A result of Kolmogorov and Arnold says that continuous functions on $\mathbb{R}^n$ can be represented as sums of the form

$$ f(x_1,\dots,x_n)=\sum_{q=0}^{2n}\Phi_q\left(\sum_{p=1}^n\phi_{p,q}(x_p)\right),$$

where $\Phi_p$ and $\phi_{p,q}$ are unary continuous functions.

I'm curious about analogous results for functions on the Hilbert cube.

Suppose I have a continuous function $f:[0,1]^\omega \rightarrow [0,1]$. Is it always possible to write this in the form

$$ f(x_0,x_1,\dots) = \sum_{q<\omega} \Phi_q\left( \sum_{p<\omega}\phi_{p,q}(x_p) \right) ,$$

where $\Phi_p$ and $\phi_{p,q}$ are continuous functions and all of the sums converge uniformly? If this fails are there known analogous results?

EDIT: What we do get immediately is this.

For each $n>\omega$ let

$$f_n(x_0,\dots,x_{n-1})=\inf_{\overline{y}\in[0,1]^\omega} f(x_0,\dots,x_{n-1},\overline{y}),$$

i.e. $f_n$ is an approximation of $f$ from below by a continuous function on $[0,1]^n$. Let $f_{0}=0$ and $g_{n}(x_0,\dots,x_{n-1})=f_n(x_0,\dots,x_{n-1})-f_{n-1}(x_0,\dots,x_{n-2})$, so that in particular

$$f(x_0,x_1,\dots)=\sum_{1\leq n < \omega}g_{n}(x_0,\dots,x_{n-1}).$$

Now we can use Kolmogorov superposition on each $g_{n}$,

$$g_n(x_0,\dots,x_{n-1}) = \sum_{q=1}^{2n} \Phi_{n,q}\left( \sum_{p=0}^{n-1} \phi_{n,p,q}(x_p) \right).$$

So then we have

$$f(x_0,x_1,\dots)=\sum_{1\leq n < \omega} \sum_{q=1}^{2n} \Phi_{n,q}\left( \sum_{p=0}^{n-1} \phi_{n,p,q}(x_p) \right) $$.

By Dini's theorem the outermost sum is uniformly convergent and we can rearrange this sum to be one of the desired form, but it's unclear if the rearranged sum converges uniformly, in particular it's not clear to me that it is monotonic.

For a related posts see:

Is there any continuous ternary function which can not be represented by composition of continuous binary functions?

Kolmogorov superposition for smooth functions

Kolmogorov-Arnold theorem for (just-)functions

Continuous functions of three variables as superpositions of two variable functions


1 Answer 1


After finally getting around to learning the proof of the Kolmogorov–Arnold representation theorem (thanks to this recorded talk by Bar-Natan) I now know that the functions can be chosen so that the rearranged sequence in my edit converges uniformly (in particular the terms are all positive, so the sum is monotonic and therefore uniformly convergent by Dini's theorem). By shifting the functions around you can remove the requirement on the range of $f$. The precise statement can be taken to be this:

There is a fixed sequence $\{\phi_p\}_{p<\omega}$ of continuous unary functions $\mathbb{R}\rightarrow\mathbb{R}$ and a fixed sequences $\{a_n\}_{n<\omega}$ and $\{b_n\}_{n<\omega}$ of intergers such that for any continuous function $f:[0,1]^\omega \rightarrow \mathbb{R}$ there exists a sequence $\{\Phi_n\}_{n<\omega}$ of continuous unary functions $\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(x_0,x_1,\dots)=\sum_{n<\omega}\Phi_n\left(\sum_{p = a_n}^{b_n} \phi_p(x_p)\right),$$ with the outermost sum converging monotonically (after the first term) and uniformly.

The key being that (after shifting so that $f$ is non-negative), each of the terms in the sum are non-negative.

I suspect that we can choose $a_n=0$ and $b_n = n$, but I haven't proven that. It also seems likely that this result or something similar to it is actually easier to prove directly than via the Kolmogorov–Arnold representation theorem. In the broadest sense we already know that every function on the Hilbert cube can be represented by an infinite sum of compositions of unary functions and addition, since this is sufficient to give polynomials in the coordinates and polynomials can uniformly approximate functions on the Hilbert cube, so this isn't really as exciting as the finite dimensional case.

  • $\begingroup$ Shouldn't you accept your answer? It seems satisfactory pertaining to your OP. $\endgroup$ Jul 31, 2020 at 14:58

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