles)

A distribution is not a function. It is a continuous linear functional on a space of functions.

Functions define distributions, but not all distributions are defined that way, like the Dirac delta or integration over a subset.

A functional is a function.
The term "function" sadly means different things in different contexts. I feel like this whole thread is evidence of a need for reform in maths education from calculus up. I wouldn't be surprised if you understood all of this, but I'm worried about students encountering this for the first time.
Don’t know if you are a mathematician or not but mathematically speaking “function” has a definition that is valid in all mathematical contexts. Functional clearly meets the criteria to be a function since being a function is part of the definition of being a functional.
Try composing two distributions.
Try composing f : A -> B with g : A -> B, for A ≠ B. Still, f and g are functions. So, what exactly is your point?
What is a delta function at a composed with a delta function at b <> a?