Kolmogorov–Arnold superposition theorem (non-universal Lorentz form)

← All problems

kolmogorov_arnold_superposition

Submitter: Kim Morrison.

Notes: Non-universal Lorentz-form existence statement of the Kolmogorov–Arnold superposition theorem on the closed cube Set.Icc (0 : Fin n → ℝ) 1. Both the outer function g : ℝ → ℝ and the inner functions φ_{k,l} : ℝ → ℝ may depend on f; this is weaker than formulations with universal inner functions. Resolves the continuous form of Hilbert's 13th problem since the two-variable combining function is addition.

Source: A. N. Kolmogorov, *On the representation of continuous functions of several variables by superposition of continuous functions of one variable and addition*, Doklady Akad. Nauk SSSR 114 (1957), 953-956. V. I. Arnold, *On functions of three variables*, Doklady Akad. Nauk SSSR 114 (1957), 679-681. G. G. Lorentz, *Metric entropy, widths, and superpositions of functions*, Amer. Math. Monthly 69 (1962), 469-485 (single-outer-function refinement). Listed as §80 in O. Knill, *Some Fundamental Theorems in Mathematics* (https://people.math.harvard.edu/~knill/graphgeometry/papers/fundamental.pdf).

Informal solution: Standard consequence of the Kolmogorov–Arnold superposition theorem in Lorentz form: for each continuous f on [0, 1]^n, the theorem provides continuous one-variable inner functions and a single continuous outer function such that f is the stated finite sum over 2n+1 layers. The construction uses rationally-independent positive reals λ_1, …, λ_n and step-approximating monotone Lipschitz functions ψ_0, …, ψ_{2n} separating cells of a nested partition of [0, 1]; the inner sums y_k(x) = ∑_l λ_l ψ_k(x_l) distinguish nearby points of the cube in at least one of the 2n + 1 indices k. The outer function is built iteratively using uniform continuity of f. Functions originally defined only on the compact sets where they are used can be extended to continuous functions on ℝ by Tietze extension.

theorem declaration uses `sorry`kolmogorov_arnold (n : ) (_hn : 1 n) (f : (Fin n ) ) (_hf : ContinuousOn f (Set.Icc 0 1)) : (g : ) (φ : Fin (2 * n + 1) Fin n ), Continuous g ( k l, Continuous (φ k l)) x Set.Icc (0 : Fin n ) 1, f x = k, g ( l, φ k l (x l)) := n:_hn:1 nf:(Fin n ) _hf:ContinuousOn f (Set.Icc 0 1) g φ, Continuous g (∀ (k : Fin (2 * n + 1)) (l : Fin n), Continuous (φ k l)) x Set.Icc 0 1, f x = k, g (∑ l, φ k l (x l)) All goals completed! 🐙

Solved by

Not yet solved.