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Comment by practal

1 month ago

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/

      5 replies →

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.

      20 replies →